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Applied mathematics for business and economics, life sciences, and social sciences [2 ed.]
 0023055901, 9780023055904

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SECOND

EOmON

APPLIED MATHEMATICS FOR BUSINESS AND ECONOMICS, LIFE SCIENCES, AND SOCIAL SCIENCES

RAYMOND

A.

BARNETT

CHARLES

|.

BURKE

MICHAEL

R.

ZSEGLER

APPLIED MATHEMATICS FOR BUSINESS AND ECONOMICS,

AND

LIFE SCIENCES,

SECOND EDITION

SOCIAL SCIENCES

RAYMOND

A.

BARNETT

CHARLES

MICHAEL

R.

ZIEGLER

Marquette University

Merritt College

J.

BURKE

City College of San Francisco

DELLEN PUBLISHING COMPANY

COLLIER MACMILLAN PUBLISHERS

San Francisco, California

London

divisions of Macmillan. Inc.

On

was executed by Los Angeles artist Charles The work, measuring 48 inches X 38 inches, is acrylic on laminated plywood and has been sculpted using a chain saw. Arnoldi's work may be seen at the Fuller Goldeen Gallery in San Francisco, the the cover; "Flintrock"

Arnoldi.

James Corcoran Gallery in Los Angeles, and the Charles Cowles Gallery in New York City. His work is included in the permanent collections of the Albright-Knox Museum in Buffalo, New York, the Chicago Art Institute, the Museum of Modern Art in New York City, the Metropolitan Museum, the Los Angeles County Museum, and the San Francisco

Museum

of

Modern

© Copyright a division of

Art.

1986 by Dellen Publishing Company, Macmillan, Inc.

Printed in the United States of America All rights reserved.

No

part of this

book may be reproduced or transmit-

ted in any form or by any means, electronic or mechanical, including

photocopying, recording, or any information storage and retrieval system, without permission in writing from the Publisher. Permissions:

Orders:

Collier

Dellen Publishing

Company

400 Pacific Avenue San Francisco, California 94133 Macmillan Publishing Company Front and

Brown

Riverside,

New

Streets

Jersey 08075

Macmillan Canada,

Inc.

Library of Congress Cataloging-in-Publication Data Harnett,

Raymond

A.

Applied mathematics life

sciences,

and

Includes index.

for business



1961Mathematics Michael R. III. QA39.2.B366 1986 1.

II.

and economics,

social sciences.

Ziegler,

.

I.

Burke, Charles

Title.

85-13322

510

ISBN 0-02-305590-1 Printing:

4 5 6 7 8

ISBN 0-05-305510-1

Year:

7 8

9

J.

Preface

PART

Algebra

I

CHAPTER

CHAPTER

1

2

Preliminaries

CHAPTER

3

4

4

1-1

Sets

1-2

Real

1-3

Inequality Statements and Line Graphs

1-4

Basic Operations on Signed

1-5

Positive Integer Exponents

41

1-6

Chapter Review

48

Numbers and

the Rules of Algebra

Numbers

Polynomials and Fractional Forms

12 21

28

55

2-1

Basic Operations on Polynomials

56

2-2

Factoring Polynomials

67

2-3

Multiplying and Dividing Fractions

79

2-4

Adding and Subtracting Fractions Chapter Review

88

2-5

CHAPTER

IX

Exponents and Radicals

96

101

3-1

Integer Exponents

3-2

Scientific Notation

111

3-3

Rational Exponents

117

3-4

Radicals

125

3-5

Basic Operations on Radicals

135

3-6

Chapter Review

141

Equations and Inequalities

102

147

4-1

Linear Equations

148

4-2

Linear Inequalities

160

111

iv

Contents

CHAPTER

PART

5

CHAPTER

Quadratic Equations

168

4-4

Nonlinear Inequalities

182

4-5

Literal Equations

190

4-6

Chapter Review

194

6

7

5-1

Cartesian Coordinate System and Straight Lines

200

5-2

Relations and Functions

213

5-3

Graphing Functions

226

5-4

Chapter Review

242

249

6-1

Simple Interest and Simple Discount

250

6-2

Compound

257

6-3

Future Value of an Annuity; Sinking Funds

6-4

Present Value of an Annuity; Amortization

273

6-5

Chapter Review

283

Interest

267

Systems of Linear Equations; Matrices 7-2

289

Review: Systems of Linear Equations Systems of Linear Equations and Augmented Matrices



290

Introduction

308 317

7-6

Gauss -Jordan Elimination Matrices Addition and Multiplication by a Number Matrix Multiplication Inverse of a Square Matrix; Matrix Equations

7-7

Leontief Input-Output Analysis (Optional)

361

7-8

Chapter Review

367

7-3

7-4 7-5

8

Mathematics

Mathematics of Finance

7-1

CHAPTER

199

Graphs and Functions

Finite

II

CHAPTER

4-3



Linear Inequalities and Linear Programming Two

8-1

Linear Inequalities in

8-2

Systems of Linear Inequalities

8-3

Linear Programming in

Approach

Two

337 347

373 374

Variables in

330

Variables

Two Dimensions — A

379

Geometric 389

Contents

8-4 8-5

PART

9

Constraints

412 432

8-7

Constraints (Optional)

448

8-8

Chapter Review

470

477

Probability

478

9-1

Introduction

9-2

The Fundamental

9-3

Permutations, Combinations, and Set Partitioning

485

9-4

Experiments, Sample Spaces, and Probability of an Event

497

9-5

Empirical Probability

515

9-6

Union, Intersection, and Complement of Events

524

9-7

Chapter Review

541

10

Principle of Counting

The Derivative

479

Introduction

550

10-2

Limits and Continuity

551

10-3

Increments, Tangent Lines, and Rates of Change

569

10-4

The

582

10-5

Derivatives of Constants,

10-7 10-8

11

549

10-1

10-6

CHAPTER

406

Calculus

III

CHAPTER

Geometric Introduction to the Simplex Method The Simplex Method: Maximization with s Problem

The Dual; Minimization with s Problem Constraints Maximization and Minimization with Mixed Problem

8-6

CHAPTER

A

Derivative

Power Forms, and Sums Derivatives of Products and Quotients Chain Rule and General Power Rule Chapter Review

Additional Derivative Topics

595

606 613 622

629 630

11-1

Implicit Differentiation

11-2

Related Rates

637

11-3

Higher-Order Derivatives

645

11-4

The

649

11-5

Marginal Analysis in Business and Economics

658

11-6

Chapter Review

663

Differential

vi

Contents

CHAPTER

CHAPTER

CHAPTER

CHAPTER

CHAPTER

12

13

14

15

16

Graphing and Optimization 12-1

Asymptotes; Limits

12-2

First Derivative

12-3

12-4

Second Derivative and Graphs Curve Sketching

12-5

Optimization; Absolute

12-6

Elasticity of

12-7

Chapter Review

at Infinity

667 and

Infinite Limits

and Graphs

Demand

668 685 704 722

Maxima and Minima

733 752

(Optional)

764

Exponential and Logarithmic Functions

— A Review — A Review

771

13-1

Exponential Functions

13-2

Logarithmic Functions

13-3

The Constant

13-4

Derivatives of Logarithmic Functions

798

13-5

Derivatives of Exponential Functions

804

13-6

Chapter Review

811

e

and Continuous Compound

772 778 Interest

791

815

Integration 14-1

Antiderivatives and Indefinite Integrals

14-2

Differential Equations

14-3

General Power Rule

839

14-4

Definite Integral

846

14-5

Area and the Definite Integral

14-6

Definite Integral as a Limit of a

14-7

Chapter Review

— Growth and Decay

816 829

854

Sum

866 879

Additional Integration Topics by Substitution

885 886

15-1

Integration

15-2

Integration by Parts

900

15-3

Integration Using Tables

907

15-4

Improper Integrals

916

15-5

Chapter Review

925

Multivariable Calculus

931

16-1

Functions of Several Variables

932

16-2

Partial Derivatives

942

16-3

Total Differentials

16-4

Maxima and Minima

and Their Applications

951

959

Contents

CHAPTER

17

APPENDIX A

APPENDIX B

vii

16-5

Maxima and Minima Using Lagrange

16-6

Method

16-7 16-8

Double Integrals over Rectangular Regions Double Integrals over More General Regions

1003

16-9

Chapter Review

1014

Multipliers

968 977

of Least Squares

990

1021

Additional Probability Topics 17-1

Random

17-2

Binomial Distributions

1033

17-3

1044

17-4

Continuous Random Variables Expected Value, Standard Deviation, and Median of Continuous Random Variables Uniform, Beta, and Exponential Distributions

1056

17-5

17-6

Normal

1077

17-7

Chapter Review

Variable, Probability Distribution,

and Expectation

Distributions

Al

A-l

Arithmetic Progressions

A-2

Geometric Progressions

A-3

The Binomial Formula

Al

A6 A12

A17

Tables I

Exponential Functions

Table

II

Common

Table

III

Natural Logarithms

Table IV Table

V

1065

1088

Special Topics

Table

1022

(e"

and

e"")

Logarithms (In

N=

log^

N)

Areas under the Standard Normal Curve

Mathematics of Finance

Answers

A18 A22 A24 A26 A29

A45

Index

II

Applications Index

18

Chapter Depenc

Preface

Many

and universities now

colleges

offer

mathematics courses that em-

phasize topics that are most useful to students in business and economics, sciences, and social sciences. Because of this trend, the authors have reviewed course outlines and college catalogs from a large number of life

and

and on the basis of this survey, selected the and emphasis found in this text. The material in this book is suitable for mathematics courses that include topics from algebra, finite mathematics, and calculus. Part provides a substantial review of the fundamentals of algebra, which may be treated colleges

universities,

topics, applications,

I

way or may be referred to as needed. In addition, certain key prerequisite topics are reviewed immediately before their use (see, for in a systematic

example. Section 8-1 and Section

Appendix

A. Part

II

13-1),

while others are discussed in

contains ample material for a finite mathematics

course covering the topics that have become standard in this area: mathematics of finance, linear systems, matrices, linear programming, and probability. Part

III

consists of a thorough presentation of calculus for functions

of one variable, including the exponential

probability topics that

make

the book readily adaptable to a variety

diagram on the facing page

Major Changes from the The second

for

chapter dependencies.)

First Edition

edition of Applied Mathematics for Business

Life Sciences,

and Social Sciences

dations of a large

fol-

to multivariable calculus

tion of topics in all three parts of courses. (See the

and logarithmic functions,

and some additional involve calculus concepts. The choice and organiza-

lowed by an introduction

number

reflects the

and Economics,

experiences and recommen-

of the users of the

first

edition. Additional

examples and exercises have been included in many sections to increase student support and to give students a better understanding of the material. In particular, a concentrated effort has ability to visualize

been made

to increase the student's

mathematical relationships and

The major changes

to deal

with graphs.

on linear programming. This chapter has been extensively rewritten; it now contains an expanded development of the basic simplex method and new sections on the dual and big IVI methods. Increased attention has been paid to the development of the simplex method and its relationship to the geometric method, which in Part

II

are in the chapter

Preface

should make the simplex method seem

The material on

much

less

mysterious

to the stu-

expanded and rearranged. Chapter 9 contains a new section on union, intersection, and complement of events, and the section on random variables has been moved to the new chapter on additional probability concepts (Chapter 17). In Part III, the most noticeable change from the first edition is the reorganization of the calculus material. The material on graphing has been expanded and rewritten and now occupies most of Chapter 12. The exponential and logarithmic functions are introduced at an earlier point so that Chapters 10-13 now deal exclusively with differential calculus and Chapters 14 and 15 deal with integral calculus. There are new sections on asymptotes, elasticity of demand, use of integral tables, continuous random variables, and binomial, uniform, beta, and normal probability distribudent.

probability has been

tions.

General Comments I of this book presents the algebraic concepts used in Parts II and III. If a minimal review is deemed desirable, then Chapter 4 could be covered before beginning Part II and Chapter 5 before beginning Part III. Part II deals with three areas that are independent of each other (see the diagram on page viii). The mathematics of finance is presented in Chapter 6. Standard angle notation is used for the compound interest factor and the present value factor. All the required exercises can be solved using either the tables in the back of the book or a hand calculator. Some optional problems have been included that require the use of a calculator. Chapters 7 and 8 cover topics from linear algebra and linear programming. Elementary row operations are used for solving systems of equations, inverting matrices, and solving linear programming problems. The material on linear programming is organized so as to provide the instructor with maximum flexibility. Those who want a good intuitive introduction to the subject can cover only the material up to the dual method. On the other hand, those who wish to emphasize the development of computational

Part

skills

those

can also cover the dual method or the big

who wish

to

M method (or both). Finally,

concentrate on problem solving (setting up problems)

and applications can cover the applications in Section 8-6 or 8-7 (or both) and omit the computational methods entirely. In order to facilitate these approaches, the answer section contains an appropriate model for each applied problem, as well as the numerical solution. Section 8-7 also con-

which lead to linear programming problems by hand. These applications provide a natural place to introduce the use of a computer program to solve linear programming problems. Such a program is available to institutions adopting this book at no charge from the publisher. tains optional applications

that are too

complex

to solve

xi

Preface

Chapter 9 covers counting techniques and the basic concepts of probability.

More advanced

In Part

topics are covered in

Chapter

17.

Chapters 10-13 present differential calculus

III,

for functions of

one variable, including the exponential and logarithmic functions. Trigonometric functions are not discussed in this book. Limits and continuity are presented in an intuitive fashion, utilizing numerical approximations

and one-sided

limits. All the rules of differentiation are

covered in Chapter

10.

Various applications of differentiation are then presented in Chapters

11

and

12,

with

a strong

emphasis on graphing concepts. Finally, the

exponential and logarithmic functions are covered in Chapter 13.

Chapters 14 and 15 deal with integral calculus. In Chapter 14, differenequations and exponential growth and decay are included as an appli-

tial

cation of antidifferentiation.

The

definite integral

is

intuitively introduced

and then is later formally defined as the limit of a sum. Techniques of integration and improper integrals are covered in Chapter 15. Since the integral table used in Section 15-3 contains formulas in terms of an area function

for a variety of rational functions, the

included

among

method

of partial fractions

is

not

the techniques of integration.

Chapter 16 introduces multivariable calculus, including partial derivatives, differentials, optimization, Lagrange multipliers, least squares, and double integrals. If desired, this chapter can be covered immediately after the diagram on page viii.) Chapter 17 presents some additional probability topics, most of which involve applications of calculus. Chapter 9 is also a prerequisite for

Chapter

14. (See

Finally,

this chapter. (See the

diagram on page

viii.)

Important Features Emphasis

Emphasis is on computational skills, ideas, and problem solving rather than on mathematical theory. Most derivations and proofs are omitted except where their inclusion adds significant insight into a particular concept. General concepts and results are usually presented only after particular cases have been discussed.

Examples and Matched Problems

This book contains over 460 completely worked out examples. Each exampie is followed by a similar problem for the student to work while reading the material.

The answers

end of each section Exercise Sets

for

to these

matched problems are included

This book contains over 5,000 exercises. Each exercise set

is

the

designed so

and

a

are mostly divided into

A

that an average or below-average student will experience success

very capable student will be challenged. They (routine, easy mechanics),

mechanics and some

at

easy reference.

B (more

theory) levels.

difficult

mechanics), and

C

(difficult

xii

Preface

Applications

Enough

applications are included in this book to convince even the most

skeptical student that mathematics

really useful.

is

The majority

of the

applications are included at the end of exercise sets and are generally

divided into business and economics,

An

groupings.

them choose

science,

life

instructor with students from

applications from their

own

all

and

social science

three disciplines can

field of interest; if

let

most students

are from one of the three areas, then special emphasis can be placed there.

Most of the applications are simplified versions of actual real-world problems taken from professional journals and books. No specialized experience is required to solve any of the applications included in this book.

Student and Instructor Aids Student Aids

Dotted "think boxes" are used

formed mentally

to

enclose steps that are usually per-

(see Section 1-2).

Examples and developments are often annotated through

A

critical stages (see

second color

Boldface type

is

is

used

used

Section

to indicate

to

to

help students

1-4).

key steps

introduce

(see Section 1-4).

new terms and

highlight important

comments.

Answers

to

odd-numbered problems

are included in the back of the

book.

Chapter review sections include a review of all important terms and symbols, a comprehensive review exercise set, and a practice test. Answers

to all

review exercises and practice

test

problems are included in

the back of the book,

A

solutions manual is available at a nominal cost through a book store. The manual includes detailed solutions to all odd-numbered problems, all

review exercises, and

all

practice test problems,

A computer applications supplement by Carolyn L.

Meitler and Michael nominal cost through a book store. The supplement contains examples, computer program listings, and exercises that demonstrate the use of a computer to solve a variety of problems in finite mathematics and calculus. No previous computing R. Ziegler is available at a

experience

Instructor Aids

a

is

test battery

necessary to use this supplement.

designed by Carolyn

publisher without charge. The

test

L.

Meitler can be obtained from the

battery contains six different tests

with varying degrees of difficulty for each chapter. The format inches for ease of reproduction.

is

8^

X

11

Preface

An

xiii

manual can be obtained from the publisher without manual contains some remarks on selection of and answers to the even-numbered problems, which are not

instructor's

charge. topics

The

instructor's

included in the

text.

A solutions manual (see Student Aids)

is

available to instructors without

charge from the publisher.

A computer applications supplement by Carolyn L. Meitler and Michael R. Ziegler (see

Student Aids)

is

available to instructors without charge

from the publisher. The programs in this supplement are also available on diskettes for APPLE II® and IBM® PC computers.* The publisher will supply one of these diskettes without charge to institutions using this book.

Acknowledgments In addition to the authors,

publication of a book.

We

many

others are involved in the successful

thank personally: Susan Boren, UniverMartin; Gary Brown, Washington State University;

wish

to

Tennessee at David Cochener, Angelo State University; Henry Decell, University of sity of

Houston; Gary Etgen, University of Houston; Sandra Gossum, University of

Tennessee

at

Martin; Freida Holly, Metropolitan State College; David lohn-

son. University of Kentucky; Stanley

Lukawecki, Clemson University; Lyle

Mauland, University of North Dakota Marquette University;

R. A.

at

Grand Forks; Carolyn

L. Meitler,

Moreland, Texas Technological University;

Marian Paysinger, University of Texas

at Arlington;

John Plachy, Metro-

politan State College; Walter Roth, University of North Carolina at Char-

Wesley Sanders, Sam Houston State College; Arthur Sparks, UniverTennessee at Martin; Martha Stewart, University of North Carolina at Charlotte; James Strain, Midwestern State College; Michael Vose, Austin Community College; Dennis Weiss, Indianapolis. Indiana; Scott Wright, Loyola Marymount University; Donald Zalewski, Northern Michigan University; Dennis Zill, Loyola Marymount University. We also wish to thank; lotte;

sity of

BoUow

Janet

for

another outstanding book design.

John Williams

for a strong

John Drooyan

for the

and

many

effective cover design.

sensitive

and beautiful photographs seen

throughout the book. Phillip Bender,

Woods

Gary Etgen, Robert Mullins, Mary Utzerath, and Caroline checking all examples and problems (a tedious but

for carefully

extremely important •

APPLE

II

is a

job).

registered trademarlc of Apple Computer, Incorporated.

registered trademark of the International Business

Machine Corporation.

IBM

is

a

xiv

Preface

Phyllis Niklas for her ability to guide the

production

Don

Dellen, the publisher,

services

book smoothly through

all

details.

who

continues

to

provide

and encouragement an author could hope

Producing

this

tent people has

new

been

edition with the help of

a

all

all

the support

for.

these extremely compe-

most satisfying experience. R. A. Barnett C.

M.

/.

Burke

R. Ziegler

Ki

^^>:-

.i*^>'^> - b b
— 25

b

(C)

?

—6

(C)

—5

the

8?2

Inequalities

now

Let us

(D)

a

left of 5

on

a

?

either


.

c

number

line

— 2 on a number line of — 10 on a number line of — 25 on a number line

to the left of

is

to the right

is

to the right

-20 ?0

(B)

?

is

Replace each question mark with (A)

mark with

replace each question

< or >.

(C)

-3? -30

(D)

0?-15

and Line Graphs

consider the inequality

X&-4 The solution

set for this inequality

is

the set of

all real

when substituted for the letter x make the statement the

number —4 and

line.

We can graph

dot at

all real

numbers

to the right of

the inequality on a real

—4 and drawing a heavy

number

line to the right of

true.

—4 line

numbers which This set includes

on

a real

number

by placing a

solid

— 4. as shown in Figures.

Solid dot indicates that

-4

is

included

-»x

-5 Figure 8

Example 15 Solution

Graph x < 4 on

a real

number

line.

The solution set is the set of all real numbers to the left of 4 on a real number line. The graph is obtained by placing an open dot at 4 and then drawing

a

heavy

line to the left of 4.

Open 4

dot indicates that

is

not included

-^x

Inequality Statements and Line Graphs

1-3

Problem 15

Graph each inequahty on

x>2

(A)

The

-3

a real

number

inequality statement

«x< 2

means

x^

Represent each Jine graph using inequaJity notation. 55.

^

——

o

-3 57.

(

X

56.

^

x

58.

(

)

o 4

5

)

-5

» 10

o

°

-4

4

)

X

)

X

Represent each pair 0/ inequalities as a single double inequality and graph. 59.

61.

63. 64.

and x^-5 and x =£ 5 -2l x3^-4 and x2

_J

Positive Integer Exponents

1



1



— y«

Simplify each expression using the properties of exponents. (A)

(xVT

(C)

(B)

(-uV)3(3u^v3

28xV 35xV

(D)

(3xV^)' (E)

(6xV^

Answers to Matched Problems

32.

(A)

64

(B)

-27

33.

(A]

uW«

(B)

pq^rs=

34.

(A)

x"

(B)

z'°

(C)

b'V

(E)

y'

(F)

1

(^^

^

(C)

-9u"v^

35.

(A)

x^V

(E)

?^

(D)

(C)

u"

23

(D)

4x^ 1°'

W'

Exercise 1-5

A

Replace each question mark with an appropriate expression. 1.

47

48

Preliminaries

13.

1-6

1-2

Chapter Review

49

Real numbers and the rules of aigebra. natural number, integer,

number, real number, real number line, symmetry property, transitive property, substitution property, commutative properties, associative properties, distributive property, addition and multiplication with the number 0, multiplication by the number 1, additive inverse, negative, multiplirational

number,

irrational

equality symbol,

cative inverse, reciprocal 1-3

Inequality statements

and

line graphs, inequality symbols, less than,

greater than, less than or equal set, line

bol,

a

1-4


2-2

^4

V4

2

/3

f^

^2



Radicals or expressions containing radicals are said to be in simplest radical form

if

the four conditions listed in the box are satisfied.

Simplest Radical 1.

A

Form

radicand contains no factor to a power greater than or equal to

the index of the radical.

vx^ violates this condition. 2.

The power

common

of the radicand

factor other

than

and the index of the radical have no 1.

Vx* violates this condition. 3.

No

radical appears in a denominator.

y/ 4x 4.

No

violates this condition.

fraction appears within a radical.

yf^

violates this condition.

3-4

In calculations

may

it

Example

21

Solutions

be desirable

The choice

simplest radical form.

to

will

129

Radicals

use radical forms other than the

depend on the circumstances.

Write each expression in simplest radical form. (A)

n/TtS

(A)

\/l75

\/^

(C)

(D)

^

(E)

J~

not in simplest radical form according to condition

is

=

175

V27^

(B)

We

52-7.

VTtS

=

Vs^

7



1,

since

1.

Note

have

=

!

Vs^ V7

1=5^7 I

I

or

=

VTtS (B)

v27x^

is

V25

7



=

I

V?

n/25

I

=

5 >/7

not in simplest radical form according to condition

that 27x3

=

33^3,

I

We

have

=N/(3xf V3x .

I

.

I

I

= 3xV3x or

727x3 (C)

"n/o^

is

=

V(9x2)(3x)

I

=

Vox^

i=3xV3x

n/3x

not in simplest radical form according to condition

a factor of both the

power

of

a**

and the index

2,

since 3

of the radical, 15.

is

We

have

(D)

—=

violates condition

3.

since the denominator contains

\/2.

We

have

V2

6x

(E)

-J—

We

6x

6xV2

^2

r-

violates condition 4, since a fraction occurs within a radical.

have '^1'= 2

I

x



2

"V 2



2

/

_

/2x

_

> 2^

n/2x

V2^

we

could write I

/x_

I

I

I

Or.

|_ >/2x

Vx|

Vl~72l

1

_

Vx

V2 {_ n/2x

~^/2* V2

r

2

2

130

Exponents and Radicals

Problem 21

Write each expression in simplest radical form. V45

(A)

In

V8?

(B)

Example

we

21,

^V^

(C)

^

(E)

j:

dealt primarily with square roots. Simplifying radical

expressions where the index n the

(D)

greater than 2

is

is

accomplished in

much

same manner. For example, we can simplify

'V54xVV as follows:

= V(27xV^j(2yz^

V54x^y^°z^

7

37„"^~

~3

I

7 V2yz2

~Q 7^

= V27xV^

I

I

Write the radicand as a product so that the

"*

.

,

,

first

factor ,.

,

,

,

can be expressed without radicals and

I

=

cube root of the

I

3x^y^V2yz^

the cube root of the second factor will

be in simplest radical form.

To

simplify the radical expression

6.\y

we would proceed 6xy

6xy

\'2^xy^

Vlx^y

'V22^

_

Vzx^y

^

as follows:

6x yV2'xy ^

I

Notice that in order

to

obtain the radical

j

V2^x^y^

I

V2Vy^

I



denominator (which

we must multiply the 3n v2^xy^. numerator and the denominator by ^



simplifies to 2xvl,

3r;

=

in the

6xyv4xv^ '—

,

2xy

,

,

,

'

= 3'V4xy2 Removing radicals from a denominator, as illustrated above, is commonly referred to as rationalizing the denominator. Keep this mind we will discuss this process again in Section 3-5.

in



Example 22

Write each expression in simplest radical form.

=

(A)

V375

(B)

v'24x5y2z'

(C)

'V64xV'

,_, (D)

—V6xy

'n/125

12x2y^'^

=

= =



3

=

^V5^



3

r=

V(4xV^z^)(6xz)

^/(32x=)(2x3y2)



12x2yv/5

-^^

V6xy

\

^=

Tbxv •

v6xy

Vt^ V3

1=5^3

= Ux*y^z^ >/32x=

=

Vbxz

^

=

2x^yz'^4%xz

V2xv"i= 2xV2xV'

12x2vV30xy 6xy

,

= 2xV30xy

3-4

_

6xy2 (E)

V3^

Vox^^ 3" 3/

(F)

_

3^ 6x^

_

V>

9v2 2x2

Bxy'Vsxp

VsVp

Vsxy^ 2x^

3

3/

V> 22x 92v

4x

Problem 22

Vsxp _ exy^Vsxy^

6xy2

3

23^3 2'x'

Radicals

131

2yV3xy2

3xy

76x2

76x2

V2^

2x

V2^x^

Write each expression in simplest radical form.

VT28

(A)

V75a^bV

(B)

Quv^-J?

V64xV

(C)

(D)

J3uv (E)

V 3y2

More about vx" So

far,

in our discussion

we have

restricted all variables to positive real

numbers in order to avoid any difficulties. If we removed we would no longer be able to say in general that

Of course,

this

continues to hold

not be true. To see

this, let

if

x

=

0;

however,

us consider Vx^. For x

=

if

x

5,

is

this restriction,

negative,

it

may

we have

^=5 = x For X

= — 5, we

have

A- 5? = V25 = n/52 = 5#x Thus, Vx^

it is

=

not true that

X

numbers.

for all real

We have

If

the following important result:

X represents n/^

a real

number, then

= |x|

Recall from Chapter

1

that

|x|

denotes the absolute value of x and

by X |x|=

Remember, x = — 5.

S

-X

if

X

is

if

x

=

if

x

is

positive

negative

is

defined

132

Exponents and Radicals

—x

Note that

=

V5^

positive

is

when

and

|5|=5

x

negative. Thus,

is

V(-5f

= |-5| =

we have

5

which is consistent with our previous calculations. Now, consider vx^. For x = 5, we have

=

Vs^

5

=X

= — 5, we

For X

V[-5f

= V-125 = -5 = X

In general,

If

have

we have

the following:

X represents a real number, then

=X

Vx^

When

variables represent real numbers,

we must

be very careful in

simplifying an expression such as

It is

common

a

error to set this expression equal to 2x. According to the

discussion above,

+ V^ =

'Vx^

When x But x

X

is

+ |x| =

+ |x|

positive or

x

when x

+ |x| =

x

we have

x

+x= is

+

0,

we have

2x

|x|

negative,

(-x)

we have

=

|x|

Thus,

^ ,^ V^ + n/^ = Example 23 Solution

Simplify:

We

r

2x

i

sVx^

-

if

X

^0

ifx/27

4 732

-

2 -Jib

= 5 79 3 - 2 725 3 = g.3^_2.5V3 = 15>/3- 1073 = 5^3 •

+ -= = 47l6

^

=4





Express each radical in



simplest radical form.

+ —=•—=

2

Write in simplest radical

^ ^

form.

872

r 472 H

2

= (C)

5\/2x^

1672

- xVl6x = =

+ 472 =

2072

sVx'

-



5x'72^

2x

x'78



Write in simplest

2x

- 2xV2^

radical form.

= 3xV2x (Di

+ Vn

f\*'-^^-il

3

$+3^3 :-V5+3V3= Problem 25

— V3

10V3

3

3

Simplify by writing in simplest radical form and combining terms when-

ever possible.

(A)

67T25-3745

(B)

—12r-2727

(C)

yty^ + 2vV32y

73 (D)

V625-10^

Multiplication

Many expressions involving radicals can be multiplied in the same manner in which we multiplied polynomials. The distributive property justifies this procedure.

radicals.

Example 26

illustrates several different

products involving

Basic Operations on Radicals

3-5

Example 26

137

Multiply and simplify whenever possible.

V5(VT5-4)

(A)

l_= Vs

n/i5- Vs •4j



= V75-4V5 = V25 3-4V5 = 5\/3-4^ •

+

(V3

(B)

4)(V3-5)

=%/3

I



V3

+ 4V3-5V3-2OJ

= 3-V3-20 = -17-V3 (3 ^/5

(C)

-

+

2 73){>/5

V3)

j=3V5-n/5-2VT5 + 3>/T5-2V3-V3

l=3-5 + VT5-2-3

= 15 + Vl5-6 = 9 + +

(Va-5)(Va

(D)

3)

r=7a



|

I

7l5

Va-sVa + sVa-lS

I

= a-2Va-15 [V^ + V^)(V^ - V^) r=

(E)

VP + 'n/xv -

V^ - Vp

J

= X + Vxy — Vx^ — y Problem 26

Example 27 Solution

Multiply and simplify whenever possible. (A)

V3(VT5-7)

(C)

(2V3-3V2)(2V3

(E)

(Va

Vb^)(V^

Evaluate x^

For x x2

Problem 27

-

-





V2,

6x

+

7

Evaluate x^

+ 7

+

3V2)

(D)

+ 4) (V^ + 6)(V^ - 3) (v^-2)(V7

Vb)

using x

=

3



V2.

we have

= (3 - ^2)2 - 6(3 - V2) + 7 = 9 - 6V2 + 2 - 18 + 6V2 + =

-

6x

Quotients In Section 3-4

+

6x

3

=

(B)

+

7

using x

=

3

+

V2.

— Rationalizing Denominators

we found that an expression such as V5/V3 can be reduced to

simplest radical form by multiplying the

by

•Is.

Thus, we

obtain

1 1

V5|_V5_V3_^ V3

~^

I

7

V3

V3^

[_ ^15 I

3

numerator and the denominator

138

Exponents and Radicals

The process by which

denominator

a

is

cleared of radicals

is

called ration-

alizing the denominator. It is

such

natural to ask

we can

if

rationalize the

denominator of an expression

as

V^-V3 That is, can we write this expression in a form where no radical appears in the denominator? The answer is yes. However, before we illustrate the procedure, (a

-

it

b){a

will be useful to recall the special product.

+ b) = a^-b^

For example, (Vs

-

>/3)(V5

+

V3)

r=

(Vf]2

we can

This suggests that

- (^y^ = 5-3 =

rationalize the

2

denominator of

4

V5-V3 by multiplying the numerator and the denominator by Vs + n/s (which obtained by changing the middle sign of the denominator). Thus,



4

—=—

V^-V3

4



n/5 •

^/H-V3

_

4V5

V5

+ V3 — + V3

.

.

is

,^. , A and Multiply numerator ,

denominator by V5

+

V3.

+ 4v^

5-3 4n/5

+ 4n/3 2

2(2\/5

+

2V3)

Simplify by factoring 2 2

^2V5

+

from the numerator and

2V3

Example 28 further

canceling.

illustrates this

procedure

for rationalizing

nators.

Example 28

Rationalize the denominator and simplify

(A)

^—= ^— -;^

3-1-



3- n/6 —

,,

whenever

possible.

J Multiply numerator and ,

.

,

denominator by

3



v6,

obtained by changing the

middle sign of Reduce.

3

+

\/^.

denomi-

Basic Operations on Radicals

3-5

Vs

2V5

2V5-3V3

V5

(B)

+ 3v^~"2V5 +

Multiply

2V5-3V3

3x/3

numerator and denominator by

10- 3VT5 20-27 10- 3715 ^7

2n/5-3>/3.

10 +

IO-3V15

(C)



Va

-Ja

\/b

a

Problem 28



-12

-fa

+ 2^fah +

+

sih

denominator by

h

-Is

,„, (B)

4^/3

/lO-

Answers to Matched Problems

3^fT5

Multiply numerator and

^h

whenever

Rationalize the denominator and simplify

(A)

+

Vx

3V2

+

Vv

(A)

-6^3

(B)

llt/2xV

(C)

zVsu^v

25.

(A)

21V5

(B)

-2V3

(C)

lly1/2y

26.

(A)

3^-7V3

(D)

X

+ 3VX-I8 V5

+

(B)

-1+2-I7

(E)

a

-

(C)

+ Vab 4-^6

Vo^b^

n/2

(A)

28.

-

possible.

^-V^

,„, (C)

24.

27.

139

(B)

l^fbuv

sVs

(D)

-6 b

2Vxy

+

(C)

10

Exercise 3-5 In the following

problems

all \'ar]ables

represent positive real numbers.

Simplify by writing in simplest radical form and combining terms whenever possible.

+ 3V3-5V7

1.

3Vx-8Vx

2.

5V7-3V7

3.

V7

4.

Vm-sVn + sVm

5.

n/i2

-

6.

V50

7.

VT8

8.

727-^3

+

2V2

V^

4-^/2

Multiply and simplify whenever possible. 9.

V5(^/5-3)

-

10.

12.

^(^fa

15.

^/5(2^/T5- 3>/2)

17.

(n/5

-

13.

5)

3)(V5

+

3)

VTl(VTT-2) ^fx[7

-

Vx)

11.

Vu(Vu

14.

Vz(5

16.

V7(3V3-4Vl4)

18.

(V3

+ 4)(V3-4)

+

+

3)

Vz)

140

Exponents and Radicals

+

19.

(V2

21.

(n/w-2)(Vvv

5)2

+

2)

20.

(V5-3)2

22.

(Va

+

5)(Va-5)

Rationalize each denominator and simplify. 1

V5-2

5

3

.. 24.

23.

^f7

-

26.

25.

4

1

+

3-

^6

V3

72 27.

28.

V6

B

+

29.

30.

vTo-2

2

yfa

+

3

Simplify by writing in simplest radical form and combining terms whenever possible. 31.

V5OX-V8X

33.

2V8

35.

+ 2xVT8x V24z^ - 3V3F + zVsT

37. 39.

+

3x/T8- V32

2\/8x'

^ + V27 — V32 + 7— 4/

42.

34.

+ 2Vl2uv 2^->/48 + 3V75

36.

3yV75y- 4^277^

38.

xVl62

V27uv

32.

Vi

40.

- 7^ 2mn

1

43.

-

V2x^

+

V40-^

41.

+

44.

^f54I

^^o^-yJY

V8 In

ProbJems 45-56 multiply and simplify the product whenever possible.

-

47.

+ 5) (2V3- 1)13^3 + 4)

49.

(3Va-5)(2N/a

+

3)

51.

(5V2-3n/3)(3V2

+

53.

(Vm

45.

55.

(Vx

3)(n/x

4V3)

- Vn)(Vm + x/n) (Vm^ - Vn)(Vm + Vi?)

57.

Evaluate x^

58.

Evaluate x^

-

lOx lOx

+ +

4 using X 4 using x

(Vz

48.

(3V5-5)(2V5-3)

50.

(5Vv

52.

(4n/3-3V5)(2\/3-2V5)

54.

(2Vx-3Vy)(2N/x

5 5

+ -

V5-

62.

65.

+

2)(3>/v

+

5)

+ 3Vy) (v^ 2W](W^ 3Vb

56.

= =

+ 4)(Vz-7)

46.

V2T. V2T.

Rationalize each denominator and simplify.

59.

3t/32x*

Chapter Review

3-6

141

- lyylZx^y + ^xyVlSxy

69.

ix^2xy^

70.

aV81ab^-bV24a^+3V3a^

Multiply and simpJify the product whenever possible. 71.

(V^ - 3v^)(V^ +

2V?2)

(2V^ + t/y)(1/^-4t/P)

72.

Rationalize each denominator and simplify.

2Vx

44u

3-6

+ 3Vy

74.

73.

+

3^lv

Chapter Review Important Terms

3-1

and Symbols

Integer exponents, positive integer exponents, zero exponents, negative integer exponents, growth, decay, a", a", a~"

powers of 10,

X

10", a

X

10""

3-2

Scientijic notation, scientific notation,

3-3

Rational exponents, root, square root, cube root, nth root, principal

nth 3-4

root, irrational

numbers, rational exponents,

a

a^'", ~, a"'^", a'""^"

Radicals, radical notation, radical, index, radicand, equivalent rational exponent form, properties of radicals, simplest radical form, rationalizing denominators, Va, Va, Va^, (Va)"'

3-5

Rasic operations on radicals, simplifying radical expressions, addition, subtraction, multiplication, rationalizing

Exercise 3-6

denominators

Chapter Review Work through

all the

problems

in this

chapter review and check your

back of the book. (Answers to all review problems are Where weaknesses show up, review appropriate sections in the text.

answers

you are

in the

satisfied that

you know the materia], take the practice

test

there.)

When

folJowing

this review.

Give the value of each expression.

3.

-49)3/2

(-27)^

4.

-8-V3

142

Exponents and Radicals

In

Problems 7-14 simplify each expression and give each answer using

positive exponents. 5

7.

m-'m^

8.

[3a^b^]°

9.

— r

11.

[x-'yT'

15.

Convert (A)

-^ a

12.

13.

53,000,000,000

17.

Convert from rational exponent form (A)

to

14.

-5

(uV3v2/5)i5

standard decimal form:

3.8X10'

(7z)=^''

(B)

(B)

5.7X10"=

M[2x^yf

(B)

to radical notation

4w3/^

Convert from radical notation (A)

u

0.000 004 9

(B)

Convert

18.

(^y

10

to scientific notation:

16.

(A)

jj-7

"

sIm'

-

to rational

exponent form

n'

Write each expression in simplest radical form. (AJJ variables represent positive real numbers.)

19.

3-6

46.

(27m^n-^)'/3

Chapter Review

a_^^^

(-i/4(i/3

47_

143

n-4/5

Problems 51 and 52 multiply and express each answer using positive

In

exponents. 51.

3xV4(5xV4-2x-3/4)

53.

Convert

54.

52.

(3xV^

- yV2)(^V2 _

gyVZ)

to correct scientific notation:

(A)

524.000,000

(B)

0.000 583

(C)

832X10'*

(D)

529X10-=

Evaluate the expression below using scientific notation, and give the answer in both scientific notation and standard decimal

form

0.000 020 8

260(0.000 04) 55.

Convert from rational exponent form -5y2/3

(A) 56.

aV3-

(B)

0-1/3

Convert from radical notation (A)

-6xt/(2xy2)3

(B)

to radical notation:

to rational

exponent form:

— \iw

Write each expression in simplest radical form. 57.

'V125XV

^

58.

^x/32^^

3^

27v^

59.

Vs^^ V4^

62.

3^==

Perform the indicated operations in Problems 63-73 and write each answer in simplest radical form. 63.

5V25-2V^ +

65.

V^+Vi

67.

2z>/T6^

+

69.

[5^ -

^f^][2^

3745

3V^ + 3s/^]

64.

3xV27^-2^/^

66.

V8U^-,/H

68.

(2^5-3721(3^5

70.

(^V^-V^)(y^ +

n/7-3

27x

71

77

+

272 73

372 74.

72.

2

37x-

+ 73 + 273

Evaluate x^

-

4x

-

9 using x

=

2

-

713.

7v

+

471)

V^)

144

Exponents and Radicals

Simplify and express each answer using positive exponents. 75.

(u-^

3-2 77.

^-3

+ v-T' + 3-^ _ q-4

76.

(w'/5

+

2w-^/=)(w-'/=

-

3^"/=)

Write each expression in Problems 78-82 in simplest radical /orm. (All variables represent positive real numbers.)

78.

716x^

2x

+4

79.

+1 -1

V4x2 80.

2nV3m^-mV3mn^+ mnV243mi?

81.

V|-Vf +

83.

Simplify st/x*

^

(A)

Practice Test:

Chapter

r-

For x

2^fa-3^/b

/

V3^

-

82.

-^_-^

3Vx^:

5^

For x

(B)




'^..*'

»

CHAPTER

4

Contents 4-1 Linear Equations

4-2 Linear Inequalities 4-3 Quadratic Equations

4-4 Nonlinear Inequalities 4-5 Literal Equations

4-6 Chapter Review

4-1

Linear Equations Linear Equations Solving Linear Equations

Equations Involving Fractions with Constant Denominators Equations Involving Fractions with Variables in the Denominators Applications

An

equation

is

a

mathematical statement obtained

when two

The expressions involve one or more

algebraic

an equation

expressions are set equal to one another.

in

may

variables.

consist of a single

number

or

The

following are examples of equations:

3x-5 =

2x-|-3y=12

5(2x-|-l)-|-4

In this section

we

+ 9x - 5 =

2x2

will only consider a special class of equations called

linear equations in one variable.

Linear Equations

An equation in one variable that involves only first-degree and zero-degree terms

is

called a linear or first-degree equation in

one variable. For

example,

=

-

=--

3x

+

5

are linear equations.

A

solution or root of an equation

2(3x-5)-|-6

when

and

3

2{x

+

substituted for the variable (wherever

3)

it

5

is

a

number which

occurs) gives the

same

numerical value on both sides of the equality sign. The set of all solutions of an equation is called the solution set. To solve an equation means to determine the solution

set.

we need to meaning of "equivalent equations." Two equations are said to be equivalent if they have exactly the same solution set. For example, the Before

we

describe the process used to solve an equation,

define the

equations x

148

+5=

15

and

x

=

10

4-1

Linear Equations

are equivalent since both have 10, and only 10, as a solution.

149

The process

of

solving an equation will involve creating a sequence of equivalent equations resulting in a final equation,

obvious.

The

such as x

=

10, in

which the solution

is

properties of equality listed in the box will serve as the basis

for solving linear equations.

Properties of Equality

150

Equations and Inequalities

3x-9 = 7x + 3 l3x — 9 + 9 = 7x + 3 + 9' I

1



3x

3x

= 7x +

7x

= 7x + 12 —

Addition property

12

7x

Subtraction property

I

I

-4x =

12

— 1-4—4 r-4x = —

12

I

I

.

.

Division property f f .

I

1

I

x

Check

^.

"I

To check

= -3 we

this solution,

tion to see

if

substitute x

= —3

back into the original equa-

the equality holds:

+3 -9-91-21 + 3 -18^-18

3(-3)-9i:7(-3)

Problem

1

Solve 2x

When

—8=

5x

+

4 and check.

symbols of grouping are present on one or both sides of an equa-

we first remove them from each expression and combine Then we can proceed as in Example 1. tion,

Example

2

Solution

8x

-

3(x

3(x

-

4)

Solve:

8x 8x



3x

+

12

5x

-I-

12

5x 3x x

The check Problem

2

Solve 3x

-

2(x

= 2(x - 6) = 2x — 12 = 2x — 6 = 2x- 18 = -8 = -6

is left

2(2x

- 4) =

-

6)

-I-

like terms.

6

-h

6

Remove

-I-

6

Combine

parentheses. like terms.

Solve as before.

to the reader.

-

5)

=

2(x

-I-

3)

-

8

and check.

Equations Involving Fractions with Constant Denominators

When X

we

an equation involves

fractions,

such as

X

can simplify the equation by "clearing" the denominators. This is done by multiplying both sides of the equation by the LCD of the fractions

4-1

present. For the above equation the

equation by

15,

we

LCD is

15.

Linear Equations

Multiplying both sides of the

obtain

Do not confuse operations on equations with operations on

Note:

151

alge-

braic expressions that are not equations. For example, in the algebraic

expression

+4 which looks very much sides" by the

have two

LCD

above equation, we cannot multiply "both

like the

15 to clear the fractions, since this expression does not

sides! In this case,

we combine

the three fractions into a single

fractional form:

X

5X-3X +

_ 111' X

which

Example

3

4

The LCD

is 6.

6-

n

/ x + 2 _ x\ |^^^^--| = 6-5 \

/x

+ -2

3(x

3/

2

2\



\

3

— namely 15 (the LCD).

Multiplying both sides of the equation by

:

I

Problem

60

3

:

^

+

=5

Solve:

B

2X

15

has a denominator

still

2

Solution

60

15

/

^

-

6



X =

3

30

6.

we

obtain

152

Equations and Inequalities

Equations involving decimal fractions can sometimes be transformed into a

form free of decimals by multiplying both sides by a power of

10.

Consider the following example.

Example

4

Solution

Solve:

To

0.4(x

-

-

30)

=

0.15x

clear all decimals,

we can

8

multiply both sides of the equation by 100:

0,4(x-30)-0.15x I

I

100

[0.4(x



-

30)]

- 100

40(x

-

40x-

Problem 4

0.25x

Solve:

+

0.4(x

(0.15X)



30)

-

1,200

=

30)

=

Multiply both sides by

8

= 100

- 15x = -15x = 25x = x =

100.

1



81

Solve as before

800 800 2,000

80

27

we have always obtained a single or unique what we normally expect when solving an equation. However, two other situations may occur. The first is when the given equation is an identity. A linear equation in one variable is called an In the

above examples

solution. This

identity

if

is

the solution set

replace x by any real

3x

we

+

7

=

3(x

+

2)

will obtain the

identity.

If

we

+

is

the set of

number

all real

numbers. For example,

if

we

in the equation

1

same value on both

sides.

Thus, this equation

an

is

attempt to solve this equation by the usual methods,

we

obtain the following result:

3x

3x

3x

+7= +7= +7= 3x = =

3(x

3x

3x

+ 2) + 1 +6+l +7

3x

Clearly, the last equation

is

true.

What

this tells us

about the original

equation can be generalized as follows:

Identities If

a linear equation in

one variable can be reduced to = using the is an identity and the of all real numbers.

properties of equality, then the given equation solution set

is

the set

4-1

The second

+

x

5

=

x

+

when an

situation occurs

example, there

is

no

real

number x

+

5

X

= = =

x X

+ +

equation has no solution. For

which

we would

obtain

10 5

5

Obviously, this

equation

last

Equations with If

153

10

Solving this in the usual manner, x

for

Linear Equations

No

is

not valid. In general,

we have the following:

Solution

one variable can be reduced

a linear equation in

properties of equality,

where

b

=5^

0,

to

=

b using the

then the given equation has no

solution.

Example

5

Solution

5(3x

Solve:

-

5)

+4=

4(5x

7)

-

5x

+ 4=4{5x + 7)-5x 15x - 25 + 4 = 20x + 28 - 5x 15X-21 = 15X + 28 15x = 15x + 49 = 49 Impossible!

5{3x-

5)

The given equation has no Problems

+

Solve:

-

3

-

4(2

3x)

=

solution.

6(2x

-

1)

+

1

Equations Involving Fractions with Variables in the

Denominators Many

equations involving fractional forms with variables in the denomi-

nators can be transformed into linear equations by clearing the denominators.

For example, consider the equation

—+— + 9

x

We

=1

,

2

5

2

can clear the denominators by multiplying both sides of

by the LCD, which

is

a solution, since this

2(x

+

2).

Before doing

would create

a

this,

note that x

this

equation

= — 2 cannot be

in the denominator. Thus,

we must

154

Equations and Inequalities

include the condition x

we

y=

— 2,

as indicated below. Solving this equation,

have

—^ + 5=x+ 2

2(*^^)



(-^^ + 2(x +

2)



5

=Z[x +

2)



+

10x 10x

cannot be a

solution.

Multiply both sides

i

by the \

18

—2

+ -1

2

\

xample 6

X

+ +

= x+ 2 =x+2 9x = -36 x = -4

LCD =

2(x

+ 2)

and simplify. I

20

Solve the linear

38

equation as before.

4-1

Linear Equations

155

Applications The methods discussed

in this section

can be utiUzed

to solve a large

variety of practical problems.

Example

7

Break-Even Analysis

It

costs a record

company $6,000

prepare a record album for production.

to

This includes recording costs, album design costs,

etc.,

which represent

one-time fixed cost. Manufacturing, marketing, and royalty costs variable costs for

to

Solution

$4 each,

— are $2.50 per album.

album

the

If

is



a

all

sold to record shops

how many albums must be produced and sold for the company

break even?

Let

= Number of records sold C = Cost for producing x records R = Revenue (return) on the sale X

of records

The company breaks even when R =

C = Fixed

costs

= $6,000 + To

1.50X

X

For X

C=

=

= = =

costs

where

and

R

we

solve

= $4x

$2. 50x

find the value of x for

4x

Check

+ Variable

C,

6,000

+

which R

=

C,

2.50x

6,000

4,000 records

4,000,

6,000

+ +

R = 4x

and

2.50x

= 4(4,000) = $16,000

= 6,000 2.50(4,000] = 6,000 + 10,000 = $16,000 Thus, the company must produce and Sales over 4,000 will result in a profit,

sell

and

4.000 records to break even.

sales

under 4,000

will result in a

loss.

Problem 7

What

is

the break-even point in

Example

7 if fixed costs are

$9,000 and

variable costs are $2.80 per record?

Since the variety of practical problems that can be solved using linear

equations will

work

is

very extensive,

for all

of the next page.

problems.

it is

difficult to

Some

describe a single procedure that

guidelines are listed in the box at the top

156

Equations and Inequalities

Guidelines for Solving 1.

2. 3.

Word Problems



several times Read the problem very carefully Write down important facts and relationships.

unknown

Identify the

if

necessary.

quantities in terms of a single letter



if

possible. 4.

Write an equation relating the

Example

8

Investment

A

5.

Solve the equation.

6.

Write

7.

Check the

down

all

quantities based

'

retired couple has $60,000 invested, part at

remainder

at

16%

on the

desired values asked for in the original problem.

solution(s).

per year. At the end of

from both investments Solution

unknown

problem.

facts in the

is

$8,100. Find the

10%

per year and the

year the total income received

1

amount invested

at

each

rate.

Let

= Amount invested X = Amount invested X

$60,000

at

10%

at

16%

The total income received is the sum of the income from the 10% investment and the income from the 16% investment. Translating this into an equation and solving the equation,

-

= lOx + 16(60,000 x) = lOx + 960,000 - 16x = -6x =

O.lOx

+

0.16(60,000

X 60,000

Check

10%

of $25,000

16%

of $35,000

= =

$60,000

Blending

Problem 8

Solve Example 8

Example

A

9

— Food Processing

8,100

Multiply both sides by 100.

810,000

Solve for

x)

-

X

we have

x.

810,000

-150,000

= $25,000 = $35,000

invested at

10%

invested at

16%

$2,500 $5,600 $8,100

if

the total income from both investments

is

$6,900.

candy company has 1,200 pounds of chocolate mix that contains 20% How many pounds of pure cocoa butter must be added to the

cocoa butter.

mix

to obtain a final

mix

that

is

25% cocoa

butter?

Solution

Let

= Amount

X

The amount

of pure cocoa butter

added

of cocoa butter in the original

mixture plus the amount of

cocoa butter added must equal the amount of cocoa butter in the mixture. That

Amount

'

vin original

of

(Amount of

'

(Amount of

^

cocoa butter

cocoa butter

added

mix,

+

0.20(1,200)

in final

/

=

X 20(1,200)

24,000

+ +

= = lOOx 75x = X = lOOx

30,000

+

+

I

mix /

0.25(1,200

25(1,200

\

+

x)

x)

25x

6,000

80 pounds of cocoa butter

must be added

Answers to Matched Problems

Solve Example 9

= -4

the original mixture contains

if

x

=4

x

=

1.

x

5.

The equation

6.

(A)

8.

$45,000

9.

112 pounds of cocoa butter must be added

X

2.

=

2

at

is

3.

No

solution

10%; $15,000

Solve each equation.

2

an identity. Every

(B)

Exercise 4-1

1.

final

is,

cocoa butter

Problem 9

157

Linear Equations

4-1

at

4.

real 7.

x

18% cocoa

= 60

number

is

a solution.

7,500 records

16%

butter.

158

Equations and Inequalities

23.

B

much

invested at each rate

the income from both investments

An

$15,000 for two paintings. She sold the first and the second for an 11% profit. If her total was $1,890, determine how much she paid for each painting.

Resale.

art dealer paid

painting for a profit

62.

if

$4,050?

totals

61.

is

ResaJe.

A

15%

profit

used car dealer paid $7,200

two

for

cars.

15% and the second for a loss of 3%. how much he paid for each car.

for a profit of

$540, find 63.

64.

Investment.

You have $12,000

how much should

12% on

amount invested?

Investment.

An investor has $20,000

$1.80,

The

the candy

If

be invested

at

(A)

Earn

(B)

Break-even analysis,

weekly

67.

how many

at

to yield

invested at

each rate

pound

8%

to yield

of

candy

week

to:

A record manufacturer has determined that its is C = 300 + 1.5x, where x is the number of If

records are sold for $4.50

to

week

for the

break even?

A fuel oil distributor has 120,000 gallons of fuel with How many gallons of fuel oil with a 0.3% sulfur

0,9% sulfur content.

content must be purchased and mixed with the 120,000 gallons obtain fuel 68.

Ecology.

reached

oil

with

a

to

0.8% sulfur content?

One day during the winter the temperature in the Antarctic a high of — 67°F. What was the temperature in Celsius

degrees? 69.

is

how

$900 per week

a profit of

records must be produced and sold each

PolJution control. a

10% and

each rate

cost equation

manufacturer Life Sciences

was

per pound, determine

for $3.30

records produced and sold each week. each,

first

small candy manufacturer has fixed costs of

be produced and sold each

Breakeven

resold the

to invest. If part is

variable cost to produce one

then sold

is

many pounds must 66.

A

Rreak-e\-en anal\'sis.

$1,200 per week.

He

his total profit

to invest. If part is invested at

the rest at 15%,

the total

If

and the rest at 12%, how much should be invested 11% on the total amount invested? 65.

159

Linear Equations

4-1

[Note:

°F

=f

°C

A

+

32]

and game department number of rainbow trout in a certain lake using the popular capture-mark-recapture technique. He netted, marked, and released 200 rainbow trout. A week later, allowing for thorough mixing, he again netted 200 trout and found eight marked ones among Wildlife

management.

naturalist for a fish

estimated the total

them. Assuming that the proportion of marked fish in the second sample was the same as the proportion of all marked fish in the total population, estimate the 70.

number

of

rainbow trout in the

lake.

Anthropology. In their study of genetic groupings, anthropologists use a ratio called the cephalic index. This

head

to its

length (looking

age. Symbolically,

down from

is

the ratio of the width of the

above) expressed as a percent-

160

Equations and Inequalities

C=

lOOW -

where C

is

the cephahc index,

W

is

the width, and L

is

the length.

If

an

Indian tribe in Baja California, Mexico, had an average cephalic index

and the average width of the Indians' heads was was the average length of their heads? of 66

Social Sciences

71.

PsychoJogy.

The

intelligence quotient (IQ)

mental age (MA), as indicated on standard

9-year-old 72.

girl

City planning.

8,

by the chronological

if

a child has a

the calculated IQ

is

mental

150.

If

a

has an IQ of 140, compute her mental age.

A city has just incorporated additional land so that the

area of the city

is

now 450

square miles. At present, only

area consists of parks and recreational areas.

have demanded that 15% of the parks and recreational areas. parks and recreational areas

4-2

what

found by dividing the

is

tests,

age (CA) and multiplying by 100. For example, age of 12 and a chronological age of

6.6 inches,

meet

this

7%

of this

voters of the city

total area of the city

How much to

The

should consist of

land must be developed for

demand?

Linear Inequalities Properties of Inequalities

Solving Inequalities Applications

Just as

we

solved linear equations in Section 4-1,

inequalities

where one

algebraic expression

is

we

will

now

solve linear

greater than or less than

another expression.

Properties of Inequalities There are four inequality symbols: , reviewed in the box.

Inequality Symbols

and

^. Their

meanings are

4-2

Formally,

we say that a < b or b >

a

if

Linear Inequalities

there exists a positive real

161

number

p such that a + p = b. Intuitively, if we add a positive real number to any real number, we would expect to make it larger. That is essentially what the formal definition of < and > states. As we have seen in Chapter 1, a < b can be interpreted geometrically by saying that a a

>

b

algebraic inequality

is

a

number number

An

line. Similarly,

means

that a

is

is

to the left of b

to the right of b

on a on a

real real

line.

mathematical statement where two expres-

sions are joined using one of the inequality symbols. For example,

2(2x ^

+

3) '


then solve for R

calculator;

^

214)

month

At $166.07 per month, the car will be yours after 36 months. That is, you have amortized the debt in 36 equal monthly payments. (Mort means "death"; you have "killed" the loan in 36 months.) In general, amortizing a debt means that the debt is retired in a given length of time by equal periodic payments that include

compound

interest.

We

are usually inter-

ested in computing the equal periodic payment. Solving the present value

formula

(5)

for

R

in

terms of the other variables,

amortization formula:

we

obtain the following

Present Value of an Annuity; Amortization

6-4

Example 16

Assume

that

you buy

a television set for

equal monthly payments

(B)

How much How much

(A)

Use formula

(A)

Solutions

1^%

at

$800 and agree

interest per

month on

to

pay

for

it

277

in 18

the unpaid balance.

are your payments? interest will

R=P- 1

+

(1

=

with P

(6)

you pay? $800,

i

=

0.015.

=

18:

or i)-"

0.015

800-

and n



l-(1.015r«

800Q^„„,,

Use Table

V

or a

calculator

= 800(0.063 806) = $51.04 per month Total interest paid

(B)

Problem 16

you

If

your car

sell

month on to

to

= Amount of all payments — = 18($51.04) -$800 = $118.72

Initial

loan

someone

for $2,400 and agree to finance it at 1% per how much should you receive each month months? How much interest will you receive?

the unpaid balance,

amortize the loan in 24

Amortization Schedules What happens

if you are amortizing a debt with equal periodic payments some point decide to pay off the remainder of the debt in one lump sum payment? This occurs each time a home with an outstanding mortgage is sold. In order to understand what happens in this situation, we must take

and

at

amortization process. We begin with an example that is simple enough to allow us to examine the effect each payment has on the a closer look at the

debt.

Example 17

you borrow $500 that you agree to repay in 6 equal monthly payments at interest per month on the unpaid balance, how much of each monthly payment is used for interest and how much is used to reduce the unpaid If

1%

balance? Solution

First,

P

we compute

= $500, = 0.01, i

fi

=P 1 =

the required monthly

and n

' ,

(1

+ ,

.,

=

or

1

-(1.01)-^

P— 1

or

500(0.172 548)

$86.27 per

(6)

a„,

I)-"

0.01

500

payment using formula

6:

month

500Qglooi

Use Table V or a calculator

with

278

Mathematics of Finance

At the end of the $500(0.01)

first

month, the interest due

is

= $5.00

The amortization payment est

is divided into two parts, payment of the interdue and reduction of the unpaid balance (repayment of principal):

Monthly payment

=

$86.27

Interest

Unpaid balance

due

reduction

+

$5.00

The unpaid balance

$81.27

for the

next month

Previous

Unpaid

New

unpaid

balance

unpaid

balance

reduction

balance

$500.00

-

=

$81.27

is

$418.73

At the end of the second month, the interest due on the unpaid balance of $418.73

is

$418.73(0.01)

= $4.19

The monthly payment $86.27

is

divided into

= $4.19 + $82.08

and the unpaid balance $418.73

- $82.08 =

for the

reduced

is

Table

which

Table

1,

1

is

is

$366.65

This process continues until

balance

next month

to zero.

all

payments have been made and the unpaid

The

calculations for each

month

referred to as an amortization schedule.

Amortization Schedule

are listed in

Present Value of an Annuity; Amortization

6-4

Notice that the

last

payment had

reduce the unpaid balance

to zero.

to

be increased by $0.03 in order

This small discrepancy

occur in the computations. In almost

off errors that

payment must be adjusted

279

is

due

to

the last

all cases,

slightly in order to obtain a final

to

round-

unpaid balance

of exactly zero.

Problem 17

Construct the amortization schedule for a $1,000 debt that is to be amortized in 6 equal monthly payments at 1.25% interest per month on the

unpaid balance.

Example 18

Solution

When a family bought their home, they borrowed $25,000 at 9% compounded monthly, which was to be amortized over 30 years in equal monthly payments. Twenty years later they decided to sell the house and pay off the loan in one lump sum. Find the monthly payment and the unpaid balance after making monthly payments for 20 years. (6) with P = $25,000, monthly payment is

Using formula 360, the

R = P1

=

i

0.09/12

=

0.0075 and n

=

30(12)

=

1

-

+

(1

i)-"

0.0075 25,0001

Use

-(1.0075)-"*°

a calculator

$201.16 per month

How

can

we

find the outstanding balance after 20 years or 20(12)

=

240

monthly payments? One way to proceed would be to construct an amortization schedule, but this would require a table with 240 lines. Fortunately, there is an easier way. The unpaid balance after 240 payments is the

amount

of a loan that can be paid off with the remaining 120 payments of $201.16. Since the bank views a loan as an annuity that they bought from you, the unpaid balance of a loan with n remaining payments is the

present value of that annuity and can be computed by using formula Substitutingfi =$201.16, after

1

=

0.0075,

240 payments have been made

=

1

and n

=

120 in

(5),

(5).

the unpaid balance

is

-(1.0075)-'2

$201.16

Use

a calculator

0.0075

= $15,879.91 Problem 18

In

Example

18,

what was the unpaid balance

after

making payments

for 5

years?

The answer after

to

Example 18 may seem

having made payments

for

amount

to

owe

20 years, but long-term amortizations

start

a surprisingly large

280

Mathematics of Finance

out with very small reductions in the unpaid balance. For example, the interest

due

at the

25,000(0.0075)

The

first

end of the very first period

=

of the loan in

Example 18 was

187.50

monthly payment was divided

into

Unpaid

Monthly payment $201.16

Interest

balance

due

reduction

=

$187.50

+

$13.66

Thus, only $13.66 was applied

Answers to Matched Problems

15. 17.

$13,577.71

16.

to the

unpaid balance.

R = $112.98 per month;

total interest

= $311.52

281

Present Value of an Annuity; Amortization

6-4

Applications Business & Economics

11.

A relative wills you 10 years.

If

money

an annuity paying $4,000 per quarter

worth

is

8% compounded

next

for the

what

quarterly,

the

is

present value of this annuity? 12.

How much

should you deposit in an account paying 12% com-

pounded monthly 13.

Parents of a college student wish

$350 per month deposit

now

annuity? 14.

in order to receive $1,000 per

A

to the

9%

at

If

(A)

If

you buy a stereo

installments at

how much

pay

should they

to establish this

for

48 months for a car, making no

the loan costs 1.5% interest per

unpaid balance, what was the original cost of the car? interest will be paid? 15.

that will

How much

compounded monthly

next 2

will the student receive in the 4 years?

person pays $120 per month

down payment.

for the

up an annuity

student for 4 years.

interest

How much

to set

month

set for

1%

$600 and agree

interest per

monthly payments?

are your

to

month on

pay

month on

How much for

it

the total

in 18 equal

the unpaid balance,

How much interest will

you pay?

16.

(B)

Repeat part

A for 1,5% interestper month on the unpaid balance.

(A)

A company

buys

it

at

12%

a large

interest

copy machine

for

$12,000 and finances

compounded monthly.

If

the loan

amortized in 6 years in equal monthly payments,

each payment? (B)

17.

A

Repeat part

A

How much

is

to

how much

be is

interest will be paid?

with 18% interest compounded monthly.

You pay 25% down and amortize

sailboat costs $16,000.

with equal monthly payments over

a 6

year period.

If

the rest

you must pay

1.5% interest per month on the unpaid balance (18% compounded monthly), what is your monthly payment? How much interest will 18.

you pay over the 6 years? A law firm buys a computerized word-processing system costing $10,000. If it pays 20% down and amortizes the rest with equal monthly payments over 5 years at 9% compounded monthly, what will be the

19.

monthly payment?

How much

interest will the firm pay?

Construct the amortization schedule for a $5,000 debt that

amortized in 8 equal quarterly payments

at

4.5%

is

to

be

interest per quarter

on the unpaid balance. 20.

amortized in 6 equal quarterly payments 21.

is to be 3.5% interest per quarter

Construct the amortization schedule for a $10,000 debt that

on the unpaid balance. A person borrows $6,000 amortized over

at

3 years in

at

12% compounded monthly, which is to be

equal monthly payments. For tax purposes,

282

Mathematics of Finance

he needs

to

know

the

amount

of interest paid during each year of the

loan. Find the interest paid during the

payments and 22.

A

after 24

year, the second year,

first

and

Find the unpaid balance after 12

the third year of the loan. [Hint:

payments.]

person establishes an annuity for retirement by depositing $50,000

an account that pays 9% compounded monthly. Equal monthly withdrawals will be made each month for 5 years, at which time the into

account will have a zero balance. Each year taxes must be paid on the

by the account during that year. How much interest was earned during the first year? [Hint: The amount in the account at the end of the first year is the present value of a 4 year annuity.] interest earned

Use a financial or

scientific calculator fo solve

each of the following

problems. 23.

Some friends tell you that they paid $25,000 down on a new house and month for 30 years. If interest is 9.8% compounded monthly, what was the selling price of the house? How much interest

are to pay $525 per

will they 24.

pay

in 30 years?

Afamily is thinking about buying a new house costing $120,000. They must pay 20% down, and the rest is to be amortized over 30 years in equal monthly payments. If money costs 9.6% compounded monthly, what will their monthly payment be? How much total interest will be paid over the 30 years?

25.

A

student receives a federally backed student loan of $6,000 at 3.5%

interest

compounded monthly.

After finishing college in 2 years, the

student must amortize the loan in the next 4 years by making equal

monthly payments. What

will the

will the student pay? [Hint:

This

payments be and what is

amount of the debt at the end of the amount over the next 4 years.] 26.

A

person establishes a sinking fund

$7,500 per year

at

first 2

for

retirement by contributing

payments are withdrawn,

the account will have a zero balance.

If

pounded annually, what yearly payments 27.

A

last

years; then amortize this

the end of each year for 20 years. For the next 20

years, equal yearly

the

total interest

a two-part problem. First find the

at the

money

end of which time worth 9% com-

is

will the person receive for

20 years?

family has a $75,000. 30 year mortgage at 13.2%

compounded

monthly. Find the monthly payment. Also find the unpaid balance after

10 years

(A) 28.

A

(B)

20 years

(C)

25 years

family has a $50,000, 20 year mortgage of 10.8%

compounded

monthly. Find the monthly payment. Also find the unpaid balance after (A)

5 years

(B)

10 years

(C)

15 years

Chapter Review

6-5

29.

A

family has a $30,000, 20 year mortgage

283

15% compounded

at

monthly. (A) (B)

Find the monthly payment and the total interest paid. Suppose the family decides to add an extra $100 to its mortgage payment each month starting with the very first payment. How long will it take the family to pay off the mortgage? How much interest will the family save?

30.

At the time they

retire, a

couple has $200,000 in an account that pays

8.4% compounded monthly. (A)

to withdraw equal monthly payments for 10 years, end of which time the account will have a zero balance, how much should they withdraw each month? If they decide to withdraw $3,000 a month until the balance in the account is zero, how many withdrawals can they make?

they decide

If

at the

(B)

6-5

Chapter Review Important Terms and

6-1

Svmbols

Simple

interest

and simple

discount, principal, interest, interest rate,

simple interest, face value, present value, future value, simple intersimple discount note, discount, proceeds, maturity value,

est note, I

6-2

= Prt, A = P(l + rt), D = Mdf,

Compound nominal P(l

6-3

+

interest,

P

compound

rate, effective rate (or

i)",

i

= M - D, P = M(l - dt)

interest, rate per

annual

yield),

compound

period,

doubling time,

A=

= r/m

Future value of an annuity; sinking funds, annuity, ordinary annuity, future value, sinking fund,

S

= R (1 + i)"

R=

= Rsh ^_S_

S (1

6-4

1

+ i)" -

1

(future value)

(sinking fund)

Shi,

Present value o/an annuity; amortization, present value, amortizing a debt, amortization schedule. 1

-

(1

R

R=P 1

(1

+

+

i)

i)

Ra^,

(present value)

:P^

(amortization)

284

Mathematics of Finance

Exercise 6-5

Chapter Review Work through

al]

the problems in this chapter review

and check your

back of the book. (Answers to ail review problems are Where weaknesses show up, review appropriate sections in the text.

answers

you are

in the

satisjied that

you know the materia], take the practice

there.)

When

following

test

this review.

Solve each problem using Table

A

V or

Find the indicated quantity, given 1.

2. 3.

4.

A = P(l + t

t

t

t

= ?, M = $5,000, d = = $4,000, M=?. d = M = $6,000. P = $5,100, M = $1,200, P = $1,080,

=

P

18%,

t

6.

P

15%,

t

8.

d d

Find the indicated quantity, given 9.

10.

A= A=

P

?,

= $1,200, P = ?,

$5,000,

i

i

= =

= =



Mil

5.

7.

rt).

A = ?, P = $100, r = 9%, = 6months A = $808, P = ?, r = 12%, = 1 month =? A = $212, P = $200, r = 8%, = 6 months A = $4,120, P = $4,000, r = ?,

Find the indicated quantify, given P

B

a calculator (or both).

= =

?,

t

10 months 8

=

10%,

months 15 months t

A = P(3 +

0.005,

n

dt].

= i)"

?

and P

= A/(J +

i)".

= 30 = 60

n

0.0075,

Find the indicated quantity, given *

11.

S

12.

S

= ?, R= $1,000, = $8,000, R = ?,

i

i

= =

0.005,

n

0.015,

n

Find the indicated quantity, given

P

=R

1

-

—+ (1

^=

i)""

and

RaHi, ^'

R

=

i

P

1

13. 14.

C

P = ?, R = P = $8,000,

Use Table 15.

2,500

V or

=

$2,500,

R

=

?,

i

i

,

l-(-l+ ,

=P

.,

l}-"

1

OHii

= 0.02, n = 16 = 0.0075, n = 60

a calculator (or bothj to solve for n to the nearest integer.

1,000(1.06)"

16.

5,000

=

100^-^

=

IOOshjooi

6-5

Chapter Review

285

Applications Business

& Economics

17.

you borrow $3,000 at 14% simple interest for 10 months, how much you owe in 10 months? How much interest will you pay? If you borrow $3,000 at 14% discount for 10 months, how much will you receive? How much will you owe when the debt comes due? How If

will

18.

19.

much will the loan cost you? How much should you deposit annually

20.

21.

22.

in an account paying 8% compounded have $20,000 in 20 years?

to

If $5,000 is invested at 10% compounded quarterly, what is the amount after 6 years? What is the value of an annuity in 8 years if $100 per month is deposited into an account earning 6% compounded monthly? Suppose you buy a stereo system costing $900. If you pay 25% down

and amortize the rest in 24 monthly payments at 1.5% interest per month on the unpaid balance, how much is each payment and how 23.

24.

much total interest will you pay? How much should you pay for an annuity that pays $1 ,000 per quarter for 10 years if money is worth 8% compounded quarterly?

A company

decides to establish a sinking fund to replace a piece of at an estimated cost of $50,000. To accomplish

equipment in 6 years they decide to

this,

make

fixed

monthly payments

into

an account

9% compounded monthly. How much should each payment

that pays

be? 25.

A

savings and loan

company pays 9% compounded monthly. What

is

the effective rate? 26.

You hold a $5,000, sell

to

it

9

month note at 10% simple interest. You decide to at 12% discount 5 months before it is due.

another investor

How much

will

you receive

for the note?

How much

will the other

investor receive in 5 months? 27.

How

long

invested 28.

(to

at

the nearest month) will

it

take

money

to

double

if it is

12% compounded monthly?

Construct the amortization schedule for a $1,000 debt that is to be amortized in 4 equal quarterly payments at 2.5% interest per quarter

on the unpaid balance. 29.

A car dealer offers to sell you a car for $500 down and $200 a month for 36 months. As required by law, he informs you that the effective rate of interest is 16%. (A) (B)

30.

What nominal rate of interest are you paying? What is the original cost of the car?

A business borrows $80,000 at 15% interest compounded monthly for 8 years. (A)

What

is

the monthly payment?

286

Mathematics of Finance

31.

(B)

What

(C)

How much

An

is

the unpaid balance at the end of the interest

was paid during the

first

first

year?

year?

individual wants to establish an annuity for retirement purposes.

He wants to make quarterly deposits for 20 years so that he can then make quarterly withdrawals of $5,000 for 10 years. The annuity earns 12% interest compounded quarterly. (A) (B)

How much will have to be in the account at the How much should be deposited each quarter

time he retires? for

20 years in

order to accumulate the required amount? (C)

What

is

the total

amount

of interest earned during the 30 year

period? 32.

Practice Test:

In order to save enough money for the down payment on a home, a young couple deposits $200 each month into an account that pays 9% interest compounded monthly. If they want $10,000 for a down payment, how many deposits will they have to make?

Chapter 6 Solve each problem using Table

2.

V or

a calculator (or both).

How much should you deposit initially in an account paying 10% compounded semiannually in order to have $25,000 in 10 years? A company decides to establish a sinking fund to replace a piece of equipment

make

in 6 years at

every 3 months,

What

an estimated cost of $15,000.

quarterly payments into an account paying

how much

If

they decide

to

10% compounded

should each payment be?

if $200 per month is compounded monthly? You decide to purchase a car costing $8,000 by paying 20% down and amortizing the rest in 4 years at 1.5% per month interest on the unpaid balance by making equal monthly payments. How much is each payment and what is your total interest?

is

the value of an annuity in 5 years

deposited into an account paying

5.

9%

interest

How much should you pay for an annuity that pays $3,000 per quarter 20 years if money is worth 8% compounded quarterly? You hold a $10,000, 10 month note at 9% simple interest. If you sell it for

6.

to

another investor

much will you

at

receive?

10%

discount 3 months before

it is

due,

how

How much will the other investor receive in 3

months? 7.

A savings and loan company pays 8% compounded quarterly. What is the effective rate?

6-5

Chapter Review

287

Each quarter a couple deposits $500 into an account that pays .merest^cotnpounded quarterly. How long will

it

take

8% the^Tsavt

10.

''7^''

''°'°°° '' ''"'" '"'^^^^ compounded be amortized over 5 years. Now you have acquired some additional funds and decide that you

monthrv which monthly whfrV" was

th^sjoan.

years?

What

is

to

the unpaid balance after

wa^t fo pay off making payments for 2

Systems of Linear Equations; Matrices

CHAPTER

7

Contents Systems of Linear Equations

7-1 Review:

7-2 Systems of Linear Equations

and Augmented Matrices

— Introduction

7-3 Gauss -Jordan Elimination

— Addition and Multiplication by

7-4 Matrices

a

Number

7-5 Matrix Multiplication 7-6 Inverse of a Square Matrix; Matrix Equations 7-7 Leontief

7-8 Chapter

Input-Output Analysis Review

In this chapter

we

will first

review

(Optional)

how

systems of equations are solved by

using techniques learned in elementary algebra. These techniques are

two or three variables, but they are not numbers of variables. After this review, we will introduce techniques that are more suitable for solving systems with larger numbers of variables. These new techniques form the basis for computer solutions of large-scale systems. suitable for systems involving

suitable for systems involving larger

7-1

Review: Systems of Linear Equations Systems

in

Two

Variables

Applications

Systems in Three Variables Applications

Systems To

in

Two

Variables

establish basic concepts, consider the following simple example:

adult tickets child tickets

Let

Then

= y= X

and one child ticket cost $8, and if one adult cost $9, what is the price of each?

ticket

If

two

and three

Price of adult ticket Price of child ticket

+ x +

2x

=8 = 9 3y y

We now have a system of two linear equations and two unknowns. To solve this system,

290

we

find all ordered pairs of real

numbers

that satisfy both

7-1

equations

same

at the

Review: Systems of Linear Equations

time. In general,

we

291

are interested in solving linear

systems of the type

ax ex

+ by = h + dy = k

where

a, b, c, d, h,

and k are

real constants.

A

pair of

numbers x

= Xq and

y = yo [also written as an ordered pair (Xq yo)] is a solution of this system if each equation is satisfied by the pair. The set of all such ordered pairs of numbers is called the solution set for the system. To solve a system is to ,

find

its

solution

set.

We

will consider three

methods of solving such

systems, each having certain advantages in certain situations. Solution by Graphing

To solve the ticket problem above by graphing, we graph both equations in the same coordinate system. The coordinates of any points that the graphs have in common must be solutions to the system, since they must satisfy both equations.

Example

1

Solve the ticket problem by graphing:

Solution

= $3 = $2 y x

Check

+y= 2(3) + 2 i 2x

8

x+

3y

=

9

8

3

+

3(2)

1

9

8^8 Problem

1

Adult ticket Child ticket

9:^ 9

Solve by graphing and check:

2xx+ It is

= -3 2y = -4 y

clear that the above

example (and problem) has exactly one solution,

since the lines have exactly one point of intersection. In general, lines in a

292

Systems of Linear Equations; Matrices

rectangular coordinate system are related to each other in one of the three

ways

Example

2

illustrated in the next

example.

Solve each of the following systems by graphing: (A)

x-2y =

2

x+y

5

=

x

(B)

2x

+ 2y = -4 + 4y=8

+ 4y = 8 x + 2y = 4

2x

(C)

Lines coincide

number Intersection at one point

Lines are parallel (each

only

has slope

— exactly one solution Problem 2

— infinite

of solutions

— |) — no solutions

Solve each of the following systems by graphing: (A)

X

2x

+y=4 -y=2

(B)

2x 6x

= y=

3y

9

(C)

3

2x- y= 4 -18 6x- 3v=

By geometrically interpreting a system of two linear equations in two unknowns, we gain useful information about solutions to the system. Since two lines in a coordinate system must intersect at exactly one point, be parallel, or coincide,

solution,

(2)

we

no solution, or

conclude that the system has [3)

infinitely

many solutions.

(1)

exactly one

In addition, graphs

of problems frequently reveal relationships that might otherwise be hid-

den. Generally, however, graphic methods only give us rough approximations of solutions.

The methods

of substitution

and elimination by addition

yield results to any decimal accuracy desired

— assuming that solutions

exist.

Solution by Substitution

Choose one of two equations in a system and solve for one variable in terms of the other. (Make a choice that avoids fractions, if possible.) Then substitute the result into the other equation and solve the resulting linear equation in one variable. Now substitute this result back into either of the original equations to find the second variable. An example should make the process clear.

7-1

Example

3

=4 3y = 5 y

Solve either equation for one variable in terms of the other; then substitute into the remaining equation. In this

choosing the

5x

first

+

y

problem we can avoid fractions by

equation and solving for y in terms of

=4

Solve

first

x.

equation for y in terms of x

Substitute into second equation

Second equation Solve for x

Now,

replace x with

= 4 — 5x y = 4-5(l) y = -l y

Check

Problem

3

293

Solve by substitution:

5x+ 2x Solution

Review: Systems of Linear Equations

5x

+

y

=

4

1

in y

=

4



5x

to find y:

294

Systems of Linear Equations; Matrices

Theorem

1

A

system of linear equations

tem

is

transformed into an equivalent sys-

if;

Two equations An equation is

(A) (B)

A

(C)

are interchanged.

multiplied by a nonzero constant.

constant multiple of another equation

is

added

to a

given

equation.

Parts

useful

B and C of Theorem

when we

theorem

Example 4

Solution

is

1

will be of

most use

to us

best illustrated

A becomes The use of the

now; part

generalize the theorem for larger systems.

by examples.

Solve the following system using elimination by addition;

3x

-

2y

2x

+

5y

We

= 8 = -l

use the theorem to eliminate one of the variables, thus obtaining a

system with an obvious solution;

3x-

2y

=

If

we

multiply the top

equation by 5 and the

bottom by

we can

2

and then add,

eliminate y

Review: Systems of Linear Equations

7-1

Let us see

what happens

in the elimination process

either no solution or infinitely

many

when

a

295

system has

solutions. Consider the following

system:

2x X

= -3 3y = 2

+ +

6y

Multiplying the second equation by

—2 and

adding,

we

obtain

= -3 -2x- 6y = -4 = -7 2x

+

6y

We have obtained a contradiction. The assumption that the original system we have proved that = — 7). Thus,

has solutions must be false (otherwise

the system has no solutions. The graphs of the equations are Systems with no solutions are said to be inconsistent.

Now

parallel.

consider the system

-

= 4 -2x+ y = -8 X

If

we

iy

multiply the top equation by 2 and add the result to the bottom we obtain

equation,

-y= 8 -2x + y = -8 2x

0= =

Obtaining is,

by addition implies that the equations are equivalent; that two equations have the same solution and the system has infinitely many solutions. If x = k, then using either

their graphs coincide. Hence, the

set,

equation,

number

we

k.

obtain y

Such

a

=

2k

system

parameter; replacing

it

is

-

8;

that

is,

(k,

2k

-

8) is a

said to be dependent.

The

solution for any real

variable k

is

called a

with any real number produces a particular solu-

tion to the system.

Applications

Many

real-world problems are readily solved by applying two-equation

two-unknown methods. We Example j-j.

5

A

shall discuss

two applications

dietitian in a hospital

is to arrange a special diet comprised of two foods, Each ounce of food M contains 8 units of calcium and 2 units of Each ounce of food N contains 5 units of calcium and 4 units of iron. many ounces of foods M and N should be used to obtain a food mix

M and N. iron.

How

in detail.

that contains 74 units of calcium

and 35 units of iron?

296

Systems of Linear Equations; Matrices

Solution

convenient

It is

to first

summarize the quantities involved

Food

M

Food

in a table:

Total Needed

Af

Calcium

74

Iron

35

= Number of ounces y = Number of ounces Calcium

/

of food

X

Let

of food

Calcium

/

N

in

in

M N \

\xozoffoodM/

\vozoffoodN/

/ironinxozN

/iron in y oz\ \ of food N /

I,

of food

M

/

8x

+

2x

+

^

/ Total calcium

needed

\

^

(Total iron\

needed

= =

5y

4y

/

/

74

35

Solve by elimination by addition:

= 74 -8x- 16v = -140 -lly= -66 y = 6 oz of food N

Check

8x

+

2x

+ 4(6) =35 2x = ll X = 5.5

8x 8(5.5)

+ +

5y

5y 5(6)

74

Problem

5

=74 1

2x

74

2(5.5)

35

5

calcium and 4 units of

Supply and

Demand

also

and the mix of M and

price, the greater the is

N must

contain 92 units

buy during some period

demand.

Similarly, the quan-

and is

some period

be willing

lower

to

of time

supply

less of a

product

a linear

model where the graphs

at

prices.

The of a

a supply equation are straight lines.

in a given city

cherries are given by

contains 6 units of

willing to sell during

demand model

demand equation and Suppose

N

price. Generally, a supplier will

of a product at higher prices

simplest supply and

contains 10 units of

price. Generally, the higher the price, the less the

product that a supplier

depends on the

more

M

of iron.

of a product that people are willing to

depends on its demand: the lower the of time

tity of a

35

each ounce of food

iron,

iron,

and 44 units

The quantity

:^

given that each ounce of food

calcium and 4 units of

Example 6

+ 4y = 35 + 4(6) 1 35

^74

Repeat Example

of calcium

M

oz of food

on

a

given day supply and

demand equations

for

= — 0.2q + 4 p = 0.07q + 0.76

Demand

p

equation (consumer)

Supply equation (supplier)

where q represents the quantity

in

pound.

On

=

(q

10)

when

the price

p

is

= — 0.2(10) +

4

=

$2 per

the other hand, suppliers will be willing to supply 17.714

thousand pounds of cherries at

thousands of pounds and p represents see that consumers will purchase 10

we

the price in dollars. For example,

thousand pounds

297

Review: Systems of Linear Equations

7-1

at

$2 per pound (solve 2

$2 per pound the suppliers are willing

to

=

0.07q

+

0.76).

Thus,

supply more cherries than

consumers are willing to purchase. The supply exceeds the demand at that price and the price will come down. At what price will cherries stabilize for the day? That is, at what price will supply equal demand? This price, if it exists, is called the equilibrium price, and the quantity sold at that price is called the equilibrium quantity. How do we find these quantities? We solve the linear system p p

We

= — 0.2q + 4 = 0.07q + 0.76

Demand

equation

Supply equation

solve this system using substitution (substituting p

= — 0.2q +

4 into

the second equation).

+ 4 = 0.07q + 0.76 -0.27q = -3.24 q = 12 thousand pounds (equilibrium quantity) Now substitute q = 12 back into either of the original equations -0.2q

system and solve

p

=

-0.2(12)

p=

+

$1.60 per

These

for

p (we choose the

first

in the

equation):

4

pound (equilibrium

price)

results are interpreted geometrically in the figure.

Equilibrium quantity

Equilibrium price

= 12 thousand pounds = $1.60 per pound

Supply curve 12, 1.6)

15

10

Thousands

of

pounds

Equilibrium point

298

Systems of Linear Equations; Matrices

the price was above the equiHbrium price of $1 .60 per pound, the supply would exceed the demand and the price would come down. If the price was below the equilibrium price of $1.60 per pound, the demand would exceed the supply and the price would rise. Thus, the price would reach equilibrium at $1.60. At this price, suppliers would supply 12 thousand pounds of cherries and consumers would purchase 12 thousand pounds. If

Problem 6

Repeat Example 6 (including drawing the graph) given: p

p

= — O.lq + = 0.08q +

3

Demand

0.66

Supply equation

equation

Systems in Three Variables Any ax

equation that can be written in the form

+

where

by

=

c

and c are constants

a, b.

(not both a

and b zero)

is

called a linear

equation in two variables. Similarly, any equation that can be written in the form

ax

+

where

by

+

=

cz

a. b, c,

k

and k are constants

(not all a, b,

and

c zero)

equation in three variables. (A similar definition holds tion in four or

Now

that

more is

no reason

QjX

+

biV

-I-

CjZ

=

QjX

-f-

bjV

+

CjZ

ajX

+

bjV

-I-

C3Z

= =

called a linear

equa-

variables.)

we know how

variables, there

is

for a linear

to solve

systems of linear equations in two

to stop there.

Systems of the form

k-i

kj

(1)

kj

encountered frequently. In fact, systems of equations are so important in solving real-world problems that whole courses are devoted to this one topic. A triplet of numbers x = x,,, y= Vo. and z = Zg [also written as an ordered triplet (x,,, Vq, Zq)] is a solution of system (1) if each equation is satisfied by this triplet. The set of as well as higher-order systems are

all If

such ordered

triplets of

numbers

is

called the solution set of the system.

operations are performed on a system and the

new system

has the same

solution set as the original, then both systems are said to be equivalent.

Linear equations in three variables represent planes in a three-dimensional space. Trying to visualize

you

insight as to

Figure

1

intersect.

what kind

shows several It

how

of the

many ways

can be shown that system

solutions, or infinitely

many

three planes can intersect will give

of solution sets are possible for system

(1)

solutions.

will

(1).

which three planes can have exactly one solution, no in

Review: Systems of Linear Equations

7-1

Figure

1

299

Three intersecting planes

In this section

we

will use

an extension of the method of elimination

discussed above to solve systems in the form of

(1).

In the next section

will consider techniques for solving linear systems that are

we

more compati-

ble with solving such systems with computers. In practice, most linear

systems involving more than three variables are solved with the aid of a computer.

Steps in Solving Systems of 1.

Form

(1)

Choose two equations from the system and eliminate one of the three variables using elimination by addition. ally

2.

Now

result

is

gener-

in

one of those used in step tion in two variables. 3.

The

two unknowns. eliminate the same variable from the unused equation and

one equation

The two equations from

1.

We

steps

(generally) obtain another equa-

1

and

2

form

a

system of two

equations and two unknowns. Solve as described in the earlier part of this section. 4.

Substitute the solution from step 3 into any of the three original

equations and solve for the third variable to complete the solution of the original system.

300

Systems of Linear Equations; Matrices

Example

7

Solve:

3x 2x 5x Solution

Step

+ 4z = 6 + 3y-5z = -8 - 4y + 3z = 7 -

1.

2y

(2) (3)

(4)

We look at the coefficients ofthe variables and choose to eliminate y from equations (2) and (4) because ofthe convenient coefficients — 2 and —4. Multiply equation (2) by —2 and add to equation (4):

-6x + 4y-8z = -12 7 5x - 4y + 3z =

2.

Novif

Equation

3.

we

From

eliminate y (the same variable) from equations

(4)

steps

and

1

2

we

3[equation

(2)]

2[equation

(3)]

(2)

and

(3):

(6)

obtain the system

-x-5z = -5

(5)

+ 2z=

(6)

13x [It

(4) (5)

9x-6y + 12z= 18 4x + 6y-10z = -16 + 2z= 2 13x Step

(2)]

-5z= -5

-X Step

-2[equation

has been

2

shown

that equations

form a system equivalent

system

(5)

and

z= equation

-x-

(5)]

5z

=

1

and

along with

(6)

(2), (3),

We

or

solve

as in the earlier part of this section:

(6)

-13x-65z = -65 2 13x+ 2z= -63z = -63

Substitute z

(5)

to the original system.]

13[equation

Equation

(5)]

(6)

1

back into either equation

(5)

or

(6)

[we choose

to find x:

= -5

(5)

-x-5(l)=-5

-x= x= Step

4.

Substitute x

=

and

z

=

1

back into any of the three original

equations [we choose equation

(2)] to

find y:

Review: Systems of Linear Equations

7-1

3x 3(0)

-

+ 4z = + 4(1) =

2y

2y

301

6

(2)

6

4= -2y=

-2y+

6 2

y = -l Thus, the solution

Check

To check the 3x 3(0)

-

solution,

+ +

2y 2(-

to the original

1)

system

is (0,

we must check each

4z

=

6

2x

4(1)

1

6

2(0)

+ +

1)

-



+

4y

3z

=

y

= — l,z =

l.

5z

= -8

5(1)

1-8

-8^-8

6^6 5x

0,

equation in the original system:

3y 3(-

— 1. l)orx =

7

5(0)-4(-l)+ 3(1)17

Problem 7

Solve:

+ 3y3x - 2y + 4x - 5y2x

= -12 2z = 1 4z = -12 5z

In the process described above,

contradiction, such as

no solution

(that

= — 2,

the system

is,

if

then is

the equations turns out to be

we encounter an equation that states a we must conclude that the system has

0,

On the other hand,

if

one of

the system has either infinitely

many

inconsistent).

=

We

must proceed further to determine which. Notice how this last result differs from the two-equation-two-unknown case. There, when we obtained = 0, we knew that there were infinitely many solutions. We shall have more to say about this in Section 7-3. solutions or none.

Applications

Now

let

us consider a real-world problem that leads to a system of three

equations and three unknowns.

Example

8

Production Scheduling

A

garment industry manufactures three

shirt styles.

Each

style shirt re-

quires the services of three departments as listed in the table on the next page.

The

many

1,560,

of each style

operate at

and packaging departments have available a and 480 labor-hours per week, respectively. How shirt must be produced each week for the plant to

cutting, sewing,

maximum of 1,160, full

capacity?

302

Systems of Linear Equations; Matrices

2x

7-1

Substitute y

=

800 into either

Review: Systems of Linear Equations

303

(10) or (11) to find x:

-X2y = -2,800 -X- 2(800) = -2,800 -X- 1,600 = -2,800

(10)

-x = -1,200 X = 1,200 Now

use either

or

(7), (8),

(9)

to find

2x+ 4y+3z= + 4(800) + 3z = 2,400+ 3,200 + 3z= 3z=

11,600

z=

2,000

2(1,200)

z:

[7)

11,600 11,600 6,000

Thus, each week, the company should produce 1,200 style style

B

shirts,

and 2,000

of the solution

Problem

8

is left

style

C shirts

shirts,

800

The check

to the reader.

Repeat Example 8 with the cutting, sewing, and packaging departments maximum of 1 ,180, 1 ,560, and 510 labor-hours per week,

having available a respectively.

Answers to Matched Problems

A

to operate at full capacity.

-2

C

304

Systems of Linear Equations; Matrices

6.

Equilibrium quantity Equilibrium price

= 13 thousand pounds = $1.70 per pound

|13, 1.7)

Equilibrium point

10

Thousands

X

7.

= - 1,

y

=

0, z

Exercise 7-1 Solve by graphing. 1.

X X

3.

+y= — v=

5 1

=

2

8.

Supply curve

15 of

pounds

900 style A; 1,300 style B; 1,600 style

C

7-1

9x

-

3y

= 24

Review: Systems of Linear Equations

305

306

Systems of Linear Equations; Matrices

Applications Business & Economics

37.

Supply and demand. Suppose the supply and demand equations printed T-shirts in a resort town for a particular

p p

= 0.7q + = — 1.7q +

where p

is

3

Supply equation

15

Demand

week

equation

the price in dollars and q

is

the quantity in hundreds.

(A)

Find the equilibrium price and quantity.

(B)

Graph the two equations

in the

same coordinate system and

identify the equilibrium point, supply curve, and 38.

demand curve.

Supply and demand. Repeat Problem 37 with the following supply and demand equations: p

p 39.

for

are

= 0.4q + = — 1.9q +

3.2

Supply equation

17

Demand

Break-even analysis.

A

small

equation

company manufactures

computers. The plant has fixed costs

$48,000 per month and variable costs (labor, $1,400 per unit produced.

portable

home

and so on] of materials, and so on) of

(leases, insurance,

The computers

are sold for $1,800 each.

Thus, the cost and revenue equations are

C

48,000

+

l,400x

l,800x

where x is the total number of computers produced and sold each month, and C and R are, respectively, monthly costs and revenue in dollars.

(A)

How many units must be manufactured and sold each month for the

(B)

40.

company

to

break even? (This

is

actually a three-equation-

three-unknown problem with the third equation R = C. It can be solved by using the substitution method.) Graph both equations in the same coordinate system and show the break-even point. Interpret the regions between the lines to the left and to the right of the break-even point.

Break-even analysis. Repeat Problem 39 with the cost and revenue equations

C=

65,000

=

l,600x

R 41.

-I-

1,100

Production scheduling.

A

small manufacturing plant makes three

types of inflatable boats: one-person, two-person, and four-person

models. Each boat requires the services of three departments as listed in the table.

The

cutting, assembly,

and packaging departments have

i

i

7-1

available a

Review: Systems of Linear Equations

maximum of 380, How many boats

respectively.

week

330, of

and 120 labor-hours per week,

each type must be produced each

for the plant to operate at full capacity?

One-Person

Two-Person

Boat

Boat

0.6 hr

1.0

Cutting

department

Assembly department

0.6 hr

Packaging department

0.2 hr

hr

307

Four-Person Boat

308

Systems of Linear Equations; Matrices

(avoidance) from the shock goal box ters

(A) (B)

(C)

from

when

the rat

is

placed d centime-

it.

Graph the two equations above in the same coordinate system. Find d when p = a by substitution. What do you think the rat would do when placed the distance d from the box found in part B?

phenomenon, see J. S. Brown, "GraApproach and Avoidance Responses and Their Relation to Motivation," Journal of Comparative and Physiologica) Psychology,

(For additional discussion of this

dients of

1948,41:450-465.)

7-2

Systems of Linear Equations and Augmented Matrices

— Introduction

Introduction

Augmented Matrices Solving Linear Systems

Introduction Most linear systems of any consequence involve large numbers of equations and unknowns. These systems are solved with computers, since hand methods would be impractical (try solving even a five-equation -five-unknown problem and you will understand why). However, even if you have a computer facility to help solve a problem, it is still important for you to know how to formulate the problem so that it can be solved by a computer. In addition, it is helpful to have at least a general idea of how computers solve these problems. Finally, it is important for you to know how to interpret the results.

Even though the procedures and notation introduced in this and the next more involved than those used in the preceding section, it is important to keep in mind that our objective is not to find an efficient hand method for solving large-scale systems [there are none), but rather to find a process that generalizes readily for computer use. It turns out that you will receive an added bonus for your efforts, since several of the processes developed in this and the next section will be of considerable value in Sections 7-6, 7-7, 8-5. 8-6, and 8-7. section are

Augmented Matrices In solving systems of equations by elimination, the coefficients of the

and the constant terms played a central role. The process can be made more efficient for generalization and computer work by the introduction of a mathematical form called a matrix. A matrix is a rectangular variables

7-2

Systems of Linear Equations and Augmented Matrices

array of

3

numbers written within

5

-2_

brackets.

— Introduction

Some examples

are

309

310

Systems of Linear Equations; Matrices

a solution exists.

The manipulative process

is

a direct

outgrowth of the

elimination process discussed in Section 7-1. Recall that two linear systems are said to be equivalent

exactly the

same

solution

equivalent linear systems?

Theorem

1

A

set.

We

How

(B)

A

(C)

is

if

they have

transform linear systems into 1,

which we

restate here.

transformed into an equivalent sys-

if:

Two equations An equation is

(A)

we

used Theorem

system of linear equations

tem

did

are interchanged.

multiplied by a nonzero constant.

constant multiple of another equation

is

added

to a given

equation.

Paralleling the discussion above,

we

say that two augmented matrices

are row-equivalent, denoted by the symbol matrices, tions.

if

(Think about

this.)

row-equivalent matrices?

quence

Theorem

2

An

of

~

placed between the two

they are augmented matrices of equivalent systems of equa-

Theorem

1:

How

We

do

we

transform augmented matrices into

use Theorem

2,

which

is

a direct conse-

Systems of Linear Equations and Augmented Matrices

7-2

SoJufion

We

— Introduction

by writing the augmented matrix corresponding

start

311

to (5)

(6)

Our (6)

objective

into the

use row operations from

to

is

Theorem

2 to try to transform

form

m (7)

n

where

m

and n are

numbers. The solution

real

obvious, since matrix

(7)

to

system

augmented matrix

will be the

(5)

will then be

of the following

system:

We now proceed Step

1.

to

To get a 1 (Theorem

use row operations in the

upper

to

transform

corner,

left

we

(6)

into

interchange

form

Rows

(7).

1

and

2

2A):

Ri^Rz [:

Now Step

2.

To

you see why we wanted Theorem lA! in the lower left corner,

get a

toR2 (Theorem 2C) it

-3

3.

4.

-2

7

3

4

1

To get a 1 (Theorem

To

shown:

-21«-

1

1

Step

multiply R^ by (— 3) and add

useful to write (— 3)Ri outside the matrix to help reduce errors

in arithmetic, as

Step

we

— this changes Rj but not R^. Some people find

get a

add the

in the

R,

+

(-3)R,

1:

-2

7

10

-20

second row, second column,

we

multiply R2 by

-

2B):

-2

7

10

20

10R2

in the first row.

result to R, 2

-2 1

~*R2

second column,

(Theorem 2C)

-4* +

2R2

7

1

-2

we

multiply Rj by 2 and

— this changes R, but not Rj:

,

R,

-2

— Ri

312

Systems of Linear Equations; Matrices

We

have accomplished our objective! The for the system

last

matrix

is

the

augmented

matrix

=

3 (8)

Since system solved

Check

3Xi 3(3)

(5);

(8) is

that

is,

equivalent to system

Xj

=

+ 4X2 = 1 + 4(-2)il

3

x,

(5),

our starting system,

we have

= — 2.

and

Xj

-

2X2

=

7

3-2(-2)J=7

9-8^1

3+4^7

The above process

is

written more compactly as follows:

R1--R2

ij

-21*

-2OJ

7]

-4*

-:] Therefore, x,

Problem 9

Example 10

=

3

and

Xj

= — 2.

Solve using augmented matrix methods: 2x,



X2

Xj

+

2X2

=— = 4

Solve using augmented matrix methods: 2x,

-

3Xi

+ 4X2

3X2

=6 =i

R2+^-^R^^R2 -

-1

^R2^R2 R,

+^2Rj-Ri '

Systems of Linear Equations and Augmented Matrices

7-2

— Introduction

Solution

.:]

— 2

1

Thus,

Problem 10

Example

11

Xi

= I and

Xj

= — 1.

Solve using augmented matrix methods: 5Xi

-

2x2

=

2Xi

+

3x2

=4

11

Solve using augmented matrix methods:



= 4 -6x, + 3x2 = -12 2Xi

Solution

X2

R2

+

Ri

+ |R,-R,

{-3)R, ->R2

313

314

Systems of Linear Equations; Matrices

This system

equivalent to the original system. Geometrically, the graphs

is

two original equations coincide and there are infinitely many solutions. In general, if we end up with a row of zeros in an augmented matrix for a two-equation -two-unknown system, the system is dependent and of the

many

there are infinitely

There are several system in

solutions.

of representing the infinitely

For example, solving the

(9).

terms of the other (we solve

+

^x,

Thus,

anv

for

(1x2

is

ways

+

that

real

Then

for

\t

t

number

Another way



and any

t,

+

is, (6,

we

obtain

to represent the infinitely

for the larger-scale

as follows:

is

systems

we

many

solutions

—a

will be solving later

We choose another variable called a param-

set the variable

on the

right of equation (11), Xj, equal to

it.

number.

real

2

+

=i(8)

That

Xj).

x.

represents a solution. For example, X,

solutions to

(10) for either variable

(U)

convenient

is

say

terms of

many

2

in this chapter eter,

for x, in

equation in

xj

2,

a solution.

way

first

2

=

8) is a

i(-3)

+

2

if

t

=

8,

then

6

solution of

=

(9). If

f

= - 3.

then

i

-3 That

is,

(|,

—3}

is

a solution of

(9).

Other solutions can be obtained in a

similar manner.

Problem 11

Solve using augmented matrix methods:

— 2Xi +

Example 12

6X2

=

6

Solve using augmented matrix methods: 2x,

-I-

6X2

Xi

+

3x2

= -3 = 2

Systems of Linear Equations and Augmented Matrices

7-2

Solution

Problem 12

2

— Introduction

315

316

Systems of Linear Equations; Matrices

Answers to Matched Problems

9.

11.

A

^ Ao

^

12.

is

X2

= 3t =

is

a solution.

X,

10.

\j

t

The system

dependent. For

f

any

real

2,

X,

number,

3

f

Inconsistent

— no solution

Exercise 7-2 Perform each of the indicated row operations on the following matrix:

3.

Ri-^Rz -4R, ^Ri

5.

2R2

7.

R2

+

(-4)R,

9.

R2 R2

+ +

(-2)R,

11.

1.

K2

*

•R,

(-l)Rr

Solve using augmented matrix methods 13.

B

Xi

+

X2

=

5

7-3

Gauss-Jordan Elimination

317

Solve using augmented matrix methods. 29.

3Xi 2x,

31.

3Xi 2Xi

=7 + 3X2 = 1 + 2X2 = 4 — X2 = 5 -

0.2X1-0.5X2

33.

0.8X1

7-3

X2

-

0.3X2

30.

2Xi-3x2 = -8

= 1 3X2 = 26 11X2 = — 0.3X1-0.6X2 = 0.18 0.5Xi - 0.2x2 = 0.54 + 4Xi + 3x, — 5Xi

32.

= =

0.07

34.

0,79

3X2

Gauss -Jordan Elimination Reduced Matrices Solving Systems by Gauss -Jordan Elimination

Application

Now that you have had some experience with row operations on simple augmented matrices, we will consider systems involving more than two variables. In addition, we will not require that a system have the same number

of equations as variables.

Reduced Matrices is to start with the augmented matrix of a linear system and by using row operations from Theorem 2 in the preceding section into a simple form where the solution can be read by inspection. The simple form so obtained is called the reduced form, and we define it as

Our

objective

transform

it

follows:

Reduced Matrix

A 1.

2. 3.

4.

matrix

is

in

reduced form

if:

Each row consisting entirely of zeros is below any row having at least one nonzero element. The leftmost nonzero element in each row is 1. The column containing the leftmost 1 of a given row has zeros above and below the 1. The leftmost 1 in any row is to the right of the leftmost 1 in the row above.

318

Systems of Linear Equations; Matrices

Example 13

The following matrices

are in reduced form.

Check each one

carefully to

convince yourself that the conditions in the definition are met.

1

7-3

(B)

X,

OXi

+ X2 + + 0X2 +

The

last

system (C)

Xi

^

+ _

= 3X3 = 0X3 =

1

inconsistent

2X3

^

= —3 _ g

a

0=1, which

and has no

We

is

a contradiction.

Hence, the

solution.

disregard the equation corresponding to the

third all

When

319

4x3

equation implies

is

Gauss-lordan Elimination

row

in the matrix, since

it is

satisfied

by

values of x, Xj. and x. ,

reduced system

(a

system corresponding

to a

reduced aug-

mented matrix) has more variables than equations, the system is dependent and has infinitely many solutions. To represent these solutions,

it

is

useful to divide the variables into two types: basic

variables and nonbasic variables. solutions to the system,

To represent the

we solve for the basic

infinitely

nonbasic variables. This can be accomplished very easily as basic variables the

first

many

variables in terms of the if

we choose

variable (with a nonzero coeflBcient) in

each equation of the reduced system. Since each of these variables occurs in exactly one equation, it is easy to solve for each in terms of the other variables, the nonbasic variables. Returning to our original

we choose

x, and Xj (the first variable in each equation) as and X3 as a nonbasic variable. We then solve for the basic variables x, and Xj in terms of the nonbasic variable X3:

system,

basic variables

Xi

= -2x3-3

320

Systems of Linear Equations: Matrices

Solve for Xj, X3, andx4 (basic variables) in terms ofxj andxs (nonbasic variables): Xi ^^

X3 X4 If

tX2

"Xc

^Xs

= -2X5 + 4

we

let X2

= s and

= -4s- 3t X2 = s X3 = 2t X4 = -2t+4 Xi

is

X5

=

f,

then

for

is

(A)

(C)

solve. 1

real

numbers

dependent and has

Write the linear system corresponding

and

any

s

and

t,

2

The system Can you find two?

a solution.

tions.

Problem 14

^

to

infinitely

many

solu-

each reduced augmented matrix

7-3

Gauss -Jordan Elimination

321

322

Systems of Linear Equations: Matrices

Steps 1-4 outlined in the solution of Example 15 are referred to as

Gauss-Jordan elimination. The steps are summarized

in the

box below

easy reference:

Gauss -Jordan Elimination 1.

Choose the leftmost nonzero column and use appropriate row operations to get a

2.

Use multiples

of the

obtained in step 3.

at the top.

1

first

get zeros in all places

Delete (mentally) the top row and

Repeat steps

1

and

2

1-3) until

(steps

below the

1

first

column

of the matrix.

with the submatrix (the matrix that remains

after deleting the top

4.

row to

1.

it is

row and

first

column). Continue this process

not possible to go further.

Consider the whole matrix obtained after mentally returning

all

the rows and columns to the matrix. Begin with the bottom

nonzero row and use appropriate multiples of it the leftmost

1.

Continue

until the matrix

Note:

If

any point

at

zeros to the

is

left

we can stop,

this process,

finally in

in the

we obtain a row with all and a nonzero number to the right,

above process

we will have a contradiction

(0

then conclude that the system has no solution.

Problem 15

Solve by Gauss-Jordan elimination: OAi

Example 16

I

"* 2

A^

^^2

^Ai

•* 2

~i

3

"

Ag

O

"'^3

*^

Solve by Gauss -Jordan elimination: 2Xi



X2

3Xi

+

2X2

+ 4x3 = — — X3 = 1

above

reduced form.

of the vertical line

since

to get zeros

moving up row by row,

=

n,

n

#

0).

We can

for

7-3

Gauss-Jordan Elimination

*R, Solution

(

Need

323

^R,

a 1 here

R,

+

(-3)R,

^R,

*R,^R, R,

+*R,

The matrix

is

now

in

reduced form. Write the corresponding system and the solution.

^X3

X2

-J

Solve for the basic variables Xj and Xj in terms of the nonbasic variable X3

— Y. — i 2X3 If

X3 X,

X2 X3 is

=

t.

+

1

then

for

t

any

real

number,

= -t-f = 2t + 4 = t

a solution.

Remark:

In general,

it

can be proved that a system with more variables

than equations cannot have a unique solution.

Problem 16

Solve by Gauss -Jordan elimination: 3x,

2Xi

Example 17

+ —

6X2

-

3X3

X2

+

2x3

= 2 =—

Solve by Gauss-Jordan elimination: 2x,

324

Systems of Linear Equations: Matrices

Solud'or.

2

7-3

Solution

First,

we summarize

the relevant manufacturing data in a table:

Labor-Hours per Sculpture

A Casting department Finishing department

Gauss-Jordan Elimination

B

325

326

Systems of Linear Equations; Matrices

(X2 = — 4t + 5), can only assume the values and 1. = 0, we have x, = 10, Xj = 5, Xj = 0; and for = 1, we have = 1, X3 = 1. These are the only possible production schedules

middle equation Thus, X]

=

for

t

11, X2

t

t

that utilize the full capacity of the plant.

Problem 18

Answers to Matched Problems

Repeat Example 18 given a casting capacity of 400 labor-hours per week and a finishing capacity of 200 labor-hours per week. 13.

(A)

Condition 2

(B)

Condition

(C)

Condition 4

The

violated:

is

3

is

3 in

violated: In the

the second

row should be

second column, the

5

a 1.

should

beaO. is

violated:

The

the right of the leftmost (D)

Condition

1 is

violated:

1

leftmost

The

all-zero

bottom. 14.

(A)

=—

X,

X2

(B)

=

3

X3=

6

Solution:

3,X3 (C)

15. 16. 17.

18.

Xi

^Xt

second row

is

not to

second row should be

at the

1

in the

in the first row.

(D)

X,

7-3

1

B

Gauss- Jordan Elimination

327

328

Systems of Linear Equations; Matrices

25.

7-3

44.

45.

46.

47.

48.

Life Sciences

49.

Gauss-Jordan Elimination

329

330

Systems of Linear Equations; Matrices

Snrial .Sripnces

55.

Two

have grant money to study school busing to conduct an opinion survey using 600 telephone contacts and 400 house contacts. Survey company A has personnel to do 30 telephone and 10 house contacts per hour; survey company B can handle 20 telephone and 20 house contacts per hour. How many hours should be scheduled for each firm to produce Sociology

sociologists

They wish

in a particular city.

exactly the 56.

number

of contacts needed?

Problem 55

SocioJogy. Repeat

if

650 telephone contacts and 350 house

contacts are needed.

7-4

Matrices

— Addition and Multiplication by a Number Basic Definitions

Sum

and Difference

Product of a

Number

k and a Matrix

M

Application

two sections we introduced the important new idea of matrices. and the following sections, we shall develop this concept further.

In the last In this

Basic Definitions Recall that

we

defined a matrix as any rectangular array of numbers

enclosed within brackets. The size or dimension of a matrix operations on matrices. to

be one with

number

of rows

m is

rows and columns,

is

important to

We define an m X n matrix (read "m by n matrix")

rows and n columns.

It

always given

matrix has the same number of

it

is

first. If

a

is

important to note that the

called a square matrix.

A

matrix with only one

column is called a column matrix, and one with only one row row matrix. These definitions are illustrated by the following:

3X2

is

called a

7-4

Sum and

Matrices

— Addition and Multiplication by a Number

Difference

The sum of two matrices of the same dimension

a

matrix with elements

sum

Addition

not defined for matrices with different dimensions.

a

Example 19

is

that are the

[A) _c

(B)

is

331

of the corresponding elements of the

two given matrices.

332

Systems of Linear Equations; Matrices

If

A and B

are matrices of the

same dimension, then we

define subtrac-

tion as follows:

A-B = A + {-B) Thus,

to subtract

matrix B from matrix A,

we

simply subtract correspond-

ing elements.

3

-2

-2

2

Example 20 3

5

Problem 20

Subtract:

[2-3

Product of a Finally, the

-Ki

4]=[:

5]

-[3

Number

-2

1]

k and a Matrix

product of a number

k and

to

equal 2M.

if

M

is

M

a matrix M, denoted by kM,

matrix formed by multiplying each element of partly motivated by the fact that

-4

5

::]=[ 2 -4

M

by

a matrix, then

a

is

This definition

is

we would like M +

M

k.

7-4

(For example, Ms.

Matrices

— Addition and Multiplication by

Smith had $18,000

in

compact

a

Number

sales in August,

333

and Mr.

Jones had $108,000 in luxury car sales in September.) (A)

What was

the

combined

dollar sales in

August and September

for

each person and each model? (B)

What was

(C)

If

the increase in dollar sales from August to September?

both salespeople receive

compute the commission

5% commissions on

for

each person

for

gross dollar sales,

each model sold in

September.

Luxury

Compact Solutions

(A)

(B)

A+

$90,000

$180.00o1

Ms. Smith

$126,000

$108.00oJ

Mr. Jones

Compact

Luxury

$54,000

$108,000l

Ms. Smith

$54,000

$108,OOoJ

Mr. Jones

B-

B-A

(0,05)($72,000) (C)

In

0.05B

-[ (0.05)($90,000)

{0.05}($144,000)

(0.05)($108,000)_

$3,600

$7,200

Ms. Smith

$4,500

$5,400

Mr. Jones

Example 22 we chose

a relatively simple

example involving an agency

with only two salespeople and two models. Consider the more

realistic

problem of an agency with nine models and perhaps seven salespeople then you can begin to see the value of matrix methods.

Problem 22

Answers to Matched Problems

Repeat Example 22 with

19.

$36,000

$36,000

$18,000

$36,000

5

1

4]

21.

$90,000

$108,000

$72,000

$108,000

13 2

1

35

$126,000

$144,000

$90,000

$144,000_

[$4,500

$5,400

$3,600

$5,400

(A) _

(C)

[-1

=

-2 2

22.

20.

B

and

(B)

$54,000

$72,000

$54,000

$72,000



334

Systems of Linear Equations; Matrices

Exercise 7-4 ProbJems 1-18

2

-1

refer to the following matrices: 2

B =

3

D=

E=[-4

-3

1

2

-3

C=

-2]

7-4

1.3

2.5

8.3

-1.4

Matrices

— Addition and Multiplication by a Number

ii

r-4.1

1.

7J

0.7

2.

21.

2.6 22.

[

335

336

Systems of Linear Equations; Matrices

Life Sciences

32.

Heredity. Gregor nist,

made

Mendel (1822-1884), an Austrian monk and

bota-

discoveries that revolutionized the science of heredity. In

one experiment, he crossed dihybrid yellow round peas (yellow and round are dominant characteristics; the peas also contained genes for the recessive characteristics green and wrinkled) and obtained 560 peas of the types indicated in the matrix:

Round

7-5

7-5

Matrix Multiplication

337

Matrix Multiplication Dot Product Application

Matrix Product Arithmetic of Matrix Products Application

In this section, that will

we are going to introduce two types of matrix

seem rather strange

at first. In spite of this

multiplication

apparent strangeness,

these operations are well founded in the general theory of matrices and, as

we

will see,

extremely useful in practical problems.

Dot Product

We

start

by defining the dot product of two special matrices,

X

matrix and an n

1

X

n

1

a,-

fa,

The

Xn

type,

Example 23

is

Qibi H-ajhs

is

number, not

a real

important.

If

the dot

is

+





=

0]

(2)(-5)

3

o

3

+

a matrix.

a„b„

+

2]

4

-1

A

real

(-3K2)

+

number

The dot between the two

will consider below.

2

[— i



omitted, the multiplication

= -10-6 + = -16

Problem 23

n row

1

-a^

which we

[2-3

X

b. •

dot product

matrices

a 1

column matrix:

(0)(-2)

is

of another

338

Systems of Linear Equations; Matrices

Application

Example 24

A

factory produces a slalom water ski that requires 4 labor-hours in the

fabricating department

and

1

labor-hour in the finishing department. Fa-

bricating personnel receive $8 per hour and finishing personnel receive $6

per hour. Total labor cost per ski

[4

Problem 24

If

!]•

the factory in

(4)(8)

+

(1)(6)

Example 24

is

=

given by the dot product:

32

+

6

=

$38 per ski

also produces a trick water ski that requires 6

labor-hours in the fabricating department and 1.5 labor-hours in the finishing department, write a dot product

between appropriate row and column

matrices that will give the total labor cost for this

Must be the same (b

=

c)

ski.

Compute

the cost.

7-5

2X3 3

1

1

2.

Matrix Multiplication

339

340

Systems of Linear Equations; Matrices

Arithmetic of Matrix Products Relative to addition, in the last section

A+B=B+A A + (B + C) = [A + B) + C where A, ties

we

noted that

Commutative Associative

and C are matrices of the same dimension. Do similar properWhat about the commutative property? Let us

B,

hold for multiplication?

compute AB and BA

for

and

B

=

2X2

2X2

[:-:] 2X2

-3

A=

14

18

4

-2

:]

2

8

10

[- 6

20

=C ]

2X2

Thus,

AB

=D

2X2 BA = D

AB = C

commute;

:]-

[1

2X2

BA. Only in some very special cases will matrix products

¥=

therefore,

we must

matrix multiplication

is

always be careful about the order in which

performed^ we cannot

order as in the arithmetic of real numbers. In defined, even though

AB

is.

Thus,

AB

BA

¥=

indiscriminately reverse

fact.

BA

is

often not even

in general.

Another important difference between matrix products and is found in the zero property for real numbers:

real

number

products

For

all real

ab For

=

A

numbers a and b, and only if a =

if

and B matrices

example,

it is

or

b

=

possible for AB

=

(or both)

and neither A nor B be

0.

For

7-5

Matrix Multiplication

341

products and sums are defined for the indicated matrices A, B, and C, then

number:

for k a real

3.

[AB]C=A(BC) A{B + C) = AB+AC (B + C)A = BA + CA

4.

k(AB)

1.

2.

=

(kA)B

Associative property

Left-hand distributive property

Right-hand distributive property

= A(kB)

Since matrix multiphcation

is

not commutative, properties 2 and 3 must be

Hsted as distinct properties.

Application Example 26

Let us

combine the time requirements

for

slalom and trick vvater skis

discussed in Example 24 and Problem 24 into one matrix:

Fabricating

Finishing

department

department

Trick ski

6hr

1.5

Slalom ski

4 hr

1

Novvf

hrl

_

J~

hr

suppose the company has two manufacturing plants X and Y in and that their hourly rates for each depart-

different parts of the country

ment

are given in the following matrix:

X

Plant

Plant

Fabricating department

$8

$7

Finishing department

$6

$4

To find the andB:

total labor costs for

2X2 AB =

2X2

$57

$48

Trick ski

$32

]

Slalom ski

plant X; the dot product of

first

row matrix

A

of A

costs, $32, for

of course, overly simplified.

items in

multiply

costs, $57, for a trick ski

column matrix of B gives us the labor ski at plant Y; and so on. is,

we

and the first column manufactured at the second row matrix of A and the second

matrix of B gives us the labor

Example 26

each factory,

= $38 [

Notice that the dot product of the

many different

ski at

X

1.5 1

each

Y

many different plants

very large numbers of rows and columns.

manufacturing a slalom

Companies manufacturing

deal with matrices that have

342

Systems of Linear Equations; Matrices

Problem 26

Repeat Example 26 with

A=

Answers to Matched Problems

23.

8

25.

(A)

no

hr

7 hr

2

5 hr

1.5

and

(B)

:]

B

hr

[-9]

26.

7-5

B

Find the dot products.

17.

[-1

19.

[-1

-2

2]



Find the matrix products. -1

1

3

-2

21.

1

23.

Matrix Multiplication

343

344

Systems of Linear Equations; Matrices

-4.5 31

32.

[2.1

3.2

-ll]

-0.8

5.7

-4.3 J

6.4

2.0

-2.8

3.9

-1.5

-2.4

[-6.3 -2.7

7-5

38.

(C)

What

(D)

Find

is

Matrix Multiplication

345

MN?

the dimension of

MN and interpret. A personal

Inventory value.

computer

retail

company sells

five differ-

ent computer models through three stores located in a large metropolitan area. The inventory of each model on hand in each store is

summarized in matrix M. Wholesale (W) and retail model computer are summarized in matrix N.

(R)

values of each

Model

BCD

A

M-

4

2

3

2

3

5

10

4

3

W

N=

(C)

39.

(A)

(B)

Life Sciences

40.

4

Store 3

$700

$840

A

$1,400

$1,800

B

$1,800

$2,400

C

$3,300

D

$4,900.

E

$2,700

(A)

1

Store 2

R

.$3,500

(B)

Store

7

What is the retail value of the inventory at store 2? What is the wholesale value of the inventory at store Compute MN and interpret.

3?

M

Multiply in Problem 38 by [1 1 1] and interpret. (The multiplication only makes sense in one direction.) Multiply in Problem 38 by [1 1 1] and interpret. (The multiplication only makes sense in one direction.)

MN

Nutrition.

A

nutritionist for a cereal

different mixes.

The amounts

company blends two

of protein, carbohydrate,

cereals in

and fat (in grams per ounce) in each cereal are given by matrix M. The amounts of each cereal used in the three mixes are given by matrix N. Cereal

N

A

346

Systems of Linear Equations; Matrices

Find the amount of protein in mix

(A)

X

by computing the dot

product

2]

[4

[':]

Find the amount of

(B)

fat in

mix

Z. Set

up

a dot

product as in part

A

and multiply. (C)

What

(D)

Find

(E)

Social Sciences

41.

is

the dimension of

MN?

MN and interpret. Find ^ MN and interpret.

Politics. In a local California election, a

firm to promote calls,

and

letters.

Cost per

M

=

its

group hired a public relations

candidate in three ways: telephone

The

cost per contact

is

calls,

given in matrix M;

house

Inverse of a Square Matrix: Matrix Equations

7-6

7-6

347

Inverse of a Square Matrix; Matrix Equations Identity Matrix for Multiplication

Inverse of a Square Matrix

Matrix Equations '

Application

Identity Matrix for Multiplication

We know la

=

al

for all real

that

=

a

numbers

a.

The number

multiplication. Does the set of

all

1 is

identity element for multiplication? ever, the set of all

an

identity,

and

called the identity for real

The answer,

in general,

square matrices of order n (dimension n

it is

number

matrices of a given dimension have an

X

is

n)

no.

How-

does have

given as follows: The identity element for multiplica-

tion for the set of all square matrices of order n

is the square matrix of order denoted by I, with I's along the main diagonal (from the upper left corner to the lower right) and O's elsewhere. For example.

n,

1

and

348

Systems of Linear Equations: Matrices

Inverse of a Square Matrix

we know that for each number a"' such that

In the set of real numbers, zero) there exists a real

a 'a

real

number

a (except

1

The number

a~'

is

called the inverse of the

number

a relative to multipli-

cation, or the multiplicative inverse of a. For example, 2"' is the multipli1. For each square matrix M, does there cative inverse of 2, since 2~^ 2 •

exist

If

M

=

an inverse matrix M~' such that the following relation

'

exists for a given matrix

M, then

M

'

is

true?

called the inverse of

relative to multiplication. Let us use this definition to find

M=

is

M~*

for

M

7-6

as

is

easily c

Inverse of a Square Matrix; Matrix Equations

349

350

Systems of Linear Equations; Matrices

Since each matrix to the

left

of the vertical bar

is

the same, exactly the

same row operations can be used on each total matrix to transform it into a reduced form. We can speed up the process substantially by combining all three augmented matrices into the single augmented matrix form 1

7-6

Inverse of a Square Matrix; Matrix Equations

351

B

I

3-1

3

1

-2

1 1

-2

1

-4 -5

2

Converting back to systems of equations equivalent to our three original systems (we don't have to do this step in practice), we have

=

a

c

And

=

d

3

=-2

b

e

= -4

= -1 = 1

3

=-2 /=-5

h

i=

these are just the elements of M"^ that

Note that

we

-2

-2

1

-4

-5

2

this

is

Inverse of a Square Matrix [M\I]

is

matrix B

Solution

Find

M 3

Hence,

M~'. However,

if

\ given:

M

^

MM"' =

1.)

M we

to the left of the vertical line,

Example 28

that

transformed by row operations into is

for!

the matrix to the right of the vertical line in the last

augmented matrix above. (You should check

If

are looking

3-1

3

M-i

2

obtain

all

[1\B],

then the resulting

zeros in one or

then M"' does not

exist.

more rows

352

Systems of Linear Equations; Matrices

"-6

1

Thus,

Inverse of a Square Matrix; Matrix Equations

353

354

Systems of Linear Equations; Matrices

xample 31

7-6

Thus.

X

-6

X,

13

32 8

-8

Inverse of a Square Matrix; Matrix Equations

355

356

Systems of Linear Equations: Matrices

Application The following application method for solving systems

Example 32

will illustrate the usefulness of the inverse

of equations.

An investment advisor currently has two types of investments available for clients; a

ment B

conservative investment

of higher risk that pays

A

that pays

20%

10%

per year and an invest-

per year. Clients

investments between the two to achieve any

may

divide their

total return desired

between

10% and 20%. However,

the higher the desired return, the higher the risk.

How

listed in the table invest to

should each client

return?

Client

Total investment

Annual return desired

$20,000

achieve the indicated

7-6

Inverse of a Square Matrix; Matrix Equations

R.

1

-1 2

-1 Thus,

A-'

1 oj

+

(-l)fl2^R:

357

358

Systems of Linear Equations; Matrices

31.

7-6

B

Given

M as indicated, find M 1

13.

15.

17.

19.

Inverse of a Square Matrix; Matrix Equations

'

and show

that

M

'M =

1.

359

360

Systems of Linear Equations; Matrices

Write

each system as a

matrix equation

[Note:

The inverses were found

27.

+

X,

Xj (A)

28.

=ki

Xj

'

k,

=

X3 2,

^

K3

k2

=

0,

in

and solve using

ProbJems 19 and

20.

inverses.

7-7

36.

Leontief Input -Output Analysis (Optional)

361

Production scheduling. Labor and material costs for manufacturing

two guitar models are given Guitar

Labor

Material

Model

Cost

Cost

A

$30 $40

$20 $30

B

in the table below:

week is allowed for labor and material, how many model should be produced each week to use exactly each of of each allocations of the $3,000 indicated in the following table? Use the decimals in computing the inverse. If

a total of $3,000 a

Weekly Allocation 1

2

3

362

Systems of Linear Equations; Matrices

— output analysis. Wassily Leontief, the primary force behind these new developments, was awarded the Nobel Prize in economics in 1973 because of the significant impact his

work had on economic planning for

Among other things,

ized countries.

of how 500 sectors of the

industrial-

he conducted a comprehensive study

American economy interacted with each other. Of

course, large-scale computers played an important role in this analysis.

Our investigation will be more modest. In fact, we will start with an economy comprised of only two industries. From these humble beginnings, ideas and definitions will evolve that can be readily generalized for more economies. Input -output analysis attempts to establish equilibrium conditions under which industries in an economy have just enough output to satisfy each other's demands in addition to final (outside) derealistic

mands. Given the internal demands within the industries for each other's output, the problem is to determine output levels that will meet various levels of final (outside)

demands.

Two-Industry Model To make the problem concrete, let us start with a hypothetical economy with only two industries electric company E and water company W. Outputs for both companies are measured in dollars. The heart of input



output analysis

a matrix, called the technology matrix, that expresses

is

how each industry uses the other industries' outputs as well as its own for its own output. (In this case, the electric company will use both electricity and water as input for its own output, and the water company will use both and water as input for its output.) Suppose that each dollar's worth of the electric company's output requires$0.10of its own output and $0.30 of the water company's output, and each dollar's worth of the water company's output requires $0.40 of the electric company's output and electricity

$0.20 of

its

own

summarized

output. These internal requirements can be conveniently

in a technology matrix:

W

E E

0.1

0.4

W

0.3

0.2

Thus, the

first

=M

column

indicates that each dollar of the electric company's

output requires $0.10 of

company's output

The

row

Technology matrix

its

as input.

own

output as input and $0.30 of the water

The second column

is

interpreted similarly.

needed

to produce a worth of electricity, and $0.40 of electricity is needed to produce a worth of water. The second row has a similar interpretation. first

tells

us that $0.10 of electricity

is

dollar's

dollar's

Now that we know how much of each output dollar must be used by each industry for input,

we

are ready to attack the

main problem.

7-7

Leontief Input -Output Analysis (Optional)

363

Basic Input -Output Problem

Given the internal demands

for

each industry's output, determine

output levels for the various industries that will meet a given (outside) level of

Suppose the

= =

d,

dj

What

final

$12 million $6 million

demand

as well as the internal

demand

(the

demand from

final

demand.

the outside sector)

is

for electricity

for

water

company and Sxj from the water demands? Before we answer this demand matrix and the output matrix:

dollar outputs, Sx, from the electric

company, are required question,

we

to

meet these

introduce the

D=

Final

final

final

demand matrix

d.

X

Output matrix

=

Given the technology matrix M and the final demand matrix D, the problem is to find the output matrix X. The following verbal equations (which summarize the discussion above) lead to a matrix equation and a solution to the problem:

Input

^

^

(Total output^

required by E

from E

/

from

W

from E

^

'

Input

^

required by E

/

from

V

Converted Xi

required by

^

(Total output\

to

W

(Input

Final

W

from E

(Input required by

from

)

W

W

outside)

'

demand

from E

J

Final

^

(outside)

demand

from

W

)

symbolic forms, these equations become

=0.1Xi

X2 =0.3Xi

+ +

0.4X2

+

di

0.2x2

+

CI2

(1)

or

fxj

fo.l

0.4irx,l

fd,]

[xjj

L0.3

O.2J [xjj

[d2j

or

X = MX + D

(2)

364

Systems of Linear Equations; Matrices

Now

our problem

to solve this

is

matrix equation for X.

We

proceed as in

the preceding section:

X - MX = D IX - MX = D - M]X = D X= -

I

(I

(I

assuming

—M

I

I-M^

M]-'D

(3)

has an inverse. Since

0.9

-0.4

-0.3

0.8

and

(I

-

3

3

1 _2

a

We

M]-' 2

this

J

convert decimals

example

to

to fractions in

work with exact forms

we have

(4)

[::]

[;

20l I5J

company must have an output of $20 million and company an output of $15 million so that each company can

Therefore, the electric the water

meet both internal and Actually,

and final

62- This

(4) is

demands

final

very useful, since

(4)

to start

If

over for

demands

dj

gives a quick solution not only for the

stated but also to the original

projected final demands.

would have

demands.

solves the original problem for arbitrary final

problem

for various other

we had solved (1) by elimination, each new set of final demands.

then

we

Suppose in the original problem that the projected final demands 5 years from now were d, = 18 and dz = 12. Determine each company's output for this projection.

We simply substitute

these values into

(4)

and multiply:

7-7

We summarize

Solution to a

these results for convenient reference.

Two-Company Input -Output Problem

Given Technology

M

IS

Leontief Input -Output Analysis (Optional)

365

366

Systems of Linear Equations; Matrices

How much

3.

worth of output for E? Find I - A and (I - A)-\

input from

and E are required

Find the output for each sector that

4.

demand

B

A

2.

is

to

needed

produce a

dollar's

to satisfy the final

D,

5.

Repeat Problem 4

for Dj.

6.

Repeat Problem 4

for D3.

Problems 7-12 pertain io the following input-output model: Assume an economy is based on three industrial sectors -agriculture [A), building (B), and energy (E). The technology matrix M and jinaJ demand matrices are (in billions of dollars)

A

B

A

0.422

0.100

0.266

B

0.089

0.350

0.134

E

0.134

0.100

0.334

^M

12

10

D,

8

How much

input from A, B, and E are required to produce a dollar's

worth of output

for

B?

How much of each of B's output dollar is required as input the three sectors?

10.

11.

12. 13.

for

each of

7-8

7-8

Chapter Review

367

Chapter Review Important Terms and

7-1

Review: systems of linear equations, graphing method, substitution

method, elimination by addition, equivalent systems, inconsistent systems, dependent systems, parameter, equilibrium price, equilibrium quantity, linear equation in two variables, linear equation in

Symbols

three variables, solution of a system, solution set 7-2



Systems of linear equations and augmented matrices introduction, augmented matrix, column, row, equivalent sys-

matrix, element,

tems, row-equivalent matrices,

+

Ri

7-3

row

operations. R, *-»R|. kR,

—»R,.

kR, -^ R;

Gauss-Jordan elimination, reduced matrix, basic variables, nonbasic Gauss -Jordan elimination

variables, submatrix,

7-4

Matrices

— addition and multiplication

sion of a matrix,

mXn

matrix, equal matrices, a matrix

matrix 7-5

by a number, size or dimen-

column matrix, row two matrices, zero matrix, negative of matrices, product of a number k and a

matrix, square matrix,

sum

M, subtraction of

of

M

Matrix multiplication, dot product, matrix product, associative property, distributive properties.

7-6

In\-erse of a

square matrix: matrix equations, inverse of a number,

multiplicative inverse of a number, identity matrix, multiplicative

inverse of a matrix, matrix equation, M~^ 7-7

Leontief input-output analysis fopfional). input-output analysis,

technology matrix,

Exercise 7-8

final

demand

matrix, output matrix

Chapter Review Work through answers

in

all

the problems in this chapter review

the back of the book. (Answers

to aJJ re\'iew

Where weaknesses shoiv up, review appropriate sections you are

satisfied that

Solve the following system by graphing:

2x- y= 4 x- 2y = -4 2.

in the text.

you know the material, take the practice

this review. 1.

and check your

problems are

Solve the system in Problem

1

by

substitution.

test

there.)

When

following

368

Systems of Linear Equations; Matrices

In

Problems 3-11 perform the operations that are defined, given the foIJow-

ing matrices:

B

=

[2 1

3.

1

1.

C=

[2

3]

-[:]

23.

Chapter Review

2X2

-

X3

3x2

+

X3

= 2 = -3

369

Solve by Gauss-Iordan elimination: 2x2

2x,

+ +

x,

+

2X2

(A)

24.

7-8

Xi

3X2

= + 4x3 = + X3 = +

3x3

(B)

1

Xi

3

2x,

3

3X1

+ + +

23A by

Solve the system in Problem

=-1

5X2

viriting the

system as

a

matrix

equation and using the inverse of the coefficient matrix (see Problem 22).

25.

Also.solve the system

the constants

Find the inverse of the matrix

A= 26.

if

and —2, respectively. By —3, —4, and

0,

4

5

4

5-6

1

1

A

1, 3.

1.

and

3

are replaced

by 0,

respectively.

given below.

Show

that

A

^A

=

1.

6

1

Solve the system 0.04Xi 0.04X, Xi

+ + +

0.05X2 0.05x2 X2

= 0.06X3 = + X3 = + -

360

0.06X3

120 7,000

by writing as a matrix equation and using the inverse of the coeffifirst two equations by 100

cient matrix. (Before starting, multiply the to

27.

eliminate decimals. Also, see Problem 25.)

Solve Problem 26 by Gauss -)ordan elimination.

Applications Business & Economics

28.

A

Resource aJJocafion.

mining company has two mines with ore How many tons of each ore should be used to obtain 4.5 tons of nickel and 10 tons of copper? Set up a system of equations and solve using Gauss-Jordan elimination.

compositions as given in the table.

Ore

Nickel (%)

Copper (%)

A B

29.

(A)

Set

up Problem 28

as a

matrix equation and solve using the

inverse of the coefficient matrix. (B)

Solve Problem 28

(as in part

copper are needed.

A)

if

2.3 tons of nickel

and

5 tons of

370

Systems of Linear Equations; Matrices

30.

Material costs.

A metal foundry wishes

to

make two

different bronze

The quantities of copper, tin, and zinc needed are indicated in matrix M. The costs for these materials in dollars per pound from two suppliers is summarized in matrix N. The company must choose one

alloys.

supplier or the other.

Copper

M=

N=

Chapter Review

7-8

5. 7.

9.

11.

B

C

and

BC

AD A+E

DA

8.

CB + 2E

10.

and

(CB)E

C(BE)

Convert the system Xt ~r

2x,

+

X-i

3X2

into a matrix equation

12.

6.

371

matrix

for:

(A)

=

A

k,

2,

k2

=

0.

person has $5,000

much

and solve using the inverse

k3

= -l

(B)

to invest, part at

should be invested

k,

=

-1,

5% and

of the coefficient

k^

k3

=

the rest at 10%.

How

each rate to yield $400 per year? Set up a system of equations and solve using augmented matrix methods. at

13.

Solve Problem 12 by using a matrix equation and the inverse of the

14.

Labor

coefficient matrix. costs. A company with manufacturing plants in California and Texas has labor-hour and wage requirements for the manufacturing of two inexpensive hand calculators as given in matrices and N

M

below:

Labor-Hours per Calculator Fabricating

M^

N--

8

Linear Inequalities and Linear

Programming

.i--

CHAPTER

8

Contents 8-1 Linear Inequalities in

Two

Variables

Two

8-2 Systems of Linear Inequalities in

Variables



A Geometric Approach Simplex Method 8-5 The Simplex Method: Maximization with « Problem Constraints 8-6 The Dual; Minimization with ^ Problem Constraints 8-3 Linear

A

8-4

Programming

in

Two

Geometric Introduction

8-7 Maximization

Dimensions

to the

and Minimization with Mixed Problem Constraints

(Optional)

8-8 Chapter

Review

In this chapter

we will discuss linear inequalities in two and more variwe will introduce a relatively new and powerful mathe-

ables; in addition,

matical tool called linear programming that will be used to solve a variety of interesting practical problems.

duced

8-1

Linear Inequalities in

in

Chapter

Two

7 will

The row operations on matrices

be particularly useful in Sections

y

=

intro-

and

8-7.

Variables

Having graphed linear equations such -2x-|-3

we now

8-5, 8-6,

2x-3y =

and

as

12

turn to graphing linear inequalities in two variables such as

y«-2x-f3

2x-3y>12

and

Graphing inequalities of

The following

this

type

is

almost as easy as graphing equations.

discussion leads to a simple solution of the problem.

A vertical line divides a

P

is

left and right half-pIanes; a nonvertiupper and lower half-planes (Fig. 1).

plane into

cal line divides a plane into

above the

line

8-1

i

R

(4, y)

y




It is


X — 2 is the upper half-plane determined by the graph of y = X — 2, and y < x — 2 is the lower half-plane. satisfy

2.

To graph y > x



2,

we show

\'

-^x

y

>

X

-

2

-5 -

\''

.'N y

< 1

X

.

.

Figure 3

2

.

.•

'

1

L

-)x

the line y

=x—

x,

2 as a

broken

line (Fig.

3),

376

Linear Inequalities and Linear Programming

indicating that the

Una

show the Hne y = x —

is

not part of the graph. In graphing y

2 as a

soHd Hne, indicating that

Four typical cases are illustrated

The above discussion without proof:

Theorem

1

The graph

of the

it is

^

x

— 2, we

part of the graph.

in Figure 3 (on the preceding page).

suggests the following theorem,

which

is

stated

8-1

true.

Linear Inequalities in

Hence, the graph of the inequality

is

1

Graph 6x

-

3y

>

18.

Variables

the upper half-plane.

^^^^

Problem

Two

377

378

Linear Inequalities and Linear Programming

— 2«0x + y ; I

-5

I 1

-5

H x

Exercise 8-1 Graph each 1.

inequality.

= -3

x

=

3

X

x

8-2

B

13.

Systems of Linear Inequalities in

Two

Variables

379

380

Linear Inequalities and Linear Programming

We wish to solve such systems graphically, that is, to find the graph of all ordered pairs of real numbers (x, y) that simultaneously satisfy all inequalities in the system. The graph is called the solution region for the system. To find the solution region, we graph each inequality in the system and then take the intersection

Example

3

of all of the graphs.

Solve the following linear system graphically:

X3=0

x«8 y=S4 Solution

Graph

all

the inequalities in the

region that satisfies

neously satisfy

Problem

3

all

four inequalities — that

The coordinates

graphs.

all

same coordinate system and shade

of

any point

in the

is,

in the

the intersection of all four

shaded region will simulta-

the original inequalities.

Solve the following linear system graphically:

xS2 x«6 y«5 y^2 Example

4

Solve the following linear system graphically:

+ 2y « 2x + 4y ^ 6x

36 32

x3=0

y^O Solution

Graph

all

same coordinate system and shade in thi The coordinates of any point in the shadei

the inequalities in the

intersection of

all

four graphs.

8-2

Systems of Linear Inequalities in

Two

Variables

381

region of the accompanying figure specify a solution to the system. For

example,

(0, 0), (3, 4),

and

(2.35, 3.87) are three of infinitely

many solutions,

as can easily be checked.

6x

J-

2v

=

36

To graph each

line, find

the x and y

intercepts, then sketch the line through

these points. point

2x

Problem 4

(4, 6)

+ 4y =

We obtain the intersection + 2y = 36 and

by solving 6x

32 simultaneously.

Solve by graphing:

+ 2y ? 24 x + 2y^l2

3x

y^O Example

5

Solve the system

+ lOx + 2x + 2x

y

«

20

y

^

36

5y

3=

36

graphically and find the coordinates of the intersection points of the bound-

ary of the solution region.

Solution

The

solution region

ties,

as

shown

is

the intersection of the graphs of the three inequali

in the figure

on the next page.

382

Linear Inequalities and Linear Programming

2x + lOx + 3'= 36

Coordinates of A

Solve the system

+ 2x +

10x

= y= y

36 20

to obtain (2, 16).

Coordinates of B

Solve the system

lOx

2x

+ +

= 5y = y

36 36

to obtain (3, 6).

Coordinates ofC

Solve the system

+ 2x + 2x

= 5y = y

20 36

to obtain (8, 4).

2x +

y=

20

5v= 36

8-2

Problem

5

Two

Systems of Linear Inequalities in

Variables

383

Solve the system

3x

+ 4y « 48

X-

y

=s

2

graphically and find the coordinates of the intersection points of the bound-

ary of the solution region.

Special Definitions Pertaining to Solution Regions

The following definitions pertain to the solution regions of systems

of linear

examples and matched

inequalities such as those found in the above

problems.

A

1.

solution region of a system of linear inequalities in

bounded

can be enclosed within a

if it

within

a circle,

ples 4

and

then

5 are

it is

circle; if

it

two variables

is

cannot be enclosed

unbounded. (The solution regions for Examfor Problem 4 is un-

bounded; the solution region

bounded.)

A corner point of a solution region is the intersection of two boundary

2.

lines.

[The corner points of the solution region in Example

(3, 6),

and

That

is

enough new terminology

be important

5 are (2, 16),

(8, 4).]

to

for the

moment. These

us in the next section. For now,

applications that will be developed

more

fully as

let

we

definitions will

us introduce two

proceed through

this

chapter.

Application Example

6

A patient in a hospital is required to have at least 84 units of drug A and 120 B each day (assume that an overdosage of either drug is gram of substance M contains 10 units of drug A and 8 units of drug B, and each gram of substance N contains 2 units of drug A and 4 units of drug B. How many grams of substances M and N can be mixed to meet the minimum daily requirements? units of drug

harmless). Each

Solution

To

clarify relationships,

we summarize

the information in the following

table:

Amount Gram

Minimum

of Drug per

Substance

Daily

Requirement

M

Substance

Drug A

10 units

2 units

DrugB

8 units

4 units

iV

84 units 120 units

384

Linear Inequalities and Linear Programming

= Number of grams of substance M used y = Number of grams of substance N used

X

Let

Then

= Number of units of drug A in x grams of substance M 2y = Number of units of drug A in y grams of substance N 8x = Number of units of drug B in x grams of substance M 4y = Number of units of drug B in y grams of substance N

lOx

The following conditions must be

(Number of units of drug in X

\

A

1

grams of substance

/

+

M/

(Number of units of

\

grams of substance

Number

of units of

drug A \ in y grams of substance

/

M/

meet daily requirements:

1

Number

+

drugB in X

satisfied to

of units of

y grams of substance

(Number

\

N/

grams of\ M used /

of

substance

(Number

1^84

N/ 1^120

drugB \ in

\

of

substance

grams of\ N used /

^

Converting these verbal statements into symbolic statements by using the variables x

and y introduced above, we obtain the system of linear

in-

equalities

+ 2y ^ 84 8x + 4y 3^120

lOx

Drug A

restriction

Drug B

restriction

X

3^

Cannot use

a negative

amount

of

M

y

^

Cannot use

a negative

amount

of

N

Graphing

this

system of linear inequalities,

solutions, or the feasible region, as

shown

we

obtain the set of feasible

in the following figure.* Thus,

any point in the shaded area (including the straight line boundaries) will meet the daily requirements; any point outside the shaded area will not. (Note that the feasible region is unbounded.)

* For problems of this type and for linear programming problems in general (Sections 8-3 through 8-7), solution regions are often referred to as feasible regions.

Systems of Linear Inequalities in

8-2

lOx

+

8x

Problem 6

A

+

2y

=

Two

385

Variables

84

4y = 120

Feasible region

manufacturing plant makes two types of inflatable boats,

a

two-person

boat and a four-person boat. Each two-person boat requires 0.9 labor-hour

department and 0.8 labor-hour in the assembly department. Each four-person boat requires 1.8 labor-hours in the cutting department and 1.2 labor-hours in the assembly department. The maximum laborin the cutting

hours available each month in the cutting and assembly departments are 864 and 672, respectively.

(A)

Summarize

(B)

X two-person boats and y four-person boats are manufactured each month, write a system of linear inequalities that reflect the conditions

this

information in a table.

If

indicated. Find the set of feasible solutions graphically.

Answers to Matched Problems

3.

"

~ ^^ Solution

^

V

=

-^

h'h

region

5

X

10

-^

3x

-I-

2y

=

24^

x

+ 2y -

X

12

386

Linear Inequalities and Linear Programming

Solution region

3x + 4y = 48

X - y = 2 _

6.

(A)

8-2

Systems of Linear Inequalities in

Two

Variables

387

Exercise 8-2

A

SoJve the foIJowing linear systems graphically; 1.

3.

x^3 x«7 y^O y «4 2x

2.

xS^O

x^4 y^3 y^ 7

+ 3y«12

4.

3x

+

+ x+

2x

y

^

2y

«

10

+ X +

2x

y

3=

2y s

6.

8

10

+ 3x + 6x

3y

«

24

6y



30

x3=0

8.

4x 3x

8

+ 3y 3= 24 + 4y ^ 8

x^O

x^O y^O

ys=0

B

24

y3=0

x^O 7.

«

x^O

x^O ySO 5.

4y

SoJve the following systems graphically and indicate whether each solution set is 9.

bounded or unbounded. Find the coordinates of each

2x+ y^lO

10.

x+ y^7 X

11.

+

2y

«

^0

y

3^

2x+ y^l6

X

12.

+ 2y s

13.

21

3y

«

X

S=

y

^ ^ 24

y

21

+

3y

S^

30

x3^

s

+ 4y«32 3x+ y€30 4x + 5y>51 x

+

3x+ X

14

xs=

y

^

x+yS16

x+y^l2 X

y

x-l-y=S9

12

X

3x+

y 14.

3=

x+ y 14 Xi X2 ^ = C 5x, + 7X2

Subject to

Xj

Minimize Subject to

lOx,

,

15.

3^

+ +

Xi x,

17.

2X2

^ ^

10

Minimize

C=

+

lOXj

7x2

+

12x3

Subject to Xj

2x,

+ +

X2 X2

+ + X2

19.

4

X2 2= 8

Xj, X2

2x3

^

7

X3 3^ 4 ,

X3

Minimize

C=

5Xi

+

2X2

+

2X3

+

X3

^

Subject to X]



\.-i

T'

4X2 "2 1

'

3

2

3

— ^^

6

Xi Xi

2x2 ^-8

-2Xi+

Subject to

2x,

+ +

+ X2

X2

^

8

2x2

5=

4

Xj, Xj 3=

^

X2

C=

8

5X2

X2

Minimize

Minimize Subject to

C=

lOx,

2Xi Xj

+

4X2

+ Xj ^ 6 -4X2^-24

446

Linear Inequalities and Linear Programming

21.

Minimize

C=

Minimize

22.

16Xi

C=

+

8X2

+ + +

2x3

3^

16

Xj

X3

3

14

Xi

X3

3

12

2x,

+

4X3

Subject to

+ + +

3Xi

4x, 5x,

3x2 3x2

2

'

3

'

+

12X3

+ 5X2 + 2x2 + 3x2

1

+ 4X2 +

5X3

+

«

12

=S

25

3

20

'

2

3X3

5X3 3X3 '

3

3=

3 3

6 4 8

—"

Repeat Problem 23 with

24.

5x,

+ + +

""^

Minimize

C=

8X2

Subject to

2X2

1

23.

+

6X1

C=

6X4

4X]

+

7X2

+

5X3

+

6X4

Subject to X,

+

Xj X3

+

+

X4

X3

Xj, X2,

+ X4315 X3, X4 3

Applications Formulate each of the following as a linear programming problem. Then method to the dual problem.

solve the problem by applying the simplex

Business & Economics

25.



Manu/acfuring production scheduling. A food processing company produces regular and deluxe ice cream at three plants. Per hour of operation, the plant in Cedarburg produces 20 gallons of regular ice cream and 10 gallons of deluxe ice cream, the Grafton plant 10 gallons of regular and 20 gallons of deluxe, and the West Bend plant 20 gallons of regular and 20 gallons of deluxe. It costs $70 per hour to operate the Cedarburg plant, $75 per hour to operate the Grafton plant, and $90 per hour to operate the West Bend plant. The company needs at least 300 gallons of regular ice cream and at least 200 gallons of deluxe ice cream each day. How many hours per day should each plant operate in order to produce the required amounts of ice cream and minimize

the cost of production? 26.

Mining

— production

What

is

minimum production cost? A mining company operates

the

scheduling.

mines, which produce three grades of ore. The West

produce 4 tons of low-grade

ore, 3 tons of

two

Summit mine can

medium-grade

ore,

and

2

The North Ridge mine medium-grade ore, and 2

tons of high-grade ore per hour of operation.

can produce

1

ton of low-grade ore,

1

ton of

tons of high-grade ore per hour of operation.

It

costs $1,000 per

hour to

Summit and $300 per hour to operate the North Ridge mine. To satisfy existing orders, the company needs at

operate the mine at West

least

48 tons of low-grade ore, 45 tons of medium-grade ore, and 31

|

|

The Dual; Minimization with ^ Problem Constraints

8-6

tons of high-grade ore.

How many

447

hours should each mine be oper-

ated to supply the required amounts of ore and minimize the cost of

27.

production?

What

Purchasing.

Acme

is

the

minimum

production cost?

Micros markets computers with single-sided and

double-sided disk drives. The disk drives are supplied by two other

companies, Associated Electronics and Digital Drives. Associated Electronics charges $250 for a single-sided disk drive and $350 for a

double-sided disk drive. Digital Drives charges $290 for a single-sided disk drive and $320 for a double-sided disk drive. Each

month Asso-

ciated Electronics can supply at most 1,000 disk drives in

nation

of

monthly drives.

single-sided

total

Acme

and double-sided

each type should the 28.

its

at least 1,200 single-sided drives

Acme

each month.

How many

and

at

disk drives of

Micros order from each supplier in order

to

What

is

monthly demand and minimize the purchase

minimum

any combi-

The combined

supplied by Digital Drives cannot exceed 2,000 disk

Micros needs

least 1,600 double-sided drives

meet

drives.

cost?

purchase cost?

Transportation.

A

feed

company

stores grain in elevators located in

Ames, Iowa, and Bedford, Indiana. Each month the grain is shipped to processing plants in Columbia, Missouri, and Danville, Illinois. The monthly supply (in tons) of grain at each elevator, the monthly demand (in tons) at each processing plant, and the cost per ton for transporting the grain are given in the table. Determine a shipping schedule that will minimize the cost of transporting the grain. What is the

minimum

Originating Elevators

cost?

Shipping per

Ton

Columbia

448

Linear Inequalities and Linear Programming

30.

Nutrition

mix

B,

— plants. A farmer can buy three types of plant food, mix A,

and mix

C.

Each cubic yard of mix

phosphoric acid, 10 pounds of nitrogen, and

A 5

contains 20 pounds of

pounds

of potash.

Each

cubic yard of mix B contains 10 pounds of phosphoric acid, 10 pounds of nitrogen,

and 10 pounds

Each cubic yard of mix C

of potash.

contains 20 pounds of phosphoric acid, 20 pounds of nitrogen, and 5

pounds of potash. The minimum monthly requirements are 480 pounds of phosphoric acid, 320 pounds of nitrogen, and 225 pounds of potash. If mix A costs $30 per cubic yard, mix B $36 per cubic yard, and mix C $39 per cubic yard, how many cubic yards of each mix should the farmer blend to meet the minimum monthly requirements at a minimal cost? What is the minimum cost? Social Sciences

Education

31.

— resource

two high schools

allocation.

A

metropolitan school district has

overcrowded and two that are under-

that are

enrolled. In order to balance the enrollment, the school board has

decided to bus students from the overcrowded schools to the underenrolled schools. North Division High School has 300 more students

than

it

should have, and South Division High School has 500 more

students than

it

should have. Central High School can accommodate

400 additional students and Washington High School can accommodate 500 additional students.

The weekly cost of busing a student from

North Division to Central is $5, from North Division to Washington is $2, from South Division to Central is $3, and from South Division to

Washington is $4. Determine the number of students that should be bused from each of the overcrowded schools to each of the underenrolled schools in order to balance the enrollment and minimize the cost of busing the students.

Education

32.

What

is

the

minimum

cost?

— resource alJocation. Repeat Problem 31 ifthe weekly cost

of busing a student from North Division to

Washington

is

$7 and

all

the other information remains the same.

8-7

Maximization and Minimization with Mixed Problem Constraints (Optional)

An

Introduction to the Big

The Big

Minimization by the Big

Summary

of

Methods

Larger Problems

An

M Method

of Solution

— A Refinery Application

Introduction to the Big

M Method

two sections, we have seen how to solve two types of programming problems: maximization problems with « problem

In the preceding linear

M Method

M Method

8-7

449

Maximization and Minimization with Mixed Problem Constraints (Optional)

constraints and nonnegative constants on the right side of each problem

and minimization problems with

constraint,

3=

problem constraints and

nonnegative coefficients in the objective function. In

this section

we

will

present a generalized version of the simplex method that will solve both

maximization and minimization problems with any combination of ^, ^, and = problem constraints. The only requirement is that each problem constraint have a nonnegative constant on the right side.

To illustrate this new method, we will consider the following problem, which has one « problem constraint and one 3= problem constraint:

= 2xi+X2 Xi + X2 =5 10 -Xi + X2^2 Xj, X2 >

Maximize

P

Subject to

To form an equation out slack variable Xj

+

+

X2

s,

s,

=

of the

as before,

,

(1)

first

and

inequality,

we

introduce a nonnegative

write:

10

How can we form an equation out of the second inequality? We introduce a second nonnegative variable

Sj

and subtract

it

from the

left

side so that

we

can write

The

variable S2

is

called a surplus variable because

by which the

(surplus)

left

it

is

the

amount

side of the inequality exceeds the right side.

Surplus variables are always nonnegative quantities.

We now

express the linear programming problem

(1)

in the standard

form. X, + X2 + S] — X1 + X2 — S2

-2x,-X2

=

10

=2 +P =

(2)

X,, Xj, Si, S2 3^

The obvious

basic solution (found by setting the nonbasic variables x^

X2 equal to zero) Xi

=

0,

X2

=

0,

Si

This basic solution

= is

10.

S2

=

-2.

not feasible.

P

=

The surplus

violation of the nonnegative requirement for

simplex method works only tableau for (2)

is

and

and

is:

feasible, so

when

we cannot

all

variable Sj

is

negative, a

variables except P.

The

the obvious basic solution for each

solve this problem by writing the tableau

starting pivot operations.

In order to use the simplex

we must modify

method on

the problem. First,

a

we

problem with mixed constraints, introduce a second nonnegative

450

Linear Inequalities and Linear Programming

variable a, in the equation involving the surplus variable Sj:

— Xi + The

Xj

— S2 +

variable a,

=

a]

2

called an artificial variable, since

is

it

has no actual

relationship to any of the variables in the original problem. Artificial

variables are always nonnegative quantities. Next

we add

the term

—Ma,

to the objective function:

P

=

2Xi

+X2-Mai

The number M is large as

we

very large positive constant whose value can be

a

We now

wish.

have a new problem, which we

made as

shall call the

modified problem:

Maximize

P

Subject to

=

+ X2 — Ma, + Xj + s, — S2 + + X2

2x,

Xj

— X1

a,

=10 =2

(3)

Xj, X2, Sj, S2, Qi 3=

Rewriting X, + — Xj +

(3)

X2

in the alternate standard form,

+

X2

-2X1-X2

Si



= 10 =2 S2 + Qi +Mai+P =

we

obtain:

(4)

Again we see that the obvious basic solution is not feasible. (Setting the nonbasic variables x,, x^. and a, equal to zero, we see that S2 is still negative.) To overcome this problem, we write the augmented matrix for (4)

and proceed Xi

as follows:

8-7

Maximization and Minimization with Mixed Problem Constraints (Optional)

The obvious

basic solution (setting the nonbasic variables x,, X2, and S2

equal to zero)

=

0,

This solution

is

also feasible, because all variables except

tive.

=

is

X2

Xj

0,

Thus,

451

Si

=

10,

S2

we can commence

=

0,

ai

=

2,

P

= -2M P are nonnega-

with pivot operations.

The pivot column is determined by the most negative element in the bottom row of the tableau. Since is a positive number, — M — 1 is certainly a negative element. What about the element M — 2? Remember that M is a very large positive number. We will assume that M is so large that any expression of the form M — k is positive. Thus, the only negative element in the bottom row is — M — 1.

M

Xi

Pivot

row

452

Linear Inequalities and Linear Programming

the original problem.

To

is true, suppose that we were able to and Sj that satisfy the original system (2) Then by using these same values in (3) along

see that this

find feasible values of Xj, Xj, Sj,

and produce with a,

=

a

value of P

P

contradicts the fact that

problem. Thus

As

(5) is

example

this

>

14.

we have found

0,

=

a feasible solution of

14

is

the

an optimal solution illustrates, if a,

(3)

with

P>

14.

This

maximum value of P for the modified for the original

=

problem.

in an optimal solution for the

modified problem, then deleting a, produces an optimal solution for the original problem.

What happens

modified problem? In this case, has no solution because

its

a,

if

it

#

in the optimal solution for the

can be shown that the original problem

feasible set

is

empty.

In larger problems, each > problem constraint will require the introduc-

and an artificial variable. If one of the problem an equation rather than an inequality, there is no need to introduce a slack or surplus variable. However, each = problem constraint will require the introduction of another artificial variable. Finally, each tion of a surplus variable

constraints

is

variable

artificial

must

also be included in the objective function for the

modified problem. The same constant variable.

method

M

can be used

Because of the role that the constant

is

often called the big

for

each

artificial

M plays in this approach, this

M method.

The Big M Method

We now summarize the key steps used in the big M method and use them to solve several problems.

The Big

M

Method

— Introducing

Slack, Surplus,

and

Artificial

Variables 1.

any problem constraints have negative constants on the right multiply both sides by —1 to obtain a constraint with a nonnegative constant. (If the constraint is an inequality, this will If

side,

reverse the direction of the inequality.)

^

2.

Introduce a slack variable in each

3.

Introduce a surplus variable and an

constraint. artificial

variable in each

^

constraint. 4.

Introduce an

5.

For each

artificial

artificial

Use the same constant

Example 17

= constraint. — add Mqj to the objective function.

variable in each

variable

Uj

,

M for all artificial variables.

Find the modified problem for the following linear programming problem. Do not attempt to solve the problem.

8-7

Maximization and Minimization with Mixed Problem Constraints (Optional)

Maximize

P

Subject to

=

+ SXj + 8X3 + 2X2 — X3 « 7

2Xj

X, Aj X,

2x,

T

+ —

'^ 2

4X2 Xj

'^ **

First,

we

+

6

"^

introduce the slack, surplus, and

2X2

—1

to

change

—5

to 5:

— X2 + 2X3 ^ 5

the rules stated in the box: Xj

3

1

+ X2-2x3)^(-l)(-5)

Xi

we

2'

^

""""

multiply the second constraint by

(-l)(-X2

Next,

3

+ 3x3 3^ + 4X3 =

1'

Solution

453



X3

+ Si

artificial

variables according to

454

Linear Inequalities and Linear Programming

The

Big Af

— Solving the Problem

Method

Form the simplex tableau for the modified problem. Use row operations to eliminate the M's in the bottom row simplex tableau in the columns corresponding

of the

to the artificial

variables. 3.

4.

Solve the modified problem by the simplex method. Relate the solution of the modified problem to the original

problem. a.

b.

the modified problem has no solution, then the original problem has no solution. If all artificial variables are zero in the solution to the modiIf

problem, then delete the

fied

artificial variables to find a

solution to the original problem. c.

If

any

are nonzero in the solution to the

artificial variables

modified problem, then the original problem has no solution.

Example 18

Solve the following linear programming problem using the big

Maximize Subject to

=

Xj

1

Solution



+ 8X3 ^ 20 X, + X2 = + X3 5 Xi X2 + X3 3= 10 P

2

'

Xj

'

'

3

State the modified problem:

Maximize

P=

Subject to

Xi

X,

Xj



+ 3X3 — Mqi — Moj + Si

Xj

+ X2

+ x. X2 + X3 Xi

,

Write the simplex tableau X2 1

x,

s,

a,

— X2

,

X3

Si

,

for the s,

,

S2

S2

,

+

Qi

,

20

02

=

02

3^

10

modified problem.

a,

P

M method:

8-7

Maximization and Minimization with Mixed Problem Constraints (Optional)

455

Xi

Eliminate the a,

M from

column

456

Linear Inequalities and Linear Programming

Since

a,

=

Max P =

Problem 18

and 10

Qj

= Xj

at

the solution to the original problem

0,

=

Xj

0,

=

5,

Xj

=

is

5

Solve the following linear programming problem using the big

Maximize

P

=

+

x^

4X2

+

M method:

2X3

Subject to

X2, X3

Example 19

Solve the following linear programming problem using the big

=

Maximize

P

Subject to

2X] + Xi +

3xi

X,

Solution

+

5x2

X2

^

4

2x2

^ ^

10

,

Xj

Introducing slack, surplus, and

artificial variables,

we

obtain the modified

problem: 2x,

+

X2

M method:

-f

Modified problem

8-7

Maximization and Minimization with Mixed Problem Constraints (Optional)

457

10

Problem 19

Solve the following linear programming problem using the big

Maximize Subject to

P

=

X]

2Xi

Sx^

+ 5X2 ^ + X2 3^ X,

>

+ 2X2

,

X2

^

5

12

M method:

458

Linear Inequalities and Linear Programming

cost requirements are listed in the table.

should be processed each day

What

Solution

is

the

minimum

cost?

to

How many gemstones of each type

minimize the

cost of the finished stones?

8-7

Xl

Maximization and Minimization with Mixed Problem Constraints (Optional)

459

460

Linear Inequalities and Linear Programming

Summary of Methods

8-7

Maximization and Minimization with Mixed Problem Constraints (Optional)

Table 4

461

462

Linear Inequalities and Linear Programming

The corresponding inequality Xo

90 ^2

+

+

_

.

_

_

110-

The

cost of the

C=

28(x,

+

110X4

+

34(x,

+

X2)

X3)

^

105(X2

is

+

+ X4)

5X43=0

components used

The revenue from

P

gasoline

^ 105

Xi

-15X2+

and the

premium

X4

90X2

R=

for

32(X3

+

is

X4)

selling all the gasoline

+ 40(X2 +

is

X4)

profit is

= R-C = 34(Xi + X3) + 40(X2 + X4) - 28(Xj + X2) - 32(X3 + X4) = (34 - 28)x, + (40 - 28)X2 + (34 - 32)x3 + (40 - 32)x4 = 6x, + 12X2 + 2X3 + 8X4

To find the maximum ming problem: Maximize

profit,

we must solve the following linear program-

8-7

Maximization and Minimization with Mixed Problem Constraints (Optional)

463

Table 5 Input to Program

NUMBER OF

'...'nR I

Output from Program

DECISIOH

flBLES = 4

NUMBER F C NSTR 2 OF THE FORM =

fl I

NTS = 6

X

X3 X4

OF THE FORM =

CONSTRfilHTS 1

'..'RR I

=

26 25

=

S750 1 1250

=

JLflCK il =

30000

1

1

'...'HR I

flBLES

flBLES

i2 = 1

1

20000

SURPLUS 1

£0 000

1

1

- 5

1

1

53 54 55 56

1

'...'HRIHBLE;

= 15 = 500 = =

5

MAXIMUM

0-15

5

3

'...'flLUE

OF OB JECT I UE FUHCTION

1

OBJECTIVE FUHCTION

Problem 21

Suppose the refinery

which $35

costs

in

Example

21 has 35.000 barrels of

$25 a barrel, and 15,000 barrels of component

a barrel. If all the

other information

is

component A, which costs

B,

unchanged, formulate

a linear

programming problem whose solution is the maximum profit. Do not attempt to solve the problem (unless you have access to a computer and a program that solves linear programming problems).

Answers to Matched Problems

17.

Maximize Subject to

P

— Ma^ — Mqj — Mqj x,-2x^+ x,-s, + a, x, + 3x2 — 4x3 —82 + 02

=

3x,



2x2

2X1+4X2 +

— 3x, +

X2

+

Xj

=10 =20 +S3 +03=15

5X3

+

X3 X,

18.

MaxP =

19.

No

20.

MinC = $86|atXi=^,

22 at Xi

=

6,

X2

=

4,

,

X3

X2, X3, Si, Qj, S2,

=

solution (empty feasible region) X2

=

=5

0,

X3

=^

2, S3,

03

s^

464

Linear Inequalities and Linear Programming

21.

Maximize

P

Subject to

=

QXj

Xj

+

+



15x2

X3

+

5X4

35,000

X2

+

Xi

X3

+ X4«

15,000

X3

5=

20,000

^

10,000

+

X2

X4

^0

+15X3

-5Xi

-

+

15x2 1'

2'

5X4

3'

4

^ ""^

Exercise 8-7 In

ProbJems 3-8:

(A)

Introduce slack, surplus, and

artificial

variables and form the modified

problem. (B)

Write the augmented

M from

(C)

1.

coejjficienf

matrix for the modified problem and

columns of the artificial variables. Solve the modified problem using the simplex method. eliminate

Maximize

P

Subject to

X,

=

the

+ 2X2 + 2x2 « 12 5Xi

2.

=

Maximize

P

Subject to

2x,

Sx,

+

Xj

+

7x2

16

8-7

B

Maximization and Minimization with Mixed Problem Constraints (Optional)

M method to solve (he following

Use (he big 9.

Minimize and maximize

P=

-

6x,

Xj

+

3x,

+

X2

^

10

2X2

^

24

X, X2

^

P = 2X] + SXj X, + 2X2 ^ 18

Maximize Subject to

problems:

Minimize and maximize

= -4x, +

P

2x2

Subject to

11.

10.

Subject to

3Xi

+

Maximize Subject to

X,

=

2Xi Xj

+

12x2

+

= 5x, +

I

^A

12 6

-1

Xj



X2

2Xi

+

X2

X^

Minimize

16.

C = -5Xi - 12x2 + 16x3

2Xi

+ +

^Aj

+ X3«

10

3X2

+X3 ^

6

^2 Xj

«

16

X2

2=

9

+ 9X3

7X2

,

+ +

X3

2^

20

3X3

=

36

X2 X3



u

1

Minimize

Subject to

2X2

I

20

X2

C = — 3x, + 15x2 — 4X3

Subject to Xi

2X2

Subject to

+ X2 + 2x3^ A9

15.

2X2

^

Maximize P

20x3

Subject to 3x,

+ + +

+

X,,X2 3=0 14.

10xi

X2

,

6x,

3=

Maximize P

=

P

X2 3=10

Xj, X2

13.

+ X2 « 28 + 2X2^ 16 X,

2xi+ X2«21 X,

16X2

Xi

12.

465

,

2Xi Xi

+ +

A

3

X2

2X2 iJ

+ 3x3^24 + X3 ^ 6

\y

X2. X3

1

2

'

"

-^3

I

3

'

"

Problems 17-24 are mixed. Some can besoJved by the methods presen(ed Sections 8-5 17.

and

8-6, vvhiie others

Minimize

18.

C = lOXi -40x2- 5x, Subject to Xi

+

3x2

mus( be solved

b;-

(he big

M method.

Maximize P = 7Xi

-

5X2

+

2x3

Subject to

«

6

x^ X,

— —

X3

3 —8

X3

^

10

Aji Aof "3



'-'

2X2 X2

+ +

'

in

466

Linear Inequalities and Linear Programming

19.

8-7

Maximization and Minimization with Mixed Problem Constraints (Optional]

467

2,000 people. Each ad in the Journal costs $200 and will be read by 500 people. Each ad in the Tribune costs $100 and will be read by 1,500 people.

many

The company wants

ads should

16.000 people to read

at least

its

ads.

How

place in each paper in order to minimize the advertising costs? What is the minimum cost? 28.

Advertising. Repeat Problem 27

more than Life Sciences

29.

it

the Tribune

if

is

unable

to

accept

4 ads from the company.



people. An individual on a high-protein, low-carbohydrate diet requires at least 100 units of protein and at most 24 units of Afutrition

carbohydrates daily. The diet will consist entirely of three special liquid diet foods. A, B, and C. The contents and cost of the diet foods are given in the table. How many bottles of each brand of diet food

should be consumed daily in order

to

drate requirements at minimal cost?

30.

31.

meet the protein and carbohy-

What

is

the

minimum

cost?

468

Linear Inequalities and Linear Programming

In

Problems 33-39. formulate each problem as a linear programming

problem. Do not solve the linear programming problem. Business & Economics

33.

Manufacturing

— production

A company manufactures Milwaukee and Racine. The Mil-

scheduling.

car and truck frames at plants in

waukee

plant has a daily operating budget of $50,000 and can produce most 300 frames daily in any combination. It costs $150 to manufacture a car frame and $200 to manufacture a truck frame at the Milwaukee plant. The Racine plant has a daily operating budget of $35,000, can produce a maximum combined total of 200 frames daily, at

34.

and produces a car frame at a cost of $135 and a truck frame at a cost of $180. Based on past demand, the company wants to limit production to a maximum of 250 car frames and 350 truck frames per day. If the company realizes a profit of $50 on each car frame and $70 on each truck frame, how many frames of each type should be produced at each plant to maximize the daily profit? Finances loan distributions. A savings and loan company has $3 million to lend. The types of loans and annual returns offered by the



company are given in the table. State laws require that at least 50% of the money loaned for mortgages must be for first mortgages and that at least 30% of the total amount loaned must be for either first or second mortgages. Company policy requires that the amount of signature and automobile loans cannot exceed 25% of the total amount loaned and that signature loans cannot exceed 15% of the total amount loaned. How much money should be allocated to each type of loan in order to maximize the company's return?

Type of Loan

Annual Return

Signature

18%

mortgage Second mortgage Automobile

12%

First

35.

14% 16%

— petroleum.

A refinery produces two grades of gasoline, premium, by blending together three components. A, B, and C. Component A has an octane rating of 90 and costs $28 a barrel, component B has an octane rating of 100 and costs $30 a barrel, and component C has an octane rating of 110 and costs $34 a barrel. The octane rating for regular must be at least 95 and the octane rating for premium must be at least 105. Regular gasoline sells for $38 a barrel and premium sells for $46 a barrel. The company has 40.000 barrels of component A, 25,000 barrels of component B, and 15,000 barrels of component C and must produce at least 30,000 barrels of regular and 25,000 barrels of premium. How should they blend the components in order to maximize their profit? Blending

regular and

8-7

Maximization and Minimization with Mixed Problem Constraints (Optional)

36.

Blending

food processing.

A company produces

mix, regular and deluxe, by mixing dried

two brands of trail and cereal. The

fruits, nuts,

recipes for the mixes are given in the table.

pounds

469

The company has

1,200

pounds of nuts, and 1,500 pounds of cereal to be used in producing the mixes. The company makes a profit of $0.40 on each pound of regular mix and $0.60 on each pound of deluxe mix. How many pounds of each ingredient should be used in each mix in order to maximize the company's profit? of dried fruits, 750

Type of Mix

470

Linear Inequalities and Linear Programming

Social Sciences

39.

Education

— resource allocation. Three towns are forming a consoli-

dated school district with two high schools. Each high school has a

maximum

capacity of 2,000 students.

town B has

Town A

has 500 high school

and town C has 1 ,800. The weekly costs of transporting a student from each town to each school are given in the table. In order to keep the enrollment balanced, the school board has decided that each high school must enroll at least 40% of the total student population. Furthermore, no more than 60% of the students in any town should be sent to the same high school. How many students from each town should be enrolled in each school in order to meet these requirements and minimize the cost of transporting the students,

1 ,200,

students?

Weekly Transportation

Town A Town B Town C

8-8

Cost per Student School I

School

$4

$8

6

4

3

9

II

Chapter Review Important Terms

8-1

Linear inequalities in two variables, graph of a linear inequality in two

8-2

Systems of linear inequalities

and Symbols

variables,

upper half-plane, lower half-plane

solution region,

bounded

in

regions,

two variables, graphical solution,

unbounded

regions, corner point,

feasible solution, feasible region

8-3

Linear programming ear

in

two dimensions

programming problem, decision

— a geometric approach,

lin-

variables, objective function,

constraints, nonnegative constraints, solution,

mathematical model, graphical maximization problem, constant-profit line, isoprofit line,

optimal solution, minimization problem, linear function, multiple optimal solutions, no feasible region, unbounded objective function 8-4

A geometric introduction

to the

simplex method, slack variables, basic

solution, basic feasible solution, nonbasic variables, basic variables,

simplex method 8-5

The simplex method: maximization with s problem

constraints, stan-

dard form of a linear programming problem, obvious basic solution, obvious basic feasible solution, simplex tableau, pivot column, pivot row, fi,

+

pivot

element,

kR,-*Ri

pivot

operation,

row

operations,

kRj^Rj,

8-8

8-6

The dual: minimizad'on

tvith 3^

Chapter Review

471

problem constrainfs. dual problem,

solution of minimization problems, transportation problem 8-7

Exercise 8-8

Maximization and minimization with mixed problem constraints. surplus variable, artificial variable, modified problem, big M method

Chapter Review Work through

all

the problems in this chapter review

and check your

back of the book. (Answers to all review problems are Where weaknesses show up, review appropriate sections in the text. answers

in the

you are satis^ed that you know the material, take the practice

test

there.)

When

following

this review.

A

1.

Solve the system of linear inequalities graphically: 3xi 2xi

2.

+ +

X2

«

9

6X2

«

18

Xj, X2

^

Solve the linear programming problem geometrically:

Maximize Subject to

=

+ 2X2 2xi + X2^8 Xi + 2x2 ^ 10 P

6X1

Xj, X2 5= 3.

Convert Problem

4.

Find

all

system of equations using slack variables.

2 into a

basic solutions for the system in Problem 3 and determine

which basic solutions are

feasible.

5.

Write the simplex tableau

for

6.

Solve Problem 2 using the simplex method.

7.

Solve the linear programming problem geometrically:

Minimize Subject to

8.

Form

9.

Solve Problem

Problem

2

and

circle the pivot element.

C = 5Xj + 2X2 X,

+

3X2

^

15

2x,

+

X2

3=

20

Xj, X2

^

the dual of Problem

7.

8by applying the simplex method to the dual problem.

472

Linear Inequalities and Linear Programming

B

10.

Solve the linear programming problem geometrically:

=

Maximize

P

Subject to

2Xi

+

3x,

4x2

+ 4X2 « 24 3x, + 3X2 'S21 4Xi + 2x2 ^ 20 Xj, X2 >

11-

Solve Problem 10 using the simplex method.

12.

Solve the linear programming problem geometrically:

Minimize

C = 3X1 + 8X2

Subject to

Xi

+ +

Xi

X2

3^

10

2x2

5^

15

X2^3 Xj, X2

3^

13.

Form

14.

Solve Problem 13 by applying the simplex method.

the dual of Problem 12.

Solve the following linear programming probJems: 15.

16.

Subject to

3X2 — 3X3 — X2 2X3 « 3 Xi =s 10 + 2x2 5X3 2Xi

Maximize

P = 5x,

Subject to

X,



X2

X,

+

Xj

Maximize

=

P

5Xj

1

C

17.

Minimize

C=

Subject to

2Xi 2Xi Xi

18.

Minimize Subject to

+



+ 3X2 — 3X3 — 2x3 ^ 3

^5 2

'

'

3

""^

2x, + 3X2 + X2 « 20 + X2^10 + 2X2 2= 8 X,, X2 ^

= 15X1 + Xi + Xj X3

+

Xi

X2

12X2

+

15X3

+

X4

« 240 « 500 ^ 400

+

X4

3^

X3

300

+

18X4

8-8

473

Chapter Review

Applications Business & Economics

19.



Manufacturing resource allocation. Set up (but do not solve) Problem 19 in Exercise 8-3 with the added restrictions that the fabricating department must operate at least 60 labor-hours per day and the finishing department at least 12 labor-hours per day. (A)

Write the linear programming problem using appropriate

(B)

(C)

Tranform part A into a system and artificial variables. Write the simplex tableau

of equations using slack, surplus,

for part B.

Do

not solve.

Formulate the following as linear programming problems but do not 20.

Transportation

in-

and objective function.

equalities

solve:

— shipping schedule. A company produces motors for

and factory B. Factory A can produce B can produce 1,000 motors a month. The motors are then shipped to one of three plants, where the washing machines are assembled. In order to meet anticipated demand, plant X must assemble 500 washing machines a month, plant Y must assemble 700 washing machines a month, and plant Z must

washing machines 1,500 motors a

at factory

month and

A

factory

assemble 800 washing machines a month. The shipping charges

one motor are given in the will

minimize the

table.

Determine

a

for

shipping schedule that

cost of transporting the motors

from the factories

to

the assembly plants.

Shipping Charges Plant

A

$5

Factory B

$9

Factory

Life Sciences

21.

Nutrition

X

Plant

$8 $7

Y

Plant

Z

$12 $ 6

— animals. A special diet for laboratory animals

is

to con-

and 900 calories. There are two feed mixes available, mix A and mix B. A gram of mix A contains 3 units of vitamins, 2 units of minerals, and 6 calories. A gram of mix B contains 4 units of vitamins. 5 units of minerals, and 10 calories. Mix A costs $0.02 per gram and mix B costs $0.04 per gram. How many grams of each mix should be used to satisfy the requirements of the diet at minimal cost? tain at least 300 units of vitamins. 200 units of minerals,

474

Linear Inequalities and Linear Programming

Practice Test:

Chapter 8 1.

Solve the system of linear inequalities graphically: 2x,

+

X2

2Xi

+

3x2

Xj, Xj 2.

« ^ ^

8

12

Convert the following maximization problem into a system of equations using slack variables:

Maximize

7.

8. 9.

8-8

Chapter Review

475

Formulate the following problems as linear programming problems bu( do not solve; 10.

South Shore Sail Loft manufactures regular and competition sails. Each regular sail takes 2 hours to cut and 4 hours to sew. Each competition sail takes 3 hours to cut and 9 hours to sew. The Loft makes a profit of $100 on each regular sail and $200 on each competition sail. If there are 150 hours available in the cutting department and 360 hours available in the sewing department, how many sails of each type should the company manufacture in order to maximize

11.

An individual requires 400 units of vitamin B and 800 units of vitamin

their profit?

C daily. The local drugstore carries two types of vitamin

tablets,

brand

X and brand Y. A brand X tablet contains 75 units of vitamin B and 100 units of vitamin C and costs $0.05. A brand Y tablet contains 50 units of vitamin B and 200 units of vitamin C and costs $0.04. How many tablets of each

brand should be taken at minimal cost?

vitamin requirements

in order to

meet the daily

9

Probability

r:^^

^^

Sb^

:'F^^«f, /— 2 and continuous on the right at x = —

and q(— 2)

= 0, / has

a vertical asymptote at x

Vx

2.

+

2

is

continuous

Since p(— 2)

= — 2.

=5

=?^

Asymptotes; Limits

12-1

Problem 4

at Infinity

and

Infinite Limits

679

Find the vertical asymptotes of each function. (A)

fix)

x^-l x

(C)

fix)

X2

+1 _

(B)

it is

=

+l +1

x^ (D)

In order to use vertical

function,

/(x)

/(x)

=

x

^

asymptotes as an aid in sketching the graph of

not enough simply to locate

all

a

of the vertical asymptotes. In

we must determine the behavior of the graph as x approaches each asymptote from the left and the right. That is. we must evaluate addition,

linix-c- fix]

Example

5

and lim,_^+

Sketch a graph of / by vertical

first

each vertical asymptote x

= c.

evaluating lim,_^^ f{x] and lim,^^, /(x) at each

asymptote of

fix]

x^lx

Then

/(x) at

find

-

4)

any horizontal asymptotes and complete the graph using

a

calculator and point-by-point plotting.

Solution

Let p(x) = x - 2 and q(x) = x^x - 4), and observe that q(0) = and q(4) = 0. Since p(0) =-2 ^ and p(4) = 2 9^ o. /has vertical asymptotes at x =Oand X = 4. It will be helpful to construct a sign chart for / relative to the real

number

line:

Sign of (x

-

2)

- - -

680

Graphing and Optimization

asymptote X =

Vertical at

asymptote X = 4

Vertical at

Figure 5

Since /

is

a rational function

we can

use

(1)

=

We

complete the graph

has a horizontal asymptote at y

0.

on page 674

to

conclude that / (see Figure 5)

using a calculator and point-by-point plotting for regions of uncertainty. (As

we

progress through this chapter, the aids to graphing that

develop will

need

Problem

5

less

tell

and

us more and more about the shape of a graph and

we will we will

less point-by-point plotting.)

Repeat Example

5 for f (x)

=

Application Example

6

Average Cost

A company estimates

that the fixed costs for manufacturing a new transisand the cost per unit produced is $2 (see Section 11-5). of producing x radios is

tor radio are $5,000

The

total cost

C(x)

=

5,000

-I-

2x

Asymptotes; Limits

12-1

and the average

C(x)

cost per unit

+

5,000

2x

and

681

Infinite Limits

is

XXX

= Cx =

at Infinity

=

5,000

,

r-

2

Since

-

,

lim C(x)

=

X—•»

the

,

/5,000

lim

I

X—•»

hne y

=

Notice that 2

defined

x

,

is

^

2

/

2 is a horizontal

The function for

\ +21 =

X

\

asymptote

for the

average cost function.

also the cost per unit.

C(x] also has a vertical asymptote at x 0,

we need

=

Since C(x)

is

not

investigate only the right-hand limit at x

=

0:

0.

limC(x)=limf^:^ + 2)=oo

X— 0+

X— 0+

\

X

/

Figure 6 shows these results graphically.

30

20

10

J

L.

^'^

1,000

Figure 6 Average cost function

Problem 6

Refer to Example function C(x)

Answers to Matched Problems

1-

X '^

=

6.

Evaluate lim,^„ C(x) and lim,_(,+

10,000

+ 4x.

C(-x) for

the cost

682

Graphing and Optimization

Vertical asymptotes

5.

at

X

= —2

lim f(x)

and x

=

=

2

—00

x-«-2-

lim /(x)

X— 2+

= 00

lim /(x)

= 00

lim /(x)

= —00

X— 2X— 2+

Horizontal asymptote

y

at

=

lim C(x)

6.

=

lim C(x)

4;

= oo

X—»0*

x-»»

Exercise 12-1 Use a caJcuJator Use the

1.

to

/(x)

=

fM

=

fix)

x

+

1

x

+

1

Use a calculator 1.0001.

evaluate each /unction at x

10, 100, 1,000.

10,000.

and

fM to

x+1 x

+

1

evaluate each function at x=l.l, 1.01. 1.001, and

at .9, .99, .999.

and

.9999.

Use the results of your calculations

estimate lim^^i+ffx) and limx^,-/(x).

5.

and

results o/ these calculations to estimate Jimx_„/(xJ.

fix)-

X-

fix) 1

11/3 (1

to

12-1

7.

/(x)=(X

-

Problems 9-12

9-

8.

If/"

re/er to the /olJowing

lim fix)

(A)

Asymptotes; Limits

=

/(x)='^

'

-

x)"/^

=

lim fix)

(B)

?

?

X—•»

lim /(x)

(A)

(1

Infinite Limits

graph oj y =fix]:

X—•— 10.

and

at Infinity

=

?

(B)

lim fix)

=

?

=

?

(B)

lim/(x)

=

?

X—*a" 11.

(A)

lim fix)

12.

(A)

Where does / have Where does /have

(B)

B

Evaluate the following

horizontal asymptotes? vertical asymptotes?

limits.

14.

lim

+ 4) (3-4 X^ X*/

» \

X

2x^ 15.

lim 3x^

17.

19.



4X''

+

16.

lim Qx" X— X-"

5

.CO

2x3

+

10

3x2 18.

lim

+7 3x^ + 5 lim x-« 4x2 + 2

lim

^_-„ X

» 4X''

20.

hm

+2

7x2

x-^» x=

+

7

Find the vertical asymptotes of each function. x2 21-

fix)

X2 23.

fix)

25.

/(x):

+

l

x2

+

4

683

684

Graphing and Optimization

For each function, find

horizonfai and vertical asymptotes. Evaluate

all

lim^^c+f(x)and lim^^^-f(x)

at

each vertical asymptote. Sketch the graph of

the function. Use a calculator

and point-by-point

plotting in regions of

uncertainty.

27.

/(x)

=

2x-h4

=

X

29.

/(x)

31.

Let/(x)

32.

v2



=

=

/(x)

28.

x-4

x

2

1

=

/(x)

30.

A

+

v2

—d

-

=

(A)

Use a calculator to evaluate x = -10, -100, and -1,000.

(B)

Evaluate lim,_„/(x) and lim,__,o/(x).

(C)

Sketch the graph

f(x) at x

10, 100.

and

1,000,

and

of/.

Repeat Problem 31 for/(x)

+

sl4x^

Each of the lim [/(x)

limits in

-

Problems 33-36

is,

33.

11m (Vx

35.

lim (Vx^

of the

is

form

g(x)]

Evaluate each limit by

That

1

multiply f/(x)

+l+

X

jirst

rationalizing the expression [f(x)

- g(x)J/l

by

[fix]

Vx)

-

x)

+ g(xM/fx) +

34.

lim(VxMM -

36.

lim (Vx^

-I-



g(xJJ/].

g(x)J.

4x

x)

- x)

Applications Business

& Economics

37.

Average C(x)

38.

=

cost.

The

3,000

-I-

cost function for

manufacturing x flashlights

2.75X

(A)

Find C(x), the average cost function.

(B)

Evaluate lim,^.. C(x).

(C)

Evaluate limx—o* C(x).

Average

pro/it.

is

Suppose the

flashlights in

Problem 37

each and that the company manufactures and (A)

Find the

(B)

Find the average

(C)

Find the limit of the average

sells

sell for

$5.25

x flashlights.

profit.

profit.

profit as

x approaches infinity.

12-2

39.

Marginal C(x)

=

cost.

The

10.000

First Derivative

cost function for a publishing

and Graphs

company

685

is

100

+

12x

H

X u'here x

is

the

number

of books

produced

in a single printing.

(A)

Find the marginal cost function.

(B)

Evaluate the limit of the marginal cost function as x approaches infinity.

40.

A company

Advertising.

estimates that

it

will sell N(x) units of a

product after spending $x thousand on advertising, as given by M, 1 N(x) ^

S.OOOx'

=

'

2.5x^

+

4,000

Evaluate the limit of N(x) as x approaches Life Sciences

41.

The

Pollution.

centimeter)

t

bacteria concentration

days after

infinity.

(number

of bacteria per cubic

a polluted lake is treated

with a bactericide

is

given by 50(2

^"^^

,2

+ 45,000 + 225

What was

(A)

the concentration at the time the lake

was

initially

treated?

What

(B)

42.

the limit of the concentration as

is

AnimaJ population. certain species

t

t

approaches infinity?

A biologist has estimated that the population of a

years from

now

will be given

by

500(2 ^^'

What Social Sciences

43.

.5(2 is

Learning.

12-2

First Derivative

is

P(t) as

t

approaches infinity?

A new worker on an assembly line performs an operation in

T minutes

What

+ 450

the limit of

after x

performances of the operation, as given by

the limit of

T

as x approaches infinity?

and Graphs Increasing and Decreasing Functions Critical

Values and Local Extrema

First-Derivative Test

Application

686

Graphing and Optimization

Since the derivative

is

associated with the slope of the graph of a function

at

we might expect that it is also associated with other properties of a As we will see in this and the next section, the first and second

a point,

graph.

derivatives can

us a great deal about the shape of the graph of a

tell

function. In addition, this investigation will lead to

absolute

maximum and minimum

methods

for finding

values for functions that do not require

graphing. Companies can use these methods to find production levels that

minimize

will

find levels of

And

drug.

cost or

maximize

profit.

Pharmacologists can use them

drug dosages that will produce

maximum

to

sensitivity to a

so on.

Increasing and Decreasing Functions Graphs of functions generally have rising or falling sections as we move from left to right. It would be an aid to graphing if we could figure out where these sections occur. Suppose the graph of a function /

Figure

7.

As we move from

graph of /is

>

[/'(x)

f[x]

is

0].

left to right,

rising, /(x) is increasing,*

On

the other hand, on the interval is

(b. c)

falling), /(x)

increasing to decreasing, the slope of the graph

(a, b)

is

the

positive

the graph of/is falling,

negative [fix)

the graph of /changes direction (from rising to

is

as indicated in

and the slope of the graph

decreasing, and the slope of the graph

tangent line

is

we see that on the interval

is

[/'(b)


/T5/3) /(— x) = — /(x): symmetry

/(O)

minimum

with respect

x

to

=

at

x

=

0)

and

(0, 2)

2

concave downward on

VS

the origin

730

Graphing and Optimization

20.

Increasing on (— x>l /(x) = 36 + x — 12x'^^ x> 1 fix) = 9x^''^ — 2x + 3,

maximum value of minimum value of minimum value of maximum value of

fix)

f(x)

maximum and minimum,

if

either exists, of each function

intervals.

= x^-6x^ + 9x-6 [-1,5]

[-1,3]

(B)

= 2x3-3x2-12x + [-3,4]

= (x-l)(x-

5)^

(B)

[0.3]

= x''-8x2 +

(A)

[-1,3]

(A)

(-00,

oc)

+

[2,5]

(C)

[-2.1]

24

[-2,3]

(B)

(C)

1

[1,7]

[C)

[3,6]

16

(B)

(B)

[0,2]

[0,00)

(C)

(C)

[-3,4]

[1.00)

748

Graphing and Optimization

16.



ffx)= '^' (A)

+

x^

+

x

l

(-00,00)

[-1,00)

(B)

[O.oo)

(C)

Preliminary Word Problems: 17.

How would lengths

18.

What

is

you divide

inch line so that the product of the two

quantity should be added to 5 and subtracted from 5 in order to

maximum

produce the 19.

a 10

maximum?

a

product of the results?

Find two numbers whose difference

30 and whose product

is

is

a

is

a

minimum. 20.

Find two positive numbers whose

sum

is

60 and whose product

maximum. 21.

Find the dimensions of

maximum

that has 22.

a rectangle

area.

Find the

with perimeter 100 centimeters

maximum

area.

Find the dimensions of a rectangle of area 225 square centimeters that has the least perimeter. What

is

the perimeter?

Applications Business

& Economics

23.

Average

costs. If the

sunglasses

where x

is

(in dollars)

per pair of

0«x«6

12

number

the

(in

thousands) of pairs manufactured,

pairs of glasses should be

manufactured

to

how

minimize the aver-

minimum average cost per pair? Maximum revenue and profit. A company manufactures and sells x television sets per month. The monthly cost and demand equations age cost per pair?

24.

average manufacturing cost

given by

= x2-6x +

C(x)

many

is

What

is

the

are

C(x)

p

=

72,000

= 200- — 30

Find the

(B)

Find the the for

Car

0«x« 6,000

each television

realize

charge

set.

company $5 for each set it company manufacture each month in order to maximize its profit? What is the maximum profit? What should the company charge for each set? If

the government decides to tax the

produces,

25.

60x

maximum revenue. maximum profit, the production level that will maximum profit, and the price the company should

(A)

(C)

+

rental.

A

how many sets should

the

car rental agency rents 100 cars per day at a rate of $10

per day. For each $1 increase in

rate, five

fewer cars are rented. At

what rate should the cars be rented to produce the maximum come? What is the maximum income?

in-

Optimization; Absolute

12-5

26.

Rental income.

every night

at

A

Agriculture. that

if

of 50

is

If

each rented room costs $3

to service

thirty trees are planted per acre,

pounds

of cherries per season.

how many

pound,

maximum to

per day, to

how

maximize

If

each tree will yield an average

for

each additional tree planted

have

reduced by

is

1

trees should be planted per acre to obtain the

yield per acre?

Agriculture.

time

capacity

What is the maximum gross profit? A commercial cherry grower estimates from past records

per acre (up to twenty) the average yield per tree

28.

filled to

should the management charge for each room

gross profit? 27.

ninety room hotel in Las Vegas

749

$25 a room. For each $1 increase in rent, three fewer

rooms are rented.

much

Maxima and Minima

What

is

the

maximum

yield?

A commercial pear grower must decide on the optimum picked and sold.

fruit

will bring 30(t per

If the pears are picked now, they pound, with each tree yielding an average of 60

pounds of salable pears. If the average yield per tree increases 6 pounds per tree per week for the next 4 weeks, but the price drops 20

(IO-

from Aj will the concentration of particulate matter be

far

at a

minimum?

P^rOstl -10

Social Sciences

43.

population N(f)

where 44.

newly incorporated

Politics. In a

= t

(in

30

is

city

it is

estimated that the voting

thousands) will increase according to

+ 12(2-1^

time in years.

0Sts8 When will

the rate of increase be most rapid?

A large grocery chain found that, on the average, a checker can memorize P% of a given price list in x continuous hours, as given

Learning.

approximately by P(x)

How

=

96x

Elasticity of

Demand

^

24x2

X

«

What

is

the

maximum?

(Optional)

and Elasticity of Demand Revenue and Elasticity of Demand Price

3

long should a checker plan to take to memorize the

percentage?

12-6

-

maximum

12-6

Price and Elasticity of In this section

demand and

we

+

500p

Demand

=

753

(Optional)

Demand

will study the effect that

changes

have on

in price

$p and the demand x for a price -demand equation:

revenue. Suppose the price

product are related by the

x

Elasticity of

certain

10,000

(1)

In problems involving revenue, cost,

demand equation

now interested

in the effects that

be more convenient

and

profit,

it is

customary

to

express price as a function of demand. Since

to

to express

use the

we

changes in price have on demand,

demand

it

are will

as a function of price. Solving (1) for

we have

X,

X

= =

- 500p 500(20 - p)

Demand

10,000

as a function of price

or

=

X =/(p)

500(20

-p)

0«pS20

(2)

Since x and p must be nonnegative quantities,

^p^

For most products, price.

we must

restrict

p so that

20.

That

is,

demand

is

assumed

to

price increases result in lower

result in higher

demand

(see Figure 19).

be a decreasing function of

demand and

Suppose the price

X = 500(20 10,000

price decreases

is

changed by an

500(20 - p) 10,000

Demand decreases

P

P +

AP

(A) Increasing price

Figure 19 Price and

demand

p + Ap (B)

p

Decreasing price

754

Graphing and Optimization

amount Ap. Then the

demand

Ap

relative

change

in price

and the relative change in

are. respectively.

Ax

and

/(p

X

P

Economists use the

+ Ap)-/(p) fip]

ratio

Ax Relative change in

X

aF

demand (3)

Relative change in price

p to

study the effect of price changes on demand. Economics texts that do not

use calculus will p.

However,

ity of

at price p,

Ax E(P)

the expression in

(3)

the elasticity of demand al price

expression obviously depends on both p and Ap. Using can let Ap —> and obtain an expression for the point elastic-

we demand

calculus,

call

this

denoted E(p):

12-6

Soiutions

E(p)

-

P-^

'P'

fip]

p(-500] 500(20

-P 20

(A)

-p

-

p)

«

p




1

E(p)

E(p)

< -1

E(p)

Thus,

we can determine where demand

elastic

by constructing a sign chart

is

for E(p)

+ 1>0 +1
-

or

22

=

i'y

Thus.

=2

3y

2

(D)

b

Problem

Example

5

6

=

Find

y, b,

(A)

y

=

Graph y

=

100

logb

2 is equivalent to

b^.

Thus.

x.

logg27

=

=

Recall that b cannot be negative.

10

or

100

log3X

(B)

logjlx

+

1)

= -l

by converting

(C)

to

log^ 1,000

=

3

an equivalent exponential form

first.

Solution

Changing y X

+

1

=

=

2>'

Even though x it is

\

log2(x

or is

+

1) to

X

=

an equivalent exponential form, 2>'

-

1

the independent variable and y

easier to assign

we have

y values and solve for

x.

is

the dependent variable,

782

Exponential and Logarithmic Functions

subtraction problems, and power and root problems into multiplication

problems. 2

=

We

will also

be able

1.06".

Logarithmic Properties

to solve

exponential equations such as

Logarithmic Functions

13-2

— A Review

The following examples and problems, though somewhat

783

artificial, will

give you additional practice in using basic logarithmic properties.

Example 8

Find x so that

- -2

3

-

logb 4

3

-

Solution

logb 4

- -2 -

4'-'^

logb

logb 8

+

logb 8

= logb X

logb 2

logb 8

+

logb 2

logb 8^/^

+ +

logb 2

-

logb 4

logb 2

8



2

logb

logb 4

x

Problem 8

9

= =

logb X

Property 4

logb X

= logb X

Properties 2 and 3

= =

Property 5

logb X

4

Find x so that 3 logb 2

Example

= logb X

+ - logb 25 - logb

Solve logio x

Solution

logio

X

+

logio(x

+

+ 1) = log,o x(x + 1) = x(x + l) = x^ + x — 6 = (x + 3)(x - 2) = x=

1)

+

logio(x

20

=

=

logb x

log^

6.

log,o 6

Property 2

log,o 6

Property 5

6

Solve by factoring,

-3,

2

We must exclude x = — 3. since negative numbers are not in the domains of logarithmic functions; hence.

x is

Problem 9

=

2

the only solution.

Solve logj x

+ logalx —

3)

= logj

10.

Calculator Evaluation of Of all

Common

possible logarithmic bases, the base e

and Natural Logarithms

and the base 10 are used almost

we can use logarithms in certain practical problems, we need to be able to approximate the logarithm of any number either to base 10 or to base e. And conversely, if we are given the logarithm of a number to exclusively. Before

784

Exponential and Logarithmic Functions

base 10 or base

we need

e,

to

such as Tables

cally, tables

be able II

and

to

III

approximate the number. HistoriAppendix B were used for this

of

now with inexpensive scientific hand calculators readily most people will use a calculator, since it is faster and far more

purpose, but available,

accurate.

Common with base

logarithms

logarithms with base "log" (or

"LOG") and

common

sents a

(also called

Natural logarithms

10.

e.

Most

Briggsian logarithms) are logarithms

(also called

Napierian logarithms] are

scientific calculators

have a button labeled The former repre-

a button labeled "In" (or "LN").

(base 10) logarithm

and the

latter a natural (base e)

"log" and "In" are both used extensively in mathemati-

logarithm. In

fact,

cal literature,

and whenever you see either used in this book without a base

indicated they will be interpreted as follows:

Logarithmic Notation

= X=

log X In

logio

X

loge X

Finding the

common or natural

very easy: you simply enter a

push the

Example 10

Use a (A)

scientific calculator to find (B)

In 0.000

(A)

3184

(B)

0.000 349

each

349

Enter

(C)

the

domain

of the function

is

and

log or In button.

log 3,184

Solutions

logarithm using a scientific calculator

number from

to six (C)

decimal places: log(-3.24)

Display

Press log

3.502973

-7.960439 log

-3.24

Error

An error is indicated in part C because — 3.24 is not in the domain of the log function.

Problem 10

Use a (A)

scientific calculator to find

log 0.013 529

(B)

each

to six

In 28.693 28

decimal places: (C)

ln(-0.438)

We now turn to the second problem to be discussed in this section: Given We make direct use of the

the logarithm of a number, find the number.

Logarithmic Functions

13-2

— A Review

785

logarithmic- exponential relationships, which follow directly from the definition of logarithmic functions at the beginning of this section.

Logarithmic - Exponential Relationships

=y =y

logx In X

Example

11

Solutions

Find x

is

equivalent to

x

is

equivalent to

x

to three significant digits,

(A)

log x

(A)

log x

=

X

= =

x

=

-9.315

(B)

In x

= IC = e*'

given the indicated logarithms:

=

2.386

Change

—9.315

to equivalent exponential form.

io-'"5 4.84

X

10^'°

The answer

is

displayed in scientific notation

in the calculator.

= = =

In X

(B)

X x

Problem 11

Find x

to equivalent exponential form.

e^^'**

10.9

to four significant digits,

= -5.062

In x

(A)

Change

2.386

(B)

log

given the indicated logarithms.

x=

12. 082

1

Application If

P

dollars are invested at

interest

money

A= The

is

in the

P(l

100i%

interest per period for n periods,

+

account

i)"

at the

end

Compound

of period n

Example 12 Doubling Time

How

is

given by

interest formula

fact that interest paid to the

account

interest during the following periods

pound

and

paid to the account at the end of each period, then the amount of

is

at

the end of each period earns

the reason this

is

called a

com-

interest formula.

long

invested

at

(to

the next whole year) will

20%

interest

it

take

compounded annually?

money

to

double

if it

is

786

Exponential and Logarithmic Functions

Solufion

Find n

for

A = 2P and =

A = P(1 + 2P

i

i)"

0.2.

13-2

rates

compounded

(1

ln(l

annually.

We

Logarithmic Functions

— A Review

787

proceed as follows:

A = P(1 +i)" 2P = P(l + i)" 2 = (1 + i)" +i)" = 2 +i)" = ln2

n ln(l

+

i)

=

"

_ ~

In 2

In 2 ln(l

+

i)

Figure 4 shows the graph of this equation (doubling times in years) for

compounded annually from 1%

interest rates

changes in doubling times from

Answers to Matched Problems

(A)

9

=

32

(B)

2

=

1%

4'^^

to

(C)

49 = 2 (B) 108,3 = = 3/2 (B) x = l/3 y = log3(x — 1) is equivalent to X = 3 + 1 (A)

log,

(A)

y

''

5

70%. Note the dramatic

to

20%. 1/9

=

b

=

1/2 (C)

3-^

(C)

10

log3(l/3)

= -l

788

Exponential and Logarithmic Functions

5.

log,8

=-

6.

logc,27=-

8.

36

RewTite in logarithmic form. 7.

9.

11.

= 72 4^/2 = 8 = A b" 49

12.

= 6^ 27^^^ = 9 = M b''

10.

Find each 0/ the following: 13.

logiolO^

14.

logiolO-^

15.

log2 2"3

16.

log3 3^

17.

log,o 1,000

18.

logeSe

Write in terms 0/ simpler logarithmic forms as in Example

19.

logb^

20.

log,FG

21.

logbL^

22.

logbW'^

23.

logb-^

24.

logbPQR

26.

log, x

28.

logj 27

7.

qrs

B

Find

X, y,

or

b.

29.

=2 =y logt,10"'' = -4

31.

log4X

=

33.

log,/3

9

35.

logb 1,000

25. 27.

log3X logj

49

30.

=2 =y logt,e-' = -2

32.

log,,x

34.

bg.g

36.

log54

^

=-

1

=y = ^

-=y

= ^

Write in terms of simpler logarithmic forms going as far as you can with logarithmic properties (see Example

4

7).

38.

logtx^y^

logbVN

40.

logfcVQ

41.

\ogt,x'V^

42.

log,

43.

log;,

44.

log,,

45.

logbP(l+r)'

46.

logeAe^"^'

47.

log,

48.

log,o (67

37.

log,

39.

(50



2"°

21)

100e-°°"

^ (100





1.06')

iQ-^'^x

Logarithmic Functions

13-2

Find

X.

49.

logb X

= -2

logb 8

50.

logb X

=-

logb 27

51.

logt,

X

=-

52.

logb

X

=

53.

logb

X

54.

logb(x

+ logb (X - 4) = + 2) + logbX =

55.

log,o(x-l)-log,o(x

56.

log,„(x

1

3

In

y

=

+

-

logb 9

2 logb 2

logb 4

- -2

logb 8

3 logb 2

+-

logb 25

+

to

logb 6

-

+

logb 3

2 logb 2

-

logb 20

logb24

+

l)

= =

l l

exponential form

log2(x-2)

first.

58.

Problems 59 and

789

logb 21

6)-log,„(x-3)

Graph by converting 57.

+-

— A Review

60, evaluate to five

y

=

log3(x

+

2)

decimal places using a

scientijic

calculator. 59.

60.

In 61.

62.

(A)

log 3,527.2

(B)

log 0.006 913 2

(C)

In 277.63

(D)

In 0.040

(A)

log 72.604

(B)

log 0.033 041

(C)

In 40,257

(D)

In 0.005

Problems 61 and

883

926

62, find x to four signijicant digits.

= 3. 128 5 = 8.776 3

= -2. 049 7 = -5.887 9

(A)

log x

(B)

log x

(C)

In x

(D)

In x

(A)

log x

2

(B)

log x

(C)

In

6

(D)

In

= 5.083 x = 10.133

63.

Find the logarithm of

64.

Why

65.

Write logm y

is 1

3

1 for

x

= -3.157 7 = -4. 328 1

any permissible base.

not a suitable logarithmic base? [Hint;



logjg c

=

Try

to find log, 8.]

0.8x in an exponential form that

is

free of

is

free of

logarithms. 66.

Write loge x



log^ 25

= 0.2t

in

an exponential form that

logarithms.

Applications Business & Economics

67.

Doubling time. double

if it is

How long (to the next whole year) will it take money to 6% interest compounded annually?

invested at

790

Exponential and Logarithmic Functions

68.

Doubling time. double

69.

if it is

How long (to the next whole year) will it take money to 3% interest compounded annually?

invested at

Tripling time. Write a formula similar to the doubling time formula in

Figure 4 for the tripling time of

money

invested

100i%

at

interest

compounded annually. 70.

Tripling time.

How long (to the

triple if invested at

Life Sciences

71.

Sound

intensity

), it

next whole year) will

interest

— decibels.

sensitivity of the 1

15%

human

ear

take

money

to

Because of the extraordinary range of (a

range of over 1,000 million millions

to

helpful to use a logarithmic scale, rather than an absolute scale,

is

measure sound intensity over

to

it

compounded annually?

this range.

The

unit of

measure

is

called the decibel, after the inventor of the telephone, Alexander

Graham

Bell. If

we

let

N be the number of decibels, I the power of the

sound

in question (in watts per square centimeter),

sound

just

and

!„

the power of

below the threshold of hearing (approximately

10""* watt

per square centimeter), then I

=

IolO~''>''

Show

that this formula

N= 72.

10 log

Sound Iq

=

can be written in the form

intensity

— decibels.

Use the formula

Problem 71 (with

in

10"'" watt/cm^) to find the decibel ratings of the following

sounds: (A)

Social Sciences

73.

Whisper: 10~" watt/cm^

(C)

Normal conversation: 3.16 X Heavy traffic: 10^" watt/cm^

(D)

Jet

(B)

10""'

watt/cm^

plane with afterburner: 10"' watt/cm^

World population. If the world population is now 4 billion (4 X lO**) people and if it continues to grow at 2% per year compounded annually,

how

long will

it

be before there

is

only

1

square yard of land

per person? (The earth contains approximately 1.68

X

lO'''

square

yards of land.) 74.



Archaeology carbon-14 dating. Cosmic-ray bombardment of the atmosphere produces neutrons, which in turn react with nitrogen to produce radioactive carbon-14. Radioactive carbon-14 enters all living tissues through carbon dioxide which is first absorbed by plants. As long as a plant or animal is alive, carbon-14 is maintained at a constant level in its tissues. Once dead, however, it ceases taking in carbon and the carbon-14 diminishes by radioactive decay according to the

equation

A = Aoe-°'''"

13-3

where

I

is

The Constant

e

if

10%

Find

present. [Hint:

The Constant

t

of the original

such that

and Continuous Compound The Constant

and Continuous Compound

Interest

791

time in years. Estimate the age of a skull uncovered in an

archaeological site

13-3

e

A=

amount

of carbon-14

is still

0.1 Aj,.]

Interest

e

Continuous Compound Interest

The Constant In the last

e

two sections we introduced the special irrational number e as a and logarithmic functions.

particularly suitable base for both exponential In this

and the following sections we

that e can be

approximated

sufficiently large.

we

why

like

this is so.

by

[1

Now we will use the limit concept

either of the following

The

will see

as closely as

two

limits:

+

We said earlier

(l/n)]"

by taking n

to formally define e as

792

Exponential and Logarithmic Functions

Compute some

of the table values with a calculator yourself

several values of s even closer to at s

0.

Note that the function

is

and

also try

discontinuous

= 0. who discovered e is still being debated. It is named after the great who computed e to twenty-

Exactly

mathematician Leonhard Euler (1707-1783), three decimal places using

[1

+ (l/n)]".

Continuous Compound Interest

Now we will see how e appears quite naturally in the important application compound interest. Let us start with simple interest, move on pound interest, and then to continuous compound interest.

of

If

A=P+

P is borrowed at an annual rate r, then after t years at simple borrower will owe the lender an amount A given by

Prt

=

P(l +rt)

the other hand,

borrower

A=P Suppose

A

com-

a principal

interest the

On

to

will

owe

if

r,

and

(

interest

is

Compound in

increase without

(2)

A

(2)

or will

it

is

increased. Will the

100

If

P

= $100, r =

0.06,

amount

tend to some limiting value?

Let us perform a calculator experiment before limit problem.

a year, then the

given by

interest

are held fixed and n

bound

(1)

compounded n times

the lender an amount

HT P,

Simple interest

and

t

=

we

2 years,

attack the general

then

(-^r

We compute A for several values of n in Table 1. The biggest gain appears in the

first step;

then the gains slow

down as n increases.

In fact,

it

appears that

A might be tending to something close to $11 2. 75 as n gets larger and larger. Now we turn back to the general problem for a moment. Keeping P, r, and t

fixed in equation

(2),

we compute

the following limit and observe an

13-3

The Constant

e

and Continuous Compound

Table 1

Compounding Frequency Annually

A=

100

(-^)"

Interest

793

794

Exponential and Logarithmic Functions

Problem 13

What amount is

invested at

(to

the nearest cent) will an account have after 5 years

8%

interest

if

$100

compounded annually? Semiannually? Contin-

uously?

Example 14

If

$100

is

amount SoJufion

We

invested at

in the

want

to

12%

compounded continuously, graph

the

time for a period of 10 years.

We construct

0«t«10

a table of values using a calculator or Table

graph the points from the

A

to

graph

A=100e°"'

1

interest

account relative

table,

and

I

join the points with a

Appendix B, smooth curve. of

13-3

Problem 15

Answers to Matched Problems

The Constant

How long will it take money compounded continuously? 13.

$146.93; $148.02; $149.18

14.

A=

1

5,000e°^'

A

e

and Continuous Compound

to triple

if it

is

invested at

Interest

12%

795

interest

796

Exponential and Logarithmic Functions

B

In

Problems 3-8 solve

for

t

or r to two decimal places.

42 =

3.

6. 8.

In

Problems 9 and 10 complete each table

hand

calculator.

e""^'

= 6°"' 3 = e""-

3

to five

decimal places using a

13-3

The Constant

e

and Continuous Compound

Interest

797

17.

How long will it take money to double if invested at compounded continuously? Doubling time. How long will it take money to double if invested at 5% interest compounded continuously? Doubling rate. At what rate compounded continuously must money

18.

be invested to double in 5 years? Doubling rate. At what rate compounded continuously must money

19.

be invested to double in 3 years? Doubling time. It is instructive to look

15.

Doubling lime.

25% 16.

interest

invested

at

at doubling times for money compounded continuously. Show 100r% interest compounded continuously is

various rates of interest

that doubling time

t

at

given by In 2

20.

Doubling time. Graph the doubling time equation from Problem 19

< Life Sciences

21

r


0

b^l,

b>0,

Iog(,x

for the derivative of

using the definition of the derivative

/'(x)=

f(x

hm

+ Ax)-/(x)



Ax

Ax—

and the two-step process discussed Step

1.

f (x

in Section 10-4.

Simplify the difference quotient

+ Ax) - f (x) _

+

logt,(x

Ax

-

Ax)

first.

logb X

Ax 1

^ =— Ax =

+

[Iogb(x

Ax)

-

logb x]

Property of logs

logk

X

Multiply by

4fe)l°^^(^+v) logb

X Step

2.

Find the

Let

=

Ax/x. For x

s

D^

logb X

AxN"/'^

/

1

=-

^^1.

1

Property of logs

"I

x /

\

limit. fixed,

f(x

Ax

if



»

0,

then



0.

Thus,

Let

s

= Ax/x.

s

>

+ Ax)-/(x)

=

lim

=

lim - logb 1 Ax— X \

Ax /

1

AxN"'''^ "I

X /

= lim - logb(l + sy/' s-*0

= —1

X

logb[Iim(l ^~'°

= —I logb e X

+

s)'/*]

Properties of limits and

continuity of log functions Definition of e

Derivatives of Logarithmic Functions

13-4

799

Thus,

D,

= -1 logb

X

logb

e

(1)

for one particuany permissible base b, then

This derivative formula takes on a particularly simple form

Which base? Since

lar base.

loge e

Thus,

=

logj,

for

1

=

X

loge

we have

=

D, In X

base

1

for the natural logarithmic function

In X

Now

=

b

D,

loge

X

11

= -1 loge e = -



1

=-

you see why we might want the complicated

— of

all

possible bases,

it

[2)

irrational

number e as

a

provides the simplest derivative formula for

logarithmic functions.

We

now

will

Recall that

y

see the

power

of the chain rule discussed in Section 10-7.

if

= /(u)

and

u

=

g(x)

then



dy

= -,dy

-r-

dx

du

Cham

;—

du dx

In particular,

y

= log(,

y

=

rule

if

and

u

u

=

u(x)

or In u

and

logb u

= - logb e

u

=

u(x)

then 1

D^

D, u

and D, In u

= — D„

u

u Let us

summarize these

and then conbox are used far more

results for convenient reference

sider several examples. Formulas

1

and

2 in the

frequently than the others; hence, they will be given more attention in the

examples and exercises that follow.

800

Exponential and Logarithmic Functions

Derivatives of Logarithmic Functions

For

b>Oand b#l:

1.

D,

1

lnx=X

2.

D^lnu=-D,

u

u 1

3.

D^

logb X

= - logb e

4.

D,

logb u

= - logb

1

Example 16

Solutions

e

D, u

Differentiate. (A)

D, In

(x2

(A)

ln(x^

+

y

=

+

1) is

D, In x"

(C)

a composite function of the

In u

Formula

D,(ln x)"

(B)

1)

=

u

u(x)

=

x^

+

form

1

2 applies: thus,

D,ln(x^

+

= ^^^D,(x2 +

l)

l)

2x x^ (B)

(In x)* is a

y

=

1

composite function of the form

uP

u

Hence, D, up D,(In x]"

+

=

=

=

u(x)

= In X

puP~' D,

u,

and

Power

4(ln x]^D^ In x

= 4(lnx)M-] _

4(ln

rule

Formula

2

xf

X (C)

We work this problem two ways. The second method takes particular advantage of logarithmic properties.

Method

Method

I.

II.

D Jn

D, In x"

4v3

4

x"

x

= — D^ x« = -— = 1

x"

x"

=

"

D^(4 In

x)

=

4D, In x

=-

Derivatives of Logarithmic Functions

13-4

Problem 16

Differentiate.

DJn{x^ +

(A)

Example 17

801

(B)

5]

D^llnx)"'

D, In x"

(C)

Find:

DAn^fxTl Solution

Using the chain rule directly results in a messy operation. (Try we first take advantage of logarithmic properties to write

(x

+

In x=

-

+

ln(x

1)'/^

=

5 In

X

-

(l/2)ln(x

+

it.)

Instead,

1)

1)'''^

Then, Dx In

,

=

,'',„,

-

Find D^ ln[(x

Example 18

Find D,[ln{2x2

Solution

Vx

1)^

-

In (x

+

1)

2(x+l)

X

Problem 17

- (1/2)D,

5D, In x

+

2].

[Hint:

Use logarithmic properties

first.]

x)]^

This problem involves two successive uses of the chain rule:

DJln(2x2

-

x)]^

=

'

3[ln(2x2

- x)YD,

3[ln(2x2

-

3[ln(2x2

_

x)f

ln(2x2

2x2

-x

-

[DA2x'

(4x

x)]2

x)

-

-

X)]

1)

2x2

3(4x-l)[ln(2x2-x)]2 2x2

Problem 18

Find D, Vln(l

-X

+ x^)

Graph Properties

of

y

=

In x

Using techniques discussed in Chapter 12. we can use the first and second derivatives of In x to give us useful information about the graph of y = In x. Using the derivative formulas given previously, we have y y

y"

=

In X

X

>

=—=X X

= -X"

-1

802

Exponential and Logarithmic Functions

We see that the first derivative is positive for all x in the domain of In x (all positive real numbers); hence, In is an increasing function for all x > 0. We also see that the second derivative

=

hence, the graph of y

In x

is

is

negative for

all

x in the

domain of In

concave dovkmward everywhere.

It

x;

can also

be shown that

=

lim In X

—°°

X— 0+ lim In X

= oo

Thus, the y axis is a vertical asymptote (there are no horizontal asymptotes) 0

Dj, ln|x|

= D,

_

1_

X

In X

Since

Ixl

=

x for x

>

Antiderivatives and Indefinite Integrals

14-1

Indefinite Integral

Formulas

1

6.

dx

e""

\

I I

8.

I

— = lnx + C dx — = +C

r

2.

D^

,

x>0 x^O

ln|x|

Formula

Note:

Case

a^O

+C

e"'

a

f dx

7.

-

==

821

7

is

a special case of formula

8.

x0

f

proportional to the

=

bacteria, etc.)

c •

Growth

money

present.

of

at

continuous

^t

compound interest •

Price -supply

curves •

Depletion of natural

resources

y

Exponential decay

'''

ce



Radioactive

decay

Rate of growth is proportional to the

=

k.

OO

y(0)

amount

=



Light absorption in

water

c •

present.

Price -demand

curves •

Atmospheric pressure

(t is

altitude]

Limited growth

dy

Rate of growth

dt

is

= k(M-y)

^c(l





l>0 y(0) =

proportional to

between the





a fixed limit.

Logistic

growth

Rate of growth

is

proportional to the

(e.g.,

Depreciation of

equipment

amount present and

Sales fads

skateboards)

k,

the difference

Learning

^ = ky{M -

the difference

between the amount present and a fixed amount.

M

y) 1

k,

-l-ce-



Learning



Long-term population

t>0

amount

present and to

Company growth

growth

M y(0)1

+c



Epidemics



Sales of

new

products •

Company growth

836

Integration

a subject that has

been extensively developed and which you are

likely to

encounter in greater depth in the future.

Answers to Matched Problems

7.

1.429 million people

8.

Approximately 22 years Approximately 5,600 years Approximately 0.27 foot per hour

9.

10.

Exercise 14-2

Applications Continuous compound

Business & Economics

f

years

dA

= 0.08 A

-3-

interest.

Find the amount

A in an account after

if

A (0) =

and

1,000

dt 2.

Continuous compound t

years

dA

=

-;-

interest.

Find the amount A in an account after

if

0.12A

and

A(0)

Continuous compound

interest.

=

5,250

at 3.

t

years

Find the amount A in an account after

if

dA

tA

'

A(0)

=

8,000

A(2)

=

9,020

dt 4.

Continuous compound t

years

dA

interest.

Find the amount

A in an account after

if

=

rA

A(0)

=

5,000

A(5)

=

7,460

dt 5.

Price -demand.

If the marginal price dp/dx at x units of demand per week is proportional to the price p, and if at $100 there is no weekly demand [p(0) = 100], and if at $77.88 there is a weekly demand of 5

units [p(5)

=

is

price-demand equation. dp/dx at x units of supply per day p, and if at a price of $10 there is no daily

77.88], find the

Price - supply.

If

the marginal price

proportional to the price

supply [p(0) = 50 units [p(50)

10],

=

and

if

at a price of

12.84], find the

$12.84 there

is

a daily supply of

price-supply equation.

Differential Equations

14-2

Advertising.

A company is

— Growth and Decay

trying to expose a

new

product

to as

837

many

people as possible through television advertising. Suppose the rate of

exposure

to

new

people

is

proportional to the

number

have not seen the product out of L possible viewers. of the product at the start of the

are

campaign and

If

of those

no one

after 10

days

is

who

aware

40%

of L

aware of the product, solve

dN

= klL-N]

N(0)

=

N{10)

=

0.4L

d(

N=

for t

8.

number of people who

N((), the

are

aware

of the product after

days of advertising.

Advertising. Repeat Problem 7 for

k{L-N]

N(0)

=

N(10)

=

0.1L

dt

Life Sciences

9.

Ecology. For relatively clear bodies of water, light intensity

according dl^

dx

= -kl

where

is

I

1(0)

reduced

= lo

the intensity of light at x feet below the surface. For the

Sargasso Sea off the West Indies, k find the

is

to

depth

at

which the

= 0.009

light is

42.

Find I in terms of x and

reduced

to half of that at the

surface. 10.

Blood pressure.

It

can be shown under certain assumptions that blood

pressure P in the largest artery in the

human body

between beats with respect

according to

dP

= -aP

P(0)

=

to

time

f

(the aorta)

changes

Po

df

where a 11.

is

a constant.

Drug concenlralions.

Find P

A

=

P{t) that satisfies

single injection of a drug

both conditions. is

administered

to a

The amount Q in the body then decreases at a rate proporto the amount present, and for this particular drug the rate is

patient.

tional

4%

per hour. Thus,

f =-""« where [Q(0)

12.

=

t

is

3],

Q{0)

= Qo

time in hours. find

Q=

If

the initial injection

is

Q(t) that satisfies both conditions.

3

milliliters

How many

milliliters of the drug are still in the body after 10 hours? Simple epidemic. A community of 1,000 individuals is assumed to be homogeneously mixed. One individual who has just returned from

another community has influenza. Assume the

home community

has

838

Integration

all are susceptible. One mathematical an influenza epidemic assumes that influenza tends to

not had influenza shots and

model

for

spread

at a rate in direct

to the

number who have

proportion to the

number who have it, N, and

not contracted

it,

in this case, 1,000

— N.

Mathematically,

dN

= kN(l,000-N)

N(0)

=

1

dt

where t

N is

the

=

days. For k

N[t]

number of people who have contracted influenza after 0.0004, it can be shown that N[t) is given by

1,000

= +

1

See Table

999e^

(logistic

1

growth) for the characteristic graph.

How many people have contracted influenza after 10 days? After

(A)

20 days?

How many

(B)

days will

it

take until half the

community has con-

tracted influenza?

Findlim,_.N(t).

(C)

Social Sciences

13.

Archaeology.

found

to

A skull

have

5%

from an ancient tomb was discovered and was

of the original

amount

14.

of radioactive carbon-14

Example

present. Estimate the age of the skull. (See

Learning. For a particular person learning to type, the

number of words per minute,

9.)

it

was found

that

N, the person was able to type after

t

hours of practice was given approximately by

N=

100(1 -e-°°")

See Table rate of

1

(limited growth) for a characteristic graph.

improvement

after 10

What

is

the

hours of practice? After 40 hours of

practice? 15.

Small group analysis. In a study on small group dynamics, sociologists

Stephan and Mischler found that, when the members of a discussion group of ten were ranked according to the number of times each participated, the number of times N(k) the kth-ranked person participated was given approximately by

Nlk]

= N,e

where N,

is

-0.11|k-l)

the

number

in the discussion.

N,

=

1

180, estimate

If,

«k «

10

of times the first-ranked person participated

in a particular discussion

how many

group of ten people,

times the sixth-ranked person partici-

The tenth-ranked person. One of the oldest laws Weber-Fechner law (discovered pated.

16.

Perception.

century).

It

concerns

a person's

in

mathematical psychology

in the

is

the

middle of the nineteenth

sensed perception of various strengths

14-3

General Power Rule

of stimulation involving weights, sound, light, shock, taste,

One form

of the

with respect

to

839

and so on.

law states that the rate of change of sensed sensation S stimulus R

is

inversely proportional to the strength of

the stimulus R. Thus,

dS^k R

dR where k

a constant. If

is

we

let

fij,

be the threshold level

stimulus R can be detected (the least amount of sound,

and so on that can be detected), then S{R„)

Rumor

which the weight,

appropriate to write

=

Find a function S in terms of R that 17.

it is

at

light,

spread.

A

waiting anxiously

satisiies

group of 400 parents,

Kennedy Airport

at

after a year in Europe.

It is

the above conditions.

relatives,

for a

and friends are

student charter to return

stormy and the plane

is late.

A

particular

parent thought he had heard that the plane's radio had gone out and

news to some friends, who in turn passed it on to others, and so on. Sociologists have studied rumor propagation and have related this

found that

a

rumor tends

to

spread

at a rate in direct

proportion to the

number who have heard it, x, and to the number who have not. P — x, where P is the total population. Mathematically, for our case, P = 400 and

dx

-p =

0.001 X (400

-x]

x(0]

=

l

df

where

(

is

time in minutes. From

this,

it

can be shown that

400 x{t] 1

See Table (A)

+ 1

3996-°'" (logistic

How many

growth) for a characteristic graph.

people have heard the rumor after 5 minutes? 20

minutes? (B)

14-3

Find lim,^.

General Power Rule Introduction

General Power Rule

Common Remarks

Errors

x(t).

840

Integration

Introduction power

Just as the general

=

D. u"

rule for differentiation

nu""' -r-

dx

significantly increases the variety of functions

we can

differentiate (see

Section 10-7), a corresponding power rule for integration will significantly increase the

number of functions we can integrate.

illustrations

and generalize from the experience.

Let us start with several

Since

1.

D,i^ = ^i^DJx^-l, = (x-ir2x then

- lY2x

X-

/

=-

dx

—-—^ + C

Since

2.

= (x-xT'(l -3x=) then

- xT'(l -

(x

/

Since, for u

3.



=

—+— = u"+'

D, '

;

n

1

dx

3x')

=

— x^l"'' +C _

fx '

'

u(x),

(n

-

+— llu" „ D,u = +1

du u"-rdx

—+— + ^C

n¥=-l

i

n

'

n^t-i

the r

du

,

u"-i-dx J

=

dx

u"+'

,

;

n

1

General Power Rule The

last illustration

This rule

is

establishes the general

the inverse of the

illustrate its use

power

with several examples.

power

rule for integration.

rule for differentiation.

We

will

General Power Rule

14-3

841

General Power Rule If

=

u

(A)

u'(x) exists, then for all real

=

U" -;-dx

/

Example 11

and

u(x)

dx

n

+

2x(x^

/

— 1)

+C

+

1

Note that

dx

5)"

numbers n(n #

=

du/dx

du dx

u

if

= x^ +

Write in

I

dx

and apply the power

D, -

+

(x-

r

(x=

+1 + x)^

dx

3X-

(B)

J

=

5)^

(x'

Note that

du/dx

dx

+

(x^ j

_

(x^

+

x)-2(3x'

X)-' ^

-1

= Check

Problem 11

D,[

-

Find:

(x^

-(x'

+

+

=

Write in

if

I

J

l)dx

u

3x-

x,

then

u"-r-dxforin dx

and apply the power

rule.

^

=

+ xP(3x^ + 3x' + l (x^ + xf

(x^

3x'(x^

/

-

1)-

dx

1)

(B)

J

(e^

+

1)^

du If

= x^ + + 1.

+ xP + C

x)-']

(A]

rule.

+ S^Zx

du

=

then

u"-;— dxform

J

Check

5,

2x.

an integrand

is

within a constant factor of u" -;—

integral to achieve this form.

dx

Example 12

,

we can

adjust the

illustrates the process.

842

Integration

Example 12

Integrate

(A)

x^-Jx^- 10 dx x^Vx^-lOdx

I

(B) ^

JI

Soiutions

\

'

J

^—— {x'-2x]

dx

Rewrite in power form:

(A)

/(X3If

10)'/2x2

= x' —

U

10,

dx

then du/dx

=

We

3x^.

are missing a factor of 3 in the

integrand to have the form

(We must have

this

form exactly in order

Recalling that a constant factor can be

we proceed

to

moved

apply the power

rule.)

across an integral sign,

as follows:

(x^-wy/^x''dx=

(x'- loy/^-x^dx

I

I

du dx

u"

=-

(x3-10)'/2(3x2)dx I

1 (x^

-

3

3/2

= - (x^ -

10)^/^

10)^/2

+C +C

9

Check

D,

-

(x^

-

10)^/^

=1 =

(B)

_

(x'

2

_

-

loy/^x^

^q^/^^^^z^

Rewrite in power form:

/ If

a

{x2-2xn(x-l)dx

2x — 2 = 2(x — 1). Again, we are within constant factor of having u" du/dx. We adjust the integrand as in u

=

x^

part A:



2x, then

du/dx

=

14-3

/

(x2-2xn(x-l)d> =

(x^J

General Power Rule

843

844

Integration

Problem 13

Solve the differential equation:

dx_ ~ dt m

7t^ (t^

+ 2f

Common

1.

Errors

2(x2-3p''2d> 2(x2-3p''2dx=

I

(x

I

-

2

3)3/^2

-

dx

X

I

dx

A

variable cannot be

moved

across an integral sign! This integral

requires techniques that are beyond the scope of this book.

3)-' 2x2

/i^-/'^'

^^

[dx No,

A

same reason

for the

as in illustration

constant factor can be

1.

moved back and

forth across

No

Yes

j

an integral

but a variable factor cannot.

sign,

kf(x)

dx

=

k

j

/(x)

lfUi^^?Fg^}^-=iW=p^x

dx

[/(x) a variable factor]

(k a constant factor)

Remarks we have touched on an

In this section

generalized in the next chapter. In

integration technique that will be

fact.

commonly used techniques of integration

Chapter 15 covers several other as well. However, even with that

chapter, our treatment will not be exhaustive.

Answers to Matched Problems

-

+C

11.

(A)

-

12.

(A)

i(3x + 5F^ + C

13.

x^

(x^

([3

12

1)^

+

2)-4

+c

(B)

(B)

--(e''

--

+

(x'

ir'

+C

+ 3xP + C

General Power Rule

14-3

845

Exercise 14-3 Find each indefinite integral and check by differentiating the

1.

I

(x^-4f2xdx

+

(x^

2.

result.

l)''3x2dx

I

V2x2

i.

-

1

4x dx

+ Sfxdx

xV3x2

+ 7dx

x2x/2x^

10.

+

l

dx

I

^'

r

J V2x^

^ dx

+

3

x'

r

12.

V4x^

J

(x-l)Vx2-2x-3dx

13.

3f dx

I

I

11.

(jx

(x'-S^x^dx

8.

I

9.

6x2

I

(x^

7.

+

(5x

6.

I

B

5

I

(3x-2)'dx

5.

+

V2x ^

4.

I

1

(x^-x)N/x^-2x2

14.

I

^ dx

-

+ 7dx

I

^ 17.

dx

,

J

(5-2x^)5

J

(e"

-

2x)^(e''

r Vl

+

In x J

19.

dx

18.

V4-x'

-

2)

dx

20.

(x^

I

-

e'')''(2x

-

e")

I

21.

r 23.

j

(x*

r (In

dx

x' + x + 2x2 +

22.

^dx

24.

ir

J

,

x^-i

r

,

dx

xf

^^

'Vx^

-

3x

- dx ,

+

7

Solve each di^erentiai equation.

25.

-^

= 7t2VF+5

26.

o^ 27.

dy

-^

=-

dt

2g

dp_ dx

^ = 10n(n2-8r dn

dt

dy

3t 28.

v't2-4 e^

+ e-"

(e'-e-'-j^

^

-

dx

dm _

^^ '

dt

5x2

(x^-7)* ln(t

- 5)

t-5

dx

846

Integration

Applications Business

& Economics

31.

Revenue function.

If

the marginal revenue in thousands of dollars of

producing x units

is

given by

fi'(x)

= x(x2 +

and no revenue

9)-'/2

results

from a zero production

function R(x). Find the revenue Life Sciences

Pollution.

32.

an

An

dR

is

60

+

Vi

dt

where R

is

new

is

t^o

the radius in feet of the circular slick after

college

and producing

losing oil

g

College enrollment.

33.

revenue

radiating outward at a rate given approximately by

the radius of the slick after 16 minutes Social Sciences

level, find the

production level of four units.

tanker aground on a reef

oil

that

oil slick

at a

is

dE 5,000(t

The projected

if

the radius

t

is

minutes. Find

when = t

0.

rate of increase in enrollment in a

estimated by

+

1]-^/^

t

^

dt

where

E(f) is the

when = t

14-4

0,

projected enrollment in

(

years.

If

enrollment

find the projected enrollment 15 years

is

2,000

from now.

Definite Integral Definite Integral

Properties

Applications

Definite Integral

We start new

this discussion

with a simple example, out of which will evolve

a

Our approach in this section these concepts will be made more precise in

integral form, called the dejinife integral.

will be intuitive

and informal;

Section 14-6.

Suppose a manufacturing company's marginal product C'(x)

is

cost equation for a given

given by

= 2-0.2x

0«x«8

where the marginal cost is in thousands of dollars and production is x units per day. What is the total change in cost per day going from a production

14-4

day

level of 2 units per

to 6 units

per day?

If

C=

Definite Integral

C(x]

is

847

the cost function,

then

(Total net change in cost \

=

between x

2

and x

=

6j

=

"

^(^'

^'2'

=

C(x)|t

(i)

The special symbol C(x)|^ is a convenient way of representing the center expression that will prove useful to us later. To evaluate (1), we need to find the antiderivative of C'(x). that is, C(x)

=

(2

-

0.2x)

dx

=

2x

-

+K

O.lx^

(2)

I

Thus,

we

are within a constant of

function. However,

problem

original

C(6)

-

C(2)

(1).

we do

We

knowing the

not need to

compute C(6)

know

-

original marginal cost

the constant

K

to solve the

C(2) for C(x) found in

= [2(6) - 0.1(6P + K] - [2(2) - 0.1(2)^ + K] = 12 - 3.6 + K - 4 + OA - K = $4,8 thousand per day increase in costs for

a

(2):

production

increase from 2 to 6 units per day

The unknown constant K canceled out! Thus, we conclude that any anti= 2 -0.2x will do, since antiderivatives of a given

derivative of C'(x)

function can differ by

do not have

at

most

a constant (see Section 14-1).

to find the original cost

Since C(x)

is

Thus,

we

really

function to solve the problem.

an antiderivative of C'(x), the above discussion suggests the

following notation:

C(6)-C(2)

The

integral

sents the

=

C(x)|^=

form on the

rC'(x)dx

(3)

right in (3) is called a dejinite integral

number found by evaluating an



it

repre-

antiderivative of the integrand at

6 and 2 and taking the difference as indicated.

Definite Integral

The X

=

definite integral of a continuous function /over an interval from

a to x

=

b

is

the net change of an antiderivative of / over the if F(x) is an antiderivative of /(x), then

interval. Symbolically,

/(x)dx

H'

Integrand:

= F(x)|„^ = F(b)-F(a) /(x)

Upper

limit:

where b

F'(x)=/(x)

Lower

limit:

a

848

Integration

In Section 14-6

we

will formally define a definite integral as a limit of a

special sum. Then the relationship in the box turns out to be the most the fundamental theorem of calculus. important theorem in calculus



Our

and the next section

intent in this

to give

is

experience with the definite integral concept and

and

better able to understand a formal definition

its

you some

use.

You

intuitive

will then be

to appreciate the signifi-

cance of the fundamental theorem.

Example 14 SoJution

Evaluate

iy-'-

2x) dx.

We choose the simplest antiderivative of (3x^ — any antiderivative

will

do (see discussion

at

2x),

namely

(x^



x^),

since

beginning of section).

(3x2-2x)dx = (x3-x2)|i,

i:

= — Problem 14

Evaluate

- 2^) - [(-1)' — 1—2] = 6

(2^ A

Be careful of

(-1)^]

^'S" errors here.

l)dx.

L'"

Remark Do

not confuse a definite integral with an indefinite integral.

definite integral /a/(x)

/ fix) dx

is

a

whole

dx

is

number; the

a real

set of functions



all

The

indefinite integral

the antiderivatives of /(x).

Properties In the next

You

box we

will note that

state several useful properties of the definite integral.

some

of these parallel the properties for the indefinite

integral listed in Section 14-1.

These properties are 1.

justified as follows:

/(x)dx

=

F(x)|2

= F(a)-F(a) =

dx

=

F(x)|S

=

If

F'(x) =/(x), then

J '

/(x)

2.

F(b)

- F(a) = -[F[a) -

F(b)]

=-

J

K/(x)dx

3.

= KF(x)|S = KF(b)-KF(a) = K[F(b)-F(a)]

I

=K and so on.

['/(xldx

f f(x] dx

14-4

Definite Integral

849

Definite Integral Properties

1.

=

f(x1dx

I

r/(x)dx = -

.

f/Wdx Jb

Ja

-2

=

.

Applications Business & Economics

29.

Consumers' and producers' surplus. Find the consumers' surplus and the producers' surplus for

30.

P

= D(x)

p

= S{x) =

-^

+

2

2

^

Consumers' and producers' surplus. Find the consumers' surplus and the producers' surplus for

31.

D[x]

50

^S(x)

X

-x^

+

2X

+

10

MarginaJ analysis. A company has a vending machine with the following marginal cost and revenue equations (in thousands of dollars per year): C'(t) R'(f)

where

= =

2

0«t «

12

C{t]

and

respectively,

10

2t

t

R(f) represent total

accumulated costs and revenues,

years after the machine

is

put into use.

The area

between the graphs of the marginal equations for the time period such that R'(t) ^ C'(t) represents the total accumulated profit for the useful

866

Integration

life

of the machine.

What

is

the useful

life

of the

machine and what

is

the total profit?

=

+

32.

Marginal analysis. Repeat Problem 31 R'(t) = 10-0.5t, 0St«20.

33.

Consumers' surplus. Supply and demand functions are given by

for

C'(t)

0.5f

2

and

= D(x) = 1006-°"^" p = S(x) = 10e''»5'
'2x

'*•

9

dx

-

1

dx

I

4(e«2'-l)dt

5.

7.

+

I

f 4 3.

x^Vx^

2.

I

6.

\-

dx

Find the equation of a function whose graph passes through the point 10)

(3,

/'(x)

and whose slope

= 6-2x

is

given by

14-7

8. 9.

883

Chapter Review

Find the area bounded by the graphs of y = x^ and y finite area bounded by the graphs of y = 1

Find the

= —

-Jx.

x^

and y

=

0.

0«xS2. 10.

Approximate /g (x^ — 4) dx using the rectangle rule with n = 3 and c^ the midpoint of the kth subinterval. (Calculate the approximation to three significant

11.

I(t)

12.

digits.) Also,

evaluate the integral exactly.

Suppose the inventory of a certain item year is given approximately by

= -2(4-36

months

after the first of the

Ost«12

What is the average inventory for the second quarter of the year? The instantaneous rate of change of production for a gold mine, thousands of ounces of gold per year, Q'(t)

where

= 40-4t Q(t)

produced during the 13.

t

is

after first

is

to

in

be given by

O^t^lO

the total quantity (in thousands of ounces) of gold t

years of operation.

2 years of operation?

do

Q(0)

=

10,000

How much

gold

During the next

Solve the differential equation;

-^ = 0.12Q

estimated

Q>0

is

produced

2 years?

i

»t

it

pi'

'" "

,

limi

^^i'-^

CHAPTER

15

Contents 15-1 Integration by Substitution

15-2 Integration by Parts 15-3 Integration Using Tables

15-4 Improper Integrals 15-5 Chapter Review

By now you should

realize that finding antiderivatives

process as finding derivatives. Indeed,

it is

is

not as routine a

not difficult to find functions

whose antiderivatives cannot be expressed in terms of the elementary functions we are familiar with. The classic example of this case is fix)



an important function

e'"',

methods of integration

certain

can integrate.

15-1

We

will

now

in statistics. Nevertheless, there are

that increase the

number

of functions

we

consider some of these methods.

Integration by Substitution Introduction Integration by Substitution Definite Integrals

Common

and Substitution

Errors

Introduction In Section 14-3

we saw

that

if

an integrand

is

of the

form

du

dx

where u

=

u(x)

a function of x,

is

then

we can

use the generalized power

rule to conclude that

du

u"-;-dx = ,

dx

/

—+— + C u"'*"'

n

n^t-i

1

For example,

du dx x-

/

+

l)''^2xdx

=

4- 1)3/2 fv2 ^ '

+C

Ifu

=

x2

du/dx

-

(x^ ^

+ IP''^

=

+

l,then

2x.

-I-

3

In this section

we

will see

how to use the relationship

the variable of integration from x to

886

u.

u

=

x-

-I-

1

to

change

This technique, called integration

by substitution,

number

large

is

a

very powerful tool that will enable us u

=

Substituting for u and du in / 2x(x'

+

= -r-

=

then the differential of u

u(x),

is

for

+

x^

l

=

is

2x dx

(x^

+

1)'-

-

we have

dx,

du

u'/^

I

if

dx

the differential

du

evaluate a

dx

Thus, u

to

of indefinite integrals.

Recall from Section 11-4 that

du

887

Integration by Substitution

15-1

l)''^2x

dx

=

'

u'

I

We

du

•'

•'

momentarily "forget" is a function of x and

that u

treat u as if

it

were the

variable of integration. u3/2 =— -,



h

Now we "remember" that we started with u = x^ + 1.

C

3/2

3

At

first

glance,

it

we

appears that

are actually

making things more compli-

cated but. as later examples will illustrate, making a substitution in order to

change the variable

problems.

in

The important

made, we can

an indefinite integral can greatly simplify many point is that, once the substitution has been

treat u as the variable of integration

We

the simplified integral directly.

will

now

and proceed

to evaluate

generalize this process of

substitution.

Integration by Substitution In general,

if

u

I

/[u(x)]

u

= u(x)

=

and du

(du/dx) dx, then

du

— dx=

I

/(u)

du

^

•^

Regarding u as the variable of integration,

we

try to find

an

antiderivative F(u) for/(u).

=

F(u)

+C

Now we

substitute u

complete the process.

=

F[u(x)]

+C

=

u(x) to

888

Additional Integration Topics

This statement D,{F[u(x)]

It is

+

convenient

is

easily verified

C}

by applying the chain rule

= F'[u(x)]^

to restate

F'

Basic Integration Formulas

+

C:

=f

some of the basic

u and du.

to F[u(x)]

integration formulas in terms of

15-1

I

(2x

+

l)(x2

+

X

+

/

(x^

I

+

X

dx

S)"

du

u^

=

Integration by Substitution

+

5)''{2x

+

1)

dx

Substitution:

u

du u''

=x' + x + 5 = (2x + 1) dx

Use formula

du

/ =— + C

(1).

Substitute:

5

u

^5

(x^

+

X

+

5)^

+C

= x^ + X +

5

Check by differentiating.

Check

dJ ^ (x^ + X + 5)M = (x^ + X + 5)''{2x +

(B)

-

-du

u

2

1)

889

890

Additional Integration Topics

e"

— du 3

(C)

/x^e^'dx =

/

e^'x^

dx

Substitution:

15-1

Example

2

Find each of the following indefinite integrals:

l^=d. ^4 +

(A)

l^^^d. X

(B)

6"

J

J

du

u->/^

Solutions

Integration by Substitution

(A)

=

dx

I ,

J

V4

(4

\

+ e^

+

e^P'^e" dx

Substitution:

J u

du

=

u"'''^

I

u 1/2 ,

=

+

4

= e^

e"'

dx

Use formula

du

+C

Substitute u

(1).

=

4

+

e\

1/2

= 2(4 + Check

D,[2(4

+

e'')"'^]

=

dx

(B)



-

(4

74

+

6"

2

=

X

J

I

e'')'''^

+C

+

eT'^'e"

u-

du

(In x)-

- dx X

J

Check by

differentiating.

Substitution:

u

du

= In

X

=—

dx

X

=

u^du

Use formula

(1).

Substitute u

=

I

u-" = -— +C

In x.

3

= -1

Check

D,

-

(In

x)M

=_



(In x)^

3(ln x)^

+C

Check by

-

(In x)^

X

Problem

2

Find each of the following indefinite integrals:

differentiating.

891

892

Additional Integration Topics

Example

3

Find each of the following indefinite

(A)

Solutions

/i^"" No obvious

(A)

u

=

X

+

h



dx

+

2

if we let we can inte-

substitution presents itself here. However,

may

the integrand

2,

integrals:

simplify to something that

grate. If

u

=X+ dx [

l^L x+2

2,

+

=

fidu J

2

dx and

= du-

J

I— x

=

then du

=u

=/

^u

To eliminate x

u

numerator, u

'

u-2

-

2 ln|u|

+

c

ln|x

Substitute u

+

2|

+

c

If

c

is

= x-2ln|x +

2[

1

-2

x

+

2



2|

+C

=x+

2.

an arbitrary

constant, so

D,(x-2ln|x +

solve

= X + 2 for x: u = x + 2 x = u -2

du

^x + 2-2

Check

in the

we

Check by

is

C=

c

+

2.

differentiating.

Integration by Substitution

15-1

893

Thus,

-x

=

dx J

+

Vx

2

= u2-22u du

'-Vx

"

J

t

-dx

= 2u

+

=

2

r

du-

u

4)du

(2u^

J

= -u3-4u + C

Substitute

3

u

-[x + 2f/^-4{x +

2y/^

+C

=

(x

+

2)V2.

Check by differentiating.

Check

°« [f''^

+

2)3/2

_

+

4(x

2)1.

^]

=

+ 2)'/2 -

2(x

(x

+2 + 2)"^

(x

(x

+ 2y/^

(x

x

+

2)-'''2

2

+

2)'/2

x

Problem

3

Find each of the following indefinite

(A)

2 rx + 2.

— — dx ;

B

,

Definite Integrals

Example 4 grals

Example 4

illustrates

by substitution.

Evaluate

I

dx

n/FTT

rx + 2

,„^

integrals:

dx

and Substitution

two

different

methods

for evaluating definite inte-

894

Additional Integration Topics

Soiut

Method

First find the indefinite integral:

1.

=

dx ,

Vx^

J

+

(x^

+

ir'-^x^ dx

Substitution:

J

l

U

du

= =

— du =

X^

+

1

3x2 ^^ x^

dx

3

= = /

u

-i/^

- du 3

1

u^ + C

3

1/2

3

Now

evaluate the definite integral. 2

dx

I VF+T

2 = -(x=' +

l)V2

+ l]V^--[(0P +

^-[(2r

l]V2

= 1(9^2 -|(ir _

^4

2

~

3

Method

3

Substitute directly in the definite integral, changing the limits

2.

of integration:

U

= (2P +

1-

r =

dx Jo

V^^HM^ u

=

u-'/^-du 3

Ji (0)3

+

Ifu

X

=

=

and X

1-

u

-Or^-gtlF' =2

2

_

4

3^3

=

x3

+

1,

then

implies u

9.

=

=

2 implies

1

15-1

Problem 4

5

Solution

895

by Substitution

Evaluate

f Example

Integration

fx^

+

dx l)

Find the area bounded by the graphs of y First

)

we

sketch a graph.

=

5/(5



x)

and y

=

0,

^

x

^

4.

896

Additional Integration Topics

Remember that only a constant can be moved across the integral sign. is now the variable of integration, it appears that x can be

Since u

considered a constant. This



the equation u

=

the integrand.

The

x

1.

not correct, since x and u are related by

is

You must

substitute for x wherever

correct procedure

j^^d.^j^-^du

=x —

u

it

occurs in

as follows:

is

= dx = x

u

+

1 l

du

|(u + 2 + l)du -u2 + 2u +

ln|u|

+C

2

= - (x 1

r=

dx

1)2

c^i

=

-

2(u

+ 2(x -

5)

1)

+ ln|x -

du

u

X

dx If a substitution is

also

made in a definite

must be changed. The new

The

+ Vx (u - 5)2 2(u - 5) 5

du

determined by the particular

correct procedure for this prob-

1

5

^dx=

+

V^

r» Je

1

-2(u-5)du

u

U

X X

=

= 5 + Vx = 1 implies = 9 implies

(2u-10ln|u|:

= (16 - 10 In 8) - (12 - 10 In 6) = 4 - 10 In 8 + 10 In 6 « 1.12 Answers to Matched Problems

+C

as follows:

is

r^ Ji

1

integral, the limits of integration

limits are

substitution used in the integral.

lem

= = =

1

+

(A)

(x^

(C)

-ln|8

(A)

-(5

2x

+

+

x^|

4)'

+C

(B)

-6"'+=

+C

+C

+ eT' + C

(B)

-[\nx]'/^

+C

u u

= =

6 8

Integration by Substitution

15-1

3.

(A)

4.

-

+

x

+ l|+C

ln|x

1

8 In 4

5.

=

(B)

|(x

+

1^==

+

2(x

+

1)'/^

+C

x^lx^

+

gj-'dx

11.1

5

Exercise 15-1

A

Find each indefinite integral.

1.

xlx^'

+ gpdx

2.

j

I

r ^-

1

J 4

+

(2x

5.

+

x

2x

+

+

f "*•

x^

J

l)e'"+''+'

dx

6.

I

x^-1 -dx x^-3x + 7 (x'2

+

2x]e''^^'"dx

I

Evaluate each definite integral.

7.

x'Vx^-l dx

I

8.

x^Vx^'

+

xe"'-'

dx

1

dx

I

— dx

9.

J-i

B

+

4x

10.

5

Jo

Find each indefinite integral. 11.

e^^tl

+

e^")'

dx

12.

I

^

I

(InxP

13. ,.

5.

r J JflBi^dx ^dx

..

14.

*

1

x(x

-

5)^

dx

^

dx

r ln(x ^— + 4)

— Jif^

16.

^ ^dx

(x I

19

.

f-^dx Vx +

20.

l^^dx x-2

22.

J 21-

I

J

^\ (x-2)

dx

V4-X

f-^i— dx

J (x-2)2

J 23.

f-^^dx J

3

24.

f-^^dx (X-2F

J

897

898

Additional Integration Topics

Find the area bounded by the graphs of the indicated equations.



8x

25.

y

26.

y

= ^-,

y

.

xM^4'

=

27.

= 4X6""', y = 0, y = x\/9^^, y =

28.

y

=

29.

y

=

30.

y=

In

J^. -

x\/2

X,

'^

-

,

,

VlO-x

y

o«x«4

o,

=

«

X

«

1

0,

0«xS3

0,

0^x^2 0«x«2

y

=

y

= 0, 6«x«9

0,

Problems 31-34, find each indefinite integral two ways: = Vx — 1 and then use the substitution u = x —

substitution u

31.

I

J 33.

I

, ,

Vx-l

dx

use the

dx

1

J

—^^ dx

34.

,

J

xVx-

32.

first 3.

Vx-1

I I

x2>/x~xVx

dx

1

J

Find each indefinite integral. 35.

r^-2^ -= dx J

-

J

Vx(l

+

^dx

J

Use the substitution u

j

is, if

f(x)

/(-x)

dx

Then show

=

I

f[x)

Use the substitution u [that

is,

= —x

to

J

x

+

J

— dx x In X

show

that

show

that

dx

=

/(x)

is

an odd function

= —x

0.

to

f /(x)dx= Jof /(u)du I

if

du

J-a

that

pdx 2n/x

;

if/(-x)=/(x)], then

Then show

dx

,

+ Vx-2

-f(x]], then

= - 1° /(u)

that

1

3

40.

x-^

[that

r J

38.

Vx)

—e-'/'dx

39.

42.

36.

1

—z

37.

41.

Vx

/(x)

dx

=

2

I

f[x)dx.

if ,f(x) is

an even function

15-1

Integration by Substitution

899

price p'(x) at x units per

week

Applications Business & Economics

43.

Price -demand equation.

The marginal

for a certain style of designer jeans is

-300,000x

=

p'(x)

(5,000

+

x^)^

demand

At a price of $30 each, the weekly

mand 44.

given by

is

100. Find the price-de-

equation.

Consumers' surplus. (Refer

to Section

14-5.)

Find the consumers'

surplus for

P ^ 45.

=

=

_,, D(x)

,

+ lOx 10 + x

400

5

p=S(x)=-X ^

;

'

'

^

2

Marginal analysis. The marginal cost and revenue equations

(in

thou-

sands of dollars per year) for a coin-operated photocopying machine are given by R'(t)

=

5te-"

C'(t)

=



*

'

11

where

time in years. The area between the graphs of the marginal

is

t

equations for the time period such that R'(t)

accumulated

total

46.

^

profit for the useful life of the

C'(f) represents

the

machine.

What is the useful Hfe of the machine? What is the total profit? Cash reser\'es. Suppose cash reserves (in thousands of dollars) are approximated by

=

C(x)

where x

+ xVl2-x

1

is

the

12

number of months after the first of the year. What is the first quarter? The fourth quarter?

average cash reserve for the Life Sciences

47.

Pollution.

A

contaminated lake

of decrease in harmful bacteria

dN _ ~

2,000t

"df"

where initial

1

N(f)

+ is



t

=5

is t

treated with a bactericide.

The

days after the treatment

given by

is

rate

10

t=^

the

number

of bacteria per milliliter of water.

count was 5,000 bacteria per

milliliter, find N(t)

If

and then

the find

the bacteria count after 10 days. 48.

Medicine.

One hour after x milligrams of a particular drug are given to

a person, the rate of T'(x),

with respect

to

change of temperature in degrees Fahrenheit, dosage x (called sensitivity) is given approxi-

mately by T'(x) ^

'

=

— x4q10

0^

900

Additional Integration Topics

What

total

change in temperature results from a dosage change from

to 5 milligrams?

Social Sciences

A

Learning.

49.

From

person learns

15t

where t number

15-2

is

+

the

N items

0« t«

N'[t]

Vl

8 to 9 milligrams? at a rate

given approximately by

10

t

number

of hours of continuous study. Find the total

of items learned from

t

=

to

t

=

8 hours of study.

Integration by Parts

we

In Section 14-1

/

In X

later,

said that

we would

return to the indefinite integral

dx

since none of the integration techniques considered up to that time

could be used to find an antiderivative

for In x.

We will now develop a very

useful technique, called integration by parts, that will not only enable us to

above

find the

integral,

but also

and

X In X dx

many

others, including integrals

such

as

dx

I

I

The

integration by parts technique

is

also

used

to derive

many

integration

formulas that are tabulated in mathematical handbooks.

The method derivatives. If/

of integration

DJ/(x)g(x)]=/(x)g'(x)

which can be written

by parts

is

based on the product formula

for

differentiable functions, then

and g are

+ g(x)nx)

in the equivalent

form

/(x)g'(x)=D,[/(x)g(x)]-g(x)nx) Integrating both sides,

/(x)g'(x)

J

The

dx

=

we

obtain

DJ/(x)g(x)]

J first

dx- j

g{x]f'(x]

integral to the right of the equal sign

/(x)g'(x)

J

dx =/(x)g(x)

-

g(x)nx) dx J

/(x)g(x)

+

C.

(Why?)

We

now, since we can add it after of the equal sign. So we have

will leave out the constant of integration for

integrating the second integral to the right

is

dx

15-2

Integration by Parts

901

form can be transformed into a more convenient form by letting and v = g(x); then du = /'(x) dx and dv = g '(x) dx. Making these substitutions, we obtain the integration by parts formula:

This u

last

= /(x)

This formula can be very useful

when

integrate using standard formulas.

the integral on the right side left.

Example

6

Solution

If

may be

the integral on the

left is difficult to

u and dv are chosen with care, then

easier to integrate than the one

on the

Several examples will demonstrate the use of the formula.

>

Find / X In X dx, x

using integration by parts.

write the integration by parts formula

First,

u dv

I

0,

=

uv



I

V du

Then

try to identify u

when

/ u dv

is

and dv

in / x In x

written in the form uv



dx

(this is

/ v du, the

the key step) so that

new

integral will be

easier to integrate.

Suppose we choose u

=

and

X

dv

= In x

dx

Then du

=

dx

V

=

?

We do not know an antiderivative of In x yet, so we change our choice for u and dv u

=

to

In X

902

Additional Integration Topics

Using the chosen

u,

du, dv,

and v in the integration by parts formula,

we

obtain

=

dv

u

I

u

~

^

V

du

I

|(lnx)xdx =

(lnx)(f)-|(f)ldx

= — In X —

I

2

J

— dx

This

new

integral

2 to integrate.

— Inx- —4 + C 2

To check

this result,

(x^ D„( Inx

— is

l-C| 4

Find / x In 2x dx.

Example

Find / xe" dx.

Solution

We

that

= xlnx

/

the integrand in the original integral.

Problem 6

7

\

x^

\2

which

show

write the integration by parts formula

u dv

=

uv



dv

=

V du

I

J

and choose u

=

e"

X dx

Then du

=

e"

dx

V

=— 2

and u dv

=

u

V



I

V

du

I

je^^dx = e''{^)-j{^)e^d x^

1

f

x^e" dx 2

2 J

This is

new

integral

more complicated

than the original one.

is

easy

15-2

Integration by Parts

This time the integration by parts formula leads to a

more complicated than the one we there

our

an error

is

first

in the formula.

It

mean

simply means that

solve. Thus,

we must make

a different selection.

Suppose we

choose u

=X

=

dv

e'

dx

Then

=

du

dx

V

=

e"

and u dv

I

=



uV

V du I

This integral

is

one we can evaluate.

= xe" — Problem

7

e"

+C

Find / xe^" dx.

Integration by Parts: Selection of u and 1.

It

must be possible

to integrate

dv

dv (preferably by using standard

formulas or simple substitutions). 2.

The new

should be simpler than the original

integral, J v du,

integral, / u dv. 3.

For integrals involving xP[ln u

4.

8

Solution

=

(In

dv

x)""

=

= xP

dv

= e"*

x)"),

try

xP dx

For integrals involving xp u

Example

e"", try

dx

Find / x^e"" dx. Following suggestion 4 in the box, u

= x^

dv

=

e""

dx

Then du

is

that

choice for u and dv did not change the original problem into one

we can

that

integral that

started with. This does not

our calculations or

in

new

903

= 2xdx

v

= — e""

we choose

904

Additional Integration Topics

and

=

dv

u

I

I'x^e-" 'x^e-''dx dx



I

= x^-e-")x^f-e"")-

I (

(-e-'')2xdx

I

xe"" dx

— The new

V

u

+

x^e~''

2

/

not one

du

V

|

we can

(1)

evaluate by standard formulas, but

it is

simpler than the original integral. Applying the integration by parts

for-

mula

we

to

integral

produce an even simpler

will

it

is

integral.

For the integral / xe"' dx,

choose

=

u

=

dv

X

dx

e""

Then du

=

dx

u

dv

= — e""

V

and

I

xe"" dx

I

=

u

=

x(— e"")

V

= — xe

"



I



I

+

(— e"") dx

e""

j

du

V

dx

= -xe-''-e-'' Substituting

/

(2)

x^e"" dx

=

Problem 8

Find / x^e^" dx.

Example

Find

9

Solution

/f In

-x^e""

+

2(-xe"''

— x^e



2X6-"

In X

=— X

"

box (with p dv

=

dx

Then du

and adding

-

a constant of integration,

-

e"")

26-"

we have

+C

+C

x dx.

tion 3 in the

=

(1)

/ In x dx; then return to the definite integral. Following sugges-

First, find

u

into

(2)

dx

V

=

X

=

0),

we choose

Integration by Parts

15-2

905

Hence,

I

=

dx

In X

(In x)(x)



(x)

— dx

J

= xlnx — x + C Thus,

r

=

dx

In x

(x In



X

x)

= (e In e - e) - (1 Inl = {e - e) - (0 - 1) =1 Problem 9

Answers to Matched Problems

Find

6.

/? in

3x dx.

— ln2x

hC

7.

4

2

X^ — 2

1)

e^"

--

X

1

2

4

9.

e^''

+C

4

2

- - e'" + - e^" + C

e'"

-

2ln6-ln 3-

1

=

1.4849

Exercise 15-2 Integrate using integration by parts.

function

/ I

B

is

Assume x >

whenever the natural log

involved.

xe'"

dx

x^ In X

dx

dx

I

xe''"

I

x^ In X

dx



Problems 5-18 are mixed some require integration by parts and others can be solved using techniques we have considered earlier. Integrate as indicated, assuming x

xe

*

xe"'

fin

whenever the natural log /unction

dx

6

dx

8

3)6"

„,

>

2x dx

dx

10

12

lie-"

II

.

I

xe""'

(x

+

dx

dx

5)e"

7x dx

dx

is

involved.

906

Additional Integration Topics

2x

r J

x^

+

r In X J

16.

X

Vx In X dx

17.

J

,

dx

15.

x^

f 1

r J

18.

I

x'

+

5

e''

,

-r-rrdx e"

+

1

^p^ dx

I

Some of these problems may require using the integration by parts formula more than once. Assume x > whenever the natural log /unction is involved.

19

21

23

25

.

1

x^e'dx

15-3

Life Sciences

37.

Pollution. (

The concentration of particulate matter

hours after

a factory ceases operation for the

+

20ln(t Clt) (t

+

in parts per million

day

is

given by

1)

If

Find the average concentration 38.

907

Integration Using Tables

for the

Medicine. After a person takes a

pill,

assimilated into the bloodstream.

time period from

t

=

to

the drug contained in the

The

rate of assimilation

t

t

=

5.

pill is

minutes

after taking the pill is

=

te-°2'

Find the

total

R(()

amount

stream during the Social Sciences

39.

first

The number

Politics.

of the

drug that

is

assimilated into the blood-

10 minutes after the

pill is

taken.

of voters (in thousands) in a certain city

is

given

by N(t)

where

==20+ t

is

41

-5te-°"

the time in years. Find the average

the time period from

15-3

t

=

to

t

=

number of voters during

5.

Integration Using Tables Introduction

Using a Table of Integrals Substitution and Integral Tables

Application

Introduction

A

table of integrals

is

a

list

of integration formulas that can be used to

who must evaluate may contain hundreds

evaluate definite integrals. Individuals

complicated

integrals often refer to a table that

of formulas.

Tables of this type can be found in mathematical handbooks; a short table illustrating the types of formulas

found

in

more extensive

tables

is

located

inside the back cover of this book. These formulas have been derived by

techniques

we have

not considered; however,

formula by differentiating the right

You may notice some for this table. 1.

We

it is

possible to verify each

side.

logical gaps in the

There are two reasons

list

of formulas

we have selected

for this:

have not included formulas for integrals that can be evaluated by the techniques we have already discussed. Thus, you will find formulas for / VuM-a^ du and J u^vuM-o^ du, but not for

908

Additional Integration Topics

/ uVu^

+

a^ du, since this last integral can be evaluated

by making a

simple substitution.

Many antiderivatives

we have

involve functions

example, a formula for / Va^

for



u^ du

is

not considered. Thus,

not included in the table

u^ involves an inverse trigonomet-

because the antiderivative of Va^ ric function.

Even though our

many new

table

is

not very large,

indefinite integrals.

We

will

it

now

will

still

permit us

to

evaluate

consider some examples that

will illustrate the use of a table of integrals.

Using Example 10

Table of Integrals

a

Use the Table

of Integrals inside the

^

f (2x

J

Solution

+

5)(3x

back cover

to find

dx

+ 4)

Since the integrand X /(x)

(2x is

+

+

5)(3x

4)

we examine

a rational function,

formulas

1

- 7 to

grand in formula

2

with

/,

we

conclude that

if any of the Comparing the inte-

determine

integrands in these formulas has the same form as

/.

formula can be used

this

to

evaluate / /(x) dx. Letting u = x and identifying the appropriate values for a, b, c, d, and A = be — ad, we have

= A= a

I

J r

=5 c = 3 d =4 = = 7 ad (5)(3) (2)(4) b

2

be -

7

(au



X

—— ;

r

(2x+t-^ttt; 5)(3x+ ;+4) 4)

7:;

J

du

TTIT +rzz b)(eu + d)

be

dx

= U- \n\au + A\a 1/5

=- -

d



Problem 10

- - ln|3x + + 5| 5|--ln|3x

+

ln|3x

5| '

'

still

I

+

5)2(x

+

dx l)

d|

C

21 is

include

Formula

)

\ 4| 4|)

7

'

2

+C

+ 4| + C '

'

not included in any of the

C

in

Use the Table of Integrals inside the back cover f J (3x

+

/

3

ln|2x

You must

ln|cu

e

4

ln|2x |2x

Notice that the constant of integration table.

--

7V2 14

formulas in the

b|

your antiderivatives.

to find

15-3

Example 11

909

Evaluate

xV25

J3

Solution

Integration Using Tables

First

we

-

x^

will use the Table of Integrals to find

f—^dx -

J

x^l25

x^

Since the integrand involves the expression V25

mulas 8-10 and

select formula 8

f-^=du = -iln

5

^ dx = J

xV25

x2

a

+

with a^ n/q^



=

+ V25-x=

In

Thus.

J3

x\/25

-X

dx 2

5

=

In 5 In

25 and a

u^

5

+ V25-x2



Formula

+C

we examine

x^,

=

5.

8

for-

910

Additional Integration Topics

SoJufion

In order to relate this integral to

we

(formulas 19-24),

u2=16x2 Thus,

we

and

observe that

-

VlGx^

=

25

one of the formulas involving Vu^ if u = 4x, then

-

Vu^

=

will use the substitution u



a^

25

4x

to

change

this integral into

one

that appears in the table.

Substitution:

-

Viex^

J

25

4x

Vu2-25

J

4

= ^f-^;=dt

X

=—

u

4

n/u2-25

64 J

dx

=—

du

4

This

can be evaluated by using formula 23 with a

last integral

=-

du

/

^lu^-25

(uVu^

— Q' +

a^ ln|u

+

-Ju^



=

5:

Formula 23

a^|)

2

Thus,

dx

/

V16x2-25

=



Vu2-25

1

(un/u^

128 1

Use formula 23 with

nil I

64 J

a

-25 +

+

25 ln|u

(4xV16x2-25 + 25

Vu^

ln|4x

+

-

25|)

+C

^116x^-25]]

128

Problem 12

Find/V9x2-16dx.

Example 13

Find

/ VxMM Solution

None u

1

=

r

1

Vx-i

Vu^

+

\ d\

+

l

4x.

if

we

dx

1

which does appear

J

=

+C

then

+

5.

Substitute u

of the formulas in the table involve fourth powers;

= x^, Vx"

=

=

in formulas 11-18. r

1

1



I

J

Vu^

+

1

dii

u

^

"^o

2 J

Subsstitution:

2

/

Vu^,

+

^"

= —

A x^

du

= 2xdx

du

=

,

1

X dx

however,

let

Integration Using Tables

15-3

We

recognize the

=

du

+

Vu^

formula 15 with a

last integral as

+

ln|u

+

Vu-

=

911

1:

Formula 15

a'l

a-

Thus,

= if-^=du f-^dx Vu^ + +

J

-Jx'

2 J

l

+

ln|u

= -1 Problem 13

Find /xVx"

Example 14

Evaluate ^'

+

Vu^

+

l|

+C

=

Substitute u

1.

x^.

+ VF + Tl + C

ln|x^

vox+b

Since none of the formulas in the table involve

would contain formulas

table

=

dx

1 Solution

a

dx.

1

+4

Vx

Use formula 15 with

1

of this type),

we

first

(a

make

more extensive

a substitution to

eliminate the square root: 21

+

Vx

4

dx

I 2u du

Substitution:

Limits:

= Vx + 4 = X u^ — 4 dx = 2u du

=5 = x 21 x

u

p (U

^Ja

Use formula

J

+

(au

=—

u

ac

3

ul_ + 2KU

with a

+

b)(cu

—A

r

I

=

du 2)

=

b

1,

c

2,

=

1,

d

= — 2,

and A

implies u

== 3

implies u =

= 4:

"

d)

~r Inlau

+

Inlcu

bl

\a^

+

dl

c^

)

Formula

3

)

Thus,

pi Vx I

Js •'^

+

4

dx

=

X

r^ 2

—+—" u-2 u^

I

J3 •"

2

^

Use formula

du

;

u

„ = b = 2, = A 4.

'

1,

= 2lu-^Yl"l" + 2|-yln|u-2|j =

2(5

=4+

-

ln|7|

2 In

+

ln|3|)

-

— = 5.5243 7

3

with a

=

d

= -2,

and

,

2(3

-

ln|5|

+

ln|l|)

1, *

5

912

Additional Integration Topics

Problem 14

Evaluate

dx

I

xVx

7

+

9

Application

One

of the

growth laws discussed

was referred to as logistic is assumed to be between y and a fixed upper

in Section 14-2

growth. In this situation, the rate of growth of a quantity y proportional both to y and to the difference limit

dy -r-

M. Hence, y must

= ky(M —

satisfy the differential

we can now

Using the Table of Integrals,

Example 15 Growth

growth equation

Logistic

y)

equation

solve this differential equation.

Solve the differential equation

Logistic

^ = ky(M Solution

-J-

y(0)

y]

= ky(M —

=

1

Multiply both sides of

y)

this

dy

y(M -

kdt



y)].

Integrate both sides of

y)

this equation.

dy

kdt

y(M-y)

J

equation by dt/[y[M

J

=

kt

+C

(1)

To evaluate dy f J

we

y(M-y)

use formula

1

with a

1

/

;au

A

dy

Substituting

(2)

y(M into

b

1,

_1,

,

+ b)(cu + d)

/

=

(1)

0,

cu

au

=_1 y)

=

M

1„

c

=

-1, d

+d +b

=

M, and A

Formula

= — M.

1

M-y

and simplifying yields

(2)

15-3

y

Integration Using Tables

913

914

Additional Integration Topics

3

I

+

3)-(2x

J

(x

J

x\/x-

+

dx

4

5)

cix

5.

+

+

(2x

J

x-Vx^- 16

+

x^

r

8.

X

:dx 2)

dx

J

.

dx ,

J

5)-(x

6.

4

fvi^^^dx

7.

I

J

+

n/x=

64

Evaluate each definite integral. Use the Table of IntegraJs

to find the anti-

derivative.

'•

11.

I

3)(x

+

"

l)'*^

dx

I

+

Vx-

Jo

B

+

(x

12.

9

1 I

+

(x

3)1x+l)'^^

-

n/x-

16 dx

J4

Use substitution techniques and the Table oflnfegrals

to find

each indejinite

integraJ.

+

r V4x^ 13.

^

I

r

dx

14.

dx

16.

^ 15.

I

J

-

,

Vx"

dx

r

+

16

,

dx

21.

20.

J dx

J

xV4-x''

J

x^Vx+l

23.

VxS

+

4

1

r

J dy

J 1

r

dx

xVx+16 ^^/FT4

X

J

16 dx

1

dx J

Vx

xVx"-

18.

J 19.

-

J

+4

xVx'^

17.

16

x-VQx-

,

X

J

dx

,

dx

22.

^

24.

J

dx

x(l +n/x)=



Problems 25-32 are mixed some require the use o/the Table 0/ Integrals and others can be solved using techniques we have considered earlier. xVx^

25.

-

9 dx

26.

dx

28.

I

x-Vx-

-

9

dx

I

71

27' J, 29.

7^1 (x^-l)^

fx+l,dx

^ + 2x J fx + l,dx ,^ J X- + 3x ,

x=

31.

J,

30.

,

.,

J-A-T^dx (x--ip

fx+l, ——— dx

+X X- + 1 ———-dx x- + 3x

J Xr 32.

J

,

915

Integration Using Tables

15-3

Applications Business

& Economics

33.

Consumers' surplus. (Refer

Find the consumers'

to Section 14-5.)

surplus for

p

360

=

D(x)

+

(x

p

34.

+

2)[x

l)

5x

=

S(x)

+

x

2

Marginal analysis. The marginal cost and revenue equations (in thousands of dollars per year) for a vending machine are given by 25t

=

R'(t)

+

(t+l)(f

= -f

C'(t) ^

'

where

6)

2

time in years. The area between the graphs of the marginal

is

f

equations for the time period such that total

accumulated

What Life Sciences

35.

the useful

is

An

Pollution.

an

oil

_

life

of the

100

where R

and producing

losing oil

the radius (in feet) of the circular slick after

is

Simple epidemic.

munity people

If

t

+

the

number

study.

is

t

minutes. Find

when =

is

t

0.

spreading through a com-

number of number who have not been

initially

and 100 were infected

will be infected after 20 days?

learns

60 Vt2

is

infected and to the

how many

A person

=

t

the radius

influenza epidemic

one individual was infected

later,

Learning.

N'(f)

if

of 1,000 people at a rate proportional both to the

infected.

where

An

who have been

10 days

total

the total profit?

2=0

the radius of the slick after 4 minutes

37.

machine.

is

radiating outward at a rate given approximately by

t

Social Sciences

is

C'(f) represents the

Vt'-l-g

df

36.

^

machine? What

tanker aground on a reef

oil slick that is

dR

fl'(t)

profit for the useful life of the

N items

at a rate

given approximately by

^0

25

number

of hours of continuous study.

of items learned in the

first

Determine the

12 hours of continuous

916

Additional Integration Topics

15-4

Improper Integrals Improper Integrals Probability Density Functions

Improper Integrals

We are now going to consider an integral form that has wide application in probability studies as well as other areas. Earlier,

when we

introduced the

idea of a definite integral,

dx

fix)

(1)

/:

we

required / to be continuous over a closed interval

going to extend the meaning of

(1)

[a, b].

so that the interval [a. h]

Now we are may become

infinite in length.

Let us investigate a particular example that will motivate several general definitions.

What would be

a reasonable interpretation for the following

expression?

r-dx x^

J,

Sketching

>

fixed b

X

=

1,

a 1,

and X

graph of /(x) /f/(x) dx

=

is

=

1/x^, x

^

1

(see Figure

1),

we

note that for any

=

1/x^, the x axis,

the area between the curve y

b.

->x

Figure

1

Let us see what happens

when we

let

b

—» oo;

that

is,

when we compute

the following limit: f'

lim

dx = ^

r

lim

l''1

(-X-')

Did you expect this result? No matter how large b is taken, the area under the curve from x = 1 to x = b never exceeds 1, and in the limit it is 1. This

15-4

suggests that

we

This integral

is

an example of an improper

f(x]dx

f{x]dx

j

is

/

917

write

J

where

Improper Integrals

integral. In general, the

forms

/(x)dx j

continuous over the indicated interval, are called improper improper integrals that will not be

integrals. (There are also other types of

considered here. These involve certain types of points of discontinuity within the interval of integration.) Each type of improper integral above is formally defined in the box:

Improper Integrals If

/

continuous over the indicated interval and the hmit

is

exists,

then:

1

/(x)dx

.

=

lim

f(x]

I

I

/(x)dx= Hm /(x)dx=

.

If

("fix) dx

f[x]dx+

j

J

dx

where

c is

on the

right exist.

any point on (-»,

/(x)d> j

oo],

provided both improper integrals

the indicated limit exists, then the improper integral is said to exist or if the limit does not exist, then the improper integral is said not to

converge;

exist or to diverge (and

Example 16 Solution

Evaluate /2 dx/x

if it

no value

is

assigned to

it).

converges.

r^=iimp^ X b-« X

Jl

J2

=

lim b— «o

(In x)

=

lim

(In b

-

as b

^

Since In b -*

00

integral diverges.

In 2)

00,

the limit does not exist. Hence, the improper

918

Additional Integration Topics

Problem 16

Evaluate /* dx/(x

Example 17

Evaluate Ji« e" dx

Solution

e"

dx

=



1)^ if it

if it

lim

I

converges.

converges.

e"

dx

J

=

lim

(e^ll)

a—*— 00

=

lim (e^-e")

Problem 17

Evaluate /li x~^ dx

Example 18

Evaluate

2x

j^ if it

(1

+ x^f

if it

=6^-0 =

62

The

integral converges.

converges.

dx

converges.

Solution

"(1

+xT'

Improper Integrals

15-4

919

At some point in time, the monthly production rate will become so low that it

no longer be economically feasible

will

to operate the well.

the purpose of estimating the total production, that the well

duced

is

it is

operated indefinitely. Thus, the

However,

convenient

total

amount

to

for

assume

of oil pro-

is

R(f)dt

=

lim

R(t)df

I

T-- Jo

Jo

lim

T— lim

T— ^

i

_

(50g-o.o5,

soe-o ") dt

Jo

(-l,000e-'"'5'-|-500e-°")

oo

lim (-1,0006-°°=^

T—

+

500e-°'T

+

500)

500 thousand barrels

Problem 19

Find the rate (in R(t)

total

amount

of oil

produced by

thousands of barrels)

=

a well

whose monthly production

given by

is

100e-°"-25e-°^'

Probability Density Functions

We

will

now

take a brief look

Hopefully,

when you

at

the use of improper integrals relative to

The approach

probability density functions.

will

be intuitive and informal.

next encounter these concepts in a more formal

you will have a better idea how calculus enters into the subject. Suppose an experiment is designed in such a way that any real number x on the interval [a, b] is a possible outcome. For example, x may represent setting,

an IQ score, the height of

a

person in inches, or the

life

of a light bulb in

hours. In certain situations it is possible to find a function / with x as an independent variable that can be used to determine the probability that x will assume a value on a given subinterval of (—='^

fy(2,l)

Using the chain rule [thinking of

(A)

we

x (part A) to obtain

6xy

4(2)-6(2)(3)

For / in Example (A)

Example

3)

+

z

= e",

u

= u(x);

y

is

held constant],

obtain

dx

dx

= 2X6"'+'" (B)

f,(x, y) fy{2, 1)

Problem

7

For (A)

Example

8

Profit

z

= /(x,

^ dy

2ye''^^'"

= =

2e=

y)

=

(B)

2(l)e^^''

(x^

2xy)=, find:

for the

surfboard company in Example 3 in Section 16-1

= 140x + 200y - 4x^ + Ixy -

Find Px(15, P,(x, y)

P,(15, 10)

PJ30,

+

/,(1,0)

The profit function was P(x, y)

Solution

=

10)

10)

= = =

and Px(30, 140 140 140

-

8x

+

8(15) 8(30)

10),

and

12y2

-

700

interpret.

2y

+ +

2(10) 2(10)

= 40 = -80

At a production level of 15 standard and 10 competition boards per week, increasing the production of standard boards by one and holding the production of competition boards fixed at 10 will increase profit by approximately $40. At a production level of 30 standard and 10 competition

boards per week, increasing the production of standard boards by one unit

16-2

Partial Derivatives

and holding the production of competition boards fixed profit by approximately $80.

Problem 8

For the

profit function in

Example

8,

find P,.(25. 10)

at

and

945

10 will decrease

P,,(25, 15),

and

interpret.

have simple geometric interpretations, as indicated in Figure 5. = a, then /y(a, y) is the slope of the curve obtained by intersecting the plane x = a with the surface z = f (x, y). A similar Partials

If

we

hold X fixed, say x

interpretation

given

is

to f^(x, b).

Surface z

=

f[x.

y

Slope of tangent line =

/„(a, b)

Curve z

=

/(a, y)

Curve

=

z

Slope of tangent line =

/(x. b)

/,,(a,

b)

Figure S

Higher-Order Partial Derivatives Just as there are higher-order ordinary derivatives, there are higher-order partials,

and we

discuss local

will be using some of these in Section 16-4 when we maxima and minima. The following second-order partials will

be useful:

Second-Order Partials If

z

= /(x,

y).

then

=

dx^~ dx \dx) dxdy ~ dx \dy)

/xx(X,

y)=/x;

" =

fyxi^, y)

= /vr

dy dx

dy'

~ dy

j=/yy(x. y)=/yy \dy}

946

Multivariable Calculus

d^z

mixed

In the

partial

d^z for

Example 9

For z

(A)

= f^y?

= /yx(x,

= /(x,

Sx^

-

(A)

= /(x,

is

we

y)

and

first

differentiate

the order of differentiation

for the functions

we

will consider,

find

(C)

/,,(2,1)

dx'

with respect

First differentiate

dz

dz — = 6x dx

to

y and then with respect

to

x and then the with respect to

to x:

d'z

= -6xy2

dx dy

with respect ct

First differentiate

a^z

=

2y3

dy dx

A ldz\ (^\ = d

d

(6x

dy \dx/

-

2y^)

y:

= - 6y2

I

Differentiate with respect to x twice

^z

=

- 2y3

6x

dx (C)

1,

(B)

dx dy dy dx

dy

(B)

+

2xy^

'

Solution

z

y).

=

y)

can be shown that

It

""

dydx

/xy{x. y)

x (holding y constant). What

to

with

start

y (holding x constant). Then

differentiate with respect to

with respect

= fyx, we '""

dxdy

dx^

First find /yx(x, y).

means to

~ dx\dx}~

Then evaluate

at (2, 1).

differentiate with respect to y

first

Again,

remember

that /y^

and then with respect to x.

Thus,

= -6xy2 = -6y2

/y(x. y) /yx(x, y)

and L,(2,

Problem 9

=

l)

-6(l)^

= -6

For the function in Example

9,

find

dH (A)

(C)

(B)

dy dx

Answers to Matched Problems

Sz (A)

dw (A)

= -3x^ +

10x(x2

Py(25, 10)

=

+

5

(B)

/y(2, 3)

2xy)^

(B)

10

10:

At

a

(D)

/,y(2,3)

= -7

production level of x

by one unit and holding x fixed

at

/yx(2,3)

=

25

andy =

10,

increasingy

25 will increase profit by approxi-

16-2

15)

15. increasing

profit 9.

=

-6y2

(A)

-12xy

(B)

(C)

-54

(D)

-54

Exercise 16-2

A

For z

3.

For z

7.

= fix,

=

10

+

3x

+

2y, find

each of the dz

dx

dy

ffx, y)

=

3x^



2xy^

+

1,

find each of the following:

dz

dz

dy

ox

/J2.

3)

y)

/,(2, 3)

8.

=

fol]oiving:

/,(1.2)

4.

f,(1.2)

For S(x,

B

yj

dz

=

5x^y^, jind each of the following:

9.

S,[x,y]

10.

Sy(x,y]

11.

Sj,(2, 1)

12.

SJ2,

For C(x,

=

y)

x^

-

2xy

+

2y^

+

6x

-

9y

+

5.

1)

find each of the following:

13.

C,(x,y)

14.

Cy[x.y)

15.

CJ2,

16.

Cj,(2. 2)

17.

C,y[x.y]

18.

Cy,[x.y]

19.

C^^(x.y)

20.

Cyy[x,y)

For z

=

947

-110: At a production level of x = 25 and y by one unit and holding x fixed at 25 will decrease by approximately $110.

mately $10; Py[25.

y=

Partial Derivatives

2)

ffx, y)

=

e^"''"^'',

jind each of the following:

dz 21-

-Tdx

23.

^-^—

dz — ay

22.

dH

d'-z

24.

dx dy

dy dx

'

25.

/,,,(!. 0)

26.

/y,(0, 1)

27.

/xx{0. 1)

28.

/,^(1,0)

Find

fj,x, y]

29.

/(x. y)

31.

/(x, y)

33.

/(x, y)

35.

/(x, y)

and

= = = =

fy(x, y) for

- y^f - 1)" ln(x2 + y^)

each /unction

f

given by:

(x^

30.

/(x,

(3x^y

32.

/(x. y)

34.

/(x, y)

y^e"'''

36.

/(x. y)

.v)

= = = =

- y^ + 2xyT ln(2x - 3y) V2x (3

x^e'''*'

948

Multivariable Calculus

=

^ — v^

x^

37-

/(-y)

Find

/xx(x, y), /xy(x, y), /yx(x, y),

39.

/(x, y)

=

41.

/(x, y)

=---

/(x, y)

45.

For

x^

+

and

y

2y^

and y such

find values of x

=

P^(x, y)

-

2xy

and

2y^v

40.

/{x, y)

=

x^y'

42.

f (x, V)

=

—y - —X

=

x In(xy)

44.

= -x^ +

^

each function /given by:

'

= xe'"'

P(x, y)

=

/(x,y)

/yy(x, y) for

x

y 43.

+

x^v'

38.

- 4x +

f[x. y]

12y

-

18y

+

+

x

+

y^

5

that

Py(x, y)

=

simultaneously. 46.

For C(x, y)

=

2x^

+

find values of x

2xy

Sy^

-

16x

-

54

and y such that

=

C,(x, y)

+

and

C,,(x, y)

=

simultaneously. In

ProbJems 47-48, show that the /unction

= =

47.

/(x, y)

48.

/{x, y)

49.

For f[x,

For/{x,

x^

y)

find:

/(x-fAxy)-/(x,y)

,.^

Ax

Ay—

=

y)

H^

= 0.

ln(x^

^.^ Ax— 50.

+ y^) - 3xy2 = x^ + 2y\

f satisfies f^Jx, y) +fyy(x, y)

/(x,

y

+ Ay)

-/(x,

y)

Ay] -/(x,

y)

Ay

2xy^find:

/(x

+ Ax.y]-/(x.y) Ax

Ax—

y ^.^ /(x. Ay-0

-f

Ay

Applications Business & Economics

51.

Cost /unction.

lem 29

C(x, y)

=

Find Cx(x, 52.

The

cost function for the surfboard

in Exercise 16-1

2,000

y)

+

70x

and Cy(x,

Advertising and sales.

company

in Prob-

was

+

lOOy

y),

and

interpret.

A company spends x thousand dollars per week

16-2

Partial Derivatives

on newspaper advertising and y thousand dollars per week on sion advertising. Its weekly sales were found to be given by

=

S(x, y)

Find 53.

S,(3, 2)

A

televi-

5x^y^

and

Projit function.

type

949

5^(3, 2),

and

interpret.

A firm produces two types of calculators, x thousand of

and y thousand

of type

B per year. The revenue and cost

functions for the year are (in thousands of dollars) R(x, y) C(x, y)

Find 54.

= =

+ 20y - 2xy +

14x x^

P^(l, 2)

Revenue and

and

q C(x, y)

bicycles.

16y

+

5

interpret.

A company manufactures ten-speed The weekly demand and cost functions are

-|-

-I-

is

the price of a ten-speed bicycle, $q

the weekly

equations.

the weekly

demand

is

the price of a

ten-speed bicycles, y for three-speed bicycles, and C(x, y) is the cost

is

demand

function. Find R,(10,

Demand

and

+

-I-

three-speed bicycle, x

55.

12x

= 230 - 9x y = 130 -l-x-4y = 200 80x 30y

where $p is

2),

+

profit functions.

and three-speed p

Py(l,

2y^

5)

and

Px(10,

5).

and

for

interpret.

A supermarket sells two brands of coffee, brand A

$p per pound and brand B at $q per pound. The daily demand A and B are, respectively,

at

equations for brands

= 200 - 5p 4q y = 300 + 2p - 4q

X

56.

-I-

Find dx/dp and by /dp, and interpret. Marginal productivity. A company has determined that its productivity (units per employee per week) is given approximately by

= 5Gxy —

z(x, y)

where x

is

x^



3y^

the size of the labor force in thousands and y

is

the

amount

of capital investment in millions of dollars. (A)

Determine the marginal productivity of labor when x

=

5

and

(B)

y = 4. Interpret. Determine the marginal productivity of capital when x

=

5

and

y Life Sciences

57.

=

4.

Interpret.

Marine biology. In using scuba diving gear, a marine mates the time of a dive according to the equation T(V,x)

=

-^ x-l-33

biologist esti-

950

Multivariable Calculus

where

T = Time

of dive in minutes

V = Volume X = Depth

58.

of air, at sea level pressure, compressed into tanks

of dive in feet

Find Tv{70, 47) and T,(70, 47), and interpret. Blood flow. Poiseuille's law states that the resistance, R,

for

blood

flowing in a blood vessel varies directly as the length of the vessel,

and inversely

may

power

of

its

radius,

r.

L,

This relationship

be stated in equation form as follows:

R(L,

r)

Find Rl 59.

as the fourth

=

—-

k

(4, 0.2)

k a constant

and R,

(4. 0.2).

and

interpret.

Physical anthropology. Anthropologists, in their study of race and

human index.

genetic groupings, often use an index called the cephalic

The cephalic index,

top). In

C(W,

W, of the head (both viewed from the

C, varies directly as the width,

head, and inversely as the length,

L,

of the

terms of an equation,

L)

=

100

W —

where

W = Width in inches L

=

Length in inches

and Cl(6, 8), and interpret. Under ideal conditions, if a person driving a car slams on the brakes and skids to a stop, the length of the skid marks (in feet) is given by the formula Find Cw(6,

60.

8)

Safety research.

L(w,

v]

=

kwv^

where k

= Constant

w— V = 61.

Weight of car in pounds Speed of car in miles per hour

For k = 0.0000133, find LJ2,500, 60) and LJ2,500, 60), and interpret. Psychology. Intelligence quotient (IQ) is defined to be the ratio of the

mental age (MA), as determined by certain cal age (CA), multiplied

Q(M,

0=^-100

tests,

and the chronologi-

by 100. Stated as an equation,

Total Differentials and Their Applications

16-3

951

where

Q = IQ M = MA C = CA Find Qm[12,

16-3

10)

and Qc(12,

10),

and

interpret.

Total Differentials and Their Applications The

Total Differential

Approximations Using Differentials

The

Total Differential

Recall (Section 11-4) that for a function defined by

y

= f[x]

the differential dx of the independent variable x variable,

which can be viewed

the dependenl \'ariable y

is

How

dent variables, x and dx. functions with two or

Suppose

z

= /(x,

Solution

y,

and

Find dz (A)

X

(B)

x

(C)

x

Since

dz

(A)

is

another independent

The

differential

dy of

dx. Thus, the differential of

a function

with two indepen-

can the differential concept be extended

more independent

to

variables?

function with the independent variables x and

y) is a

= f^(x, y) dx + fy(x,

Notice that dz

Example 10

= f'[x)

is

in x.

y.

define the total differential of the dependent variable z to be

dz

and

given by dy

change

with one independent variable

a function

We

as Ax, the

is

y)

dy

a function of /our variables: the

their differentials

for f(x, y)

=

x^y^. Evaluate

= 2. y = -l, dx = 0.1, = l, y = 2. dx = -0.1. = -2, y=l, dx = 0.3,

=

f^(x. y)

2xy^ and

fy[x, y)

= /x(x, y) dx -f fy[x, y) = 2xy^ dx + 3x2y2 dy When dz

=

x

=

2,

y

=

2(2)(- l)'(O.l)

dz

for:

and and and

=

= = dy = dy dy

0.2

0.05

-0.1

3x^y^,

dy

-1, dx

+

independent variables x

dx and dy.

=

0.1.

and dy

3(2)-(-l)2(0.2)

=

=

2

0.2,

952

Multivariable Calculus

When

(B)

dz

dz Find dz X

(A) (B)

x

(C)

x

If

=

When

(C)

Problem 10

x

=

=

x

=

=

-2, y

=

y

3,

=

w = f(x,

dx

=

l,

y, z),

dw = f,[x,

y, z)

dw

dx

11

Solution

Find

dw

(A)

x

(B)

x

(C)

x

Since

yz^ dx

dw =

When

x

dw =

dw= Problem 11

Find

dw

(A)

x

(B)

x

0.2, 0.1,

(C)

x

dz

independent variables: the original

and their differentials dx, dy, and dz. functions with more than three independent variables and

x, y,

=

z,

xz^

y

1,

+

dy

=

=

+

-2,

y=

(1)(2)2(0.2)

= y = y =

=

l,

z

2,

z

-3,

for:

= 0.1, dy = -0.2, = -0.1, dy = 0.1, =

= xz^,

=

dy

0.2,

and

0.3,

/^(x, y, z)

and dz = 0.05 and dz = and dz = -0.4

=

2xyz,

2xyz dz

z

3,

dx dx dx

y, z)

+

= 0,

+ z

1,

-1, dx

=

dx

=

=

2,

dx

=

+

yz

= =

-2,

z

=

l,

+

zx.

=

dy

+

+

0.2,

+

-0.2, and dz

2(2)(3)[-l)(0.05)

-0.1, dy

(1)(0)2(0.1)

(-1)(2)2(0.3)

xy

0.1,

(2)(-l)2(-0.2)

z

(-2)(0)2(-0.1)

-1,

dw

xyz^. Evaluate

(3)(- 1)2(0.1)

y

dy + f^(x,

y, z)

a function of six

for /(x, y, z)

= l, = 2, = 4,

for:

pattern.

+

Whenx =

(C)

-2.4

and dy = -0.1 and dy = 0.1 and dy = -0.04

0.05,

= yz^, /^(x,

=

=

y, z)

Whenx = 2,y=

(B)

Evaluate dz

= 2, y=3, z = -l, = l, y = -2, z = 0, = -l, y=l, z = 2,

dw = (A)

x^.

-0.1,

dx + fy(x,

for /(x, y, z)

/^(x, y, z)

=

is

Generalizations to

Example

= -1

and dy

0.3,

0.05,

then the total differential

is

same

=

=

3(-2)2(l)2(-0.1)

= =

dx

independent variables follow the

+

+

-0.1, and dy

3(1)2(2)2(0.05)

dx

1,

xy^

-2, y = 2, l, y = -2,

This time,

+

2(-2)(l)3(0.3)

=

=

dx

2,

2(1)(2)3(-0.1)

for f(x, y)

= = =

y

1,

=

0.1,

and dz

2(l)(-2)(0)(0)

dy

=

0.3,

dw

0.05,

=

-0.7

=

0,

=

and dz

2(-l)(l)(2)(-0.4)

Evaluate

=

=

=

-0.4,

1.2

for:

dx = 0.1, dy = 0.1, and dz = 0.1 dx = 0.5, dy = 0.5, and dz = dx = 0.1, dy = -0.2. and dz = l,

0.4

Approximations Using Differentials If

z

= /(x,

and Ax and Ay represent the changes in the independent and y, then the corresponding change in the dependent variable

y)

variables x

z is

953

Total Differentials and Their Applications

16-3

given exactly by

Az = fix + Ax, y + Ay) — fix, For small values of

Ax and

y)

Ay, the differential dz can be used to approxi-

mate the change Az.

Example 12

Solution

Find Az and dz for/(x, Ay = dy = - 0.02.

Az = f[x + Ax, y +

y)

A3']

= —

x^

+ y^ when x =

+

2y dy

2(3){0.01)

-0.1

Problem 12

+

=

4,

Ax = dx =

0.01,

and

Note that dz is a good approximation for Az, the exact change in z, and dz was easier to calculate

2(4)(-0.02)

«

Repeat Example 12 In addition to

approximate

y

f(x, y)

= f(3.01, 3.98) -/(3. 4) = [(3.01)2 + (3.98P]-[3^ + 42] = 24.9005 - 25 = -0.0995 * dz = /„(x, y) dx + fy(x, y) dy 2x dx

3,

f(x

for

x

=

2,

y

=

5,

Ax = dx =

-0.01, and

0.05.

approximating Az, the differential can also be used

+

Ax, y

+

Ay).

These approximations are summarized

the box and illustrated in the examples that follow.

Diiferential

Ay = dy =

Approximation

to

in

954

Multivariable Calculus

What What

(A)

is

the cost of producing 100 boards of each type?

if one fewer standard and two more competition boards are produced? Approximate the change

(B)

is

the approximate change in the cost

using differentials. Solution

C(100, 100)

(A)

We

(B)

will use

= = =

700 700

+ +

70,000

+

+

-

100(100)^/^

100,000

-

20(100)'/2(ioo)i/2

2,000

$168,700

dC

to

approximate

changes from 100

dx

70(100)^/^

= Ax = — 1

,

AC

We must dy = Ay = 2:

to 102.

and

as x changes

evaluate

dC

from 100 for

x

=

to

100,

99 and y

y

=

100,

AC-dC = = =

C,(x, y]

dx

[105x^/2 _

+ Cy(x,

y]

dy

iox-V2yV2] dx

[105(100)^/2

_

[150y'/2

-

IQx'/'y"'/^]

dy

io(iOO)-'/2(100)'/2](- 1)

+ = -1,040 + = $1,940

+

[150(100)^/2

_

io(100)'''2(;ioO)-^/2)](2)

2,980

Thus, decreasing the production of standard boards by one and increasing the production of competition boards by two will increase the cost by approximately $1,940.

Problem 13

For the cost function in Example

What

(A)

is

13:

the cost of producing 25 standard boards and 100 competition

boards?

What

(B)

is

the approximate change in the cost

boards and

Example 14

five

if

three

more standard

fewer competition boards are produced?

Approximate the hypotenuse of a

right triangle

with legs of length 6.02 and

7.97 inches.

Solution

X and y are the lengths of the legs of a right triangle, then from the Pythagorean theorem we find the hypotenuse z to be If

z

We

= /(x,

y)

=

Vx^

+

y2

could use a calculator to compute the value of

however, our purpose here

to illustrate the

is

/(6.02, 7.97) directly,

use of the differential to

we will proceed as though a we must select values of x and y

approximate the value of a function. Thus, calculator

is

that satisfy

second,

not available. This

two conditions:

we must be

means

First,

n/62

+

8^

=

v/36

+

they must be near 6.02 and 7.97; and

able to evaluate

Since 64

=

^100

=

10

that

VxM-y^ without

using a calculator.

\

= 6 and y = 8 satisfy both of these conditions. So, we dx = Ax = 0.02, and dy = Ay = —0.03, and then we use X

f(x

+ Ax,

+

y

= f[x,

Ay)

y]

955

Total Differentials and Their Applications

16-3

=

x

let

6,

y

=

8,

+ Az + dz

= /(x,y) = /(x,y)+/,(x,y)dx+/,(x,y)dy Now we

can obtain an approximation to/(6.02,

we can

7.97) that

evaluate

bv hand: fix

V(x

+ Ax,

+ AxY +

y

[y

+

Ay)

- /(x,

+ Ayf ^

-Jx^

y)

+ /Jx,

X

+ y^ +

+

+

0.02)2

+

[8

(-0.03)]'

V(6.02)2

Problem 14

+

(7.97)2

= =

Approximate the hypotenuse

Ve^

10

+ 8^ +

+

0.012

.

dx

+

Vx2 V(6

+ f,(x, y)

dx

y)

0.024

y

.

(0.02) '~'~

dv

-

+

Vx'

v'

VeM^ -

+

dy

+

y2 8 -0.03)

-

'

Ve'

+

8'

= 9.988

of a right triangle with legs of length 2.95

and

4.02.

Answers to Matched Problems

= (y2 + 2x) dx + 2xy dy: (A) -0.25 (y + z) dx + (x + z) dy + (y + x) dz:

10.

dz

11.

dw =

12.

Az

14.

4.986

-0.8

(B)

(A)

(C)

0.6

(B)

-0.8

(C)

= 0.4626,

dz

=

0.46

$108,450

(A)

13.

-$5,960

(C)

Exercise 16-3 Find dz for each function. 1.

z

3.

z

= x2 + = x^y^

y2

= 2x + xy + = V2x + 6y z = xVl +y

2.

z

4.

z

5

r-

6.

3y

yly

Find 7. 9.

B

dw for each

w= w=

x^

xy

function.

+ y^ + z^ + 2xz + 3yz

Evaluate dz and Az 11.

z

= /(x,

y)

Ay = dy = 12.

Z

= /(X,

y)

Ay = dy =

for

w= w=

8.

10.

each function

= x2-2xy +

y2.

x

xy'z^ sl2x

+

3v

-

z

at the indicated values.

=

3,

y

=

l,

Ax = dx =

0.1,

0.2

= 2X2 + 0.05

xy

-

3y2,

x

=

2,

y

= 4, Ax =

dx

=

0.1,

0.76

956

Mullivariable Calculus

13.

= /(x,

z

=

y)

Ay = dy= 14.

z

= /(x,

=

y)

(

3--j,

x

=

2,

+ — j,

x

=

3,

y=l,

Ax = dx =

0.05,

Ax = dx =

-0.1,

0.1

Ay = dy = In

100

50

(

1

=

y

9,

0.2

Problems 15-38 evaluate

dw and A w for each

/unction at the indicated

values. 15.

16.

17.

w = /(x, y, z) = x^ + yz, x = 2, y=3, z = 5, Ax = Ay = dy = 0.2, Az = dz = 0.1 w = /(x, y, z) = 2xz + y2 - z^ x = 4, y = 2, z = 3, Ax = dx = 0.2, Ay = dy = 0.1, Az = dz = -0.1 H'

= /(x,

,

y, z)

+

10x

=

20y X

,

=

=

y

4,

z

3.

=

dx

=

0.1,

5,

z

Ax = dx = 18.

w = /(x,

19.

20.

21.

=

y, z)

Ax = dx =

Ay = dy = -0.05,

0.05,

50

(xH

^~)'

Ay = dy =

0.2,

2,

y=2,

Az = dz =

0.1.

A can in

z

=

l,

0.1

the shape of a right circular cylinder with radius 5 inches and is

coated with ice 0.1 inch thick. Use differentials to

approximate the volume of the

23.

=

0.1

Approximate the hypotenuse of a right triangle with legs of length 3.1 and 3.9 inches. Approximate the hypotenuse of a right triangle with legs of length 4.95 and 12.02 inches. height 10 inches

22.

x

Az = dz =

A box with edges of length 10,

ice (V

=

;rr^h).

and 20 centimeters is covered with a 1 centimeter thick coat of fiberglass. Use differentials to approximate the volume of the fiberglass shell. A plastic box is to be constructed with a square base and an open top.

The

plastic material

15,

used in construction

is 0.1

inside dimensions of the box are 10 by 10 differentials to

centimeter thick. The

by

5 centimeters.

approximate the volume of the plastic required

for

Use one

box. 24.

The

surface area of a right circular cone with radius

r

and

altitude h

is

given by S

Use

=

n^^l^^

+

h'

approximate the change in S when and h changes from 8 to 8.05 inches.

differentials to

6 to 6.1 inches 25.

Find dz

if

z

26.

Find dz

if

z

= =

xye'"+>".

x In(xy)

+

y

In(xy).

r

changes from

16-3

27.

Find

28.

Find

Total Differentials and Their Applications

dw if w = xyze"''^. dw if w = xy In(xz) +

957

yz In(xy).

Applications Business & Economics

29.

A

microcomputer company manufactures two types of and model II. The cost in thousands of dollars of producing x model I's and y model II's per month is given by Cost /unction.

computers, model

=

C(x, y)

X

I

+ 2v

Vx^

+

y2

company manufactures 30 model I computers and 40 model II computers each month. Use differentials to approximate the change in the cost function if the company decides to produce 5 more model I and 3 more model II computers each month. Ad\'ertisiiig and sales. A company spends x thousand dollars per week on newspaper advertising and y thousand dollars per week on television advertising. Its weekly sales were found to be given by Currently, the

30.

=

S(x, y)

Use

5x^y^

differentials to

approximate the change in sales

spent on newspaper advertising

week and

the

from $2,000 31.

to

amount spent on

A

supermarket

$x per pound and brand B

equations for brands u

V

= =

200 300

-

television advertising

5x

-I-

4y

4y

+

2x

A

and B

at

sells

two brands of

$y per pound.

the

is

amount

increased

coffee:

The

daily

brand

A

demand

are, respectively,

(both in pounds). Thus, the daily revenue equation

is

= xu + yv = x(200 - 5x + 4y) + y(300 - 4y + 2x) = -5x^ + 6xy - 4y2 + 200x + 300y

fi(x, y)

32.

if

increased from $3,000 to $3,100 per

$2,200 per week.

fie\enue function. at

is

Use differentials to approximate the change in revenue if the price of brand A is increased from $2.00 to $2.10 per pound and the price of brand B is decreased from $3.00 to $2.95 per pound. Marginal productivity. A company has determined that its productivity (units

per employee per week)

-

is

given approximately by

z(x, y)

=

50xy

where x

is

the size of the labor force in thousands and y

x^

-

3y^

of capital investment in millions of dollars.

The current

is

the

amount

labor force

is

958

Multivariable Calculus

5,000 workers. differentials to

The current capital investment is $4 million. Use approximate the change in productivity if both the

labor force and the capital investment are increased by 10%. Life Sciences

33.

Blood flow. Poiseuille's law states that the resistance, R,

for

blood

flowing in a blood vessel varies directly as the length of the vessel,

and inversely

may be R(L,

Use

as the fourth

power

of

its

radius,

r.

L,

This relationship

stated in equation form as follows:

k a constant

r)

approximate the change in the resistance

differentials to

if

the

length of the vessel decreases from 8 to 7.5 centimeters and the radius

34.

decreases from 1 to 0.95 centimeter. Drug concentration. The concentration of a drug after having been injected into a vein is given by

C(x, y)

= 1

where x

is

/ + + Vx^

the time passed since the injection and y

concentration C(3.1, 35.

Safety research.

bloodstream

y2

from the point of injection. Use

Social Sciences

in the

differentials to

is

the distance

approximate the

4.1).

Under

ideal conditions,

if

a person driving a car slams

on the brakes and skids to a stop, the length of the skid marks is

(in feet)

given by the formula

Hw,

v)

=

kwv^

where

= w= V =

Constant

k

For k

Weight of car

in

pounds

Speed of car in miles per hour

= 0.000

013

3,

use differentials to approximate the change in

the length of the skid marks

if

the weight of the car

2,000 to 2,200 pounds and the speed

is

is

increased from

increased from 40 to 45 miles

per hour. 36.

Psychology, Intelligence quotient (IQ)

is

defined to be the ratio of the

mental age (MA), as determined by certain cal age (CA) multiplied

Q(M,C) where

= ^-100

by

100. Stated as

tests,

and the chronologi-

an equation,

16-4

Maxima and Minima

959

Q = IQ M = MA C = CA Use

differentials to

approximate the change in IQ as

a person's

mental

age changes from 12 to 12.5 and chronological age changes from 10 to 11.

16-4

Maxima and Minima We are now ready to undertake a brief but useful analysis of local maxima for functions of the type z = f{x, y). Basically, we are going to

and minima

extend the second-derivative

pendent variable. To for the function / in

that the surface z

words,

we

start,

some

test

developed

we assume

for functions of a single inde-

that all second-order partials exist

circular region in the

xy plane. This guarantees

= f[x, y) has no sharp points, breaks, or ruptures. In other

are dealing only with

edge of a box); or breaks

(like

smooth surfaces with no edges

an earthquake

bottom point of a golf tee). See Figure

fault); or

sharp points

(like

the

(like

the

6.

NO

Figure 6

we will not concern maxima-minima theory. In

In addition,

absolute

ourselves with boundary points or spite of these restrictions, the proce-

960

Multivariable Calculus

dure

we

now

are

going to describe will help us solve a large

number

of

useful problems.

What does it mean for/(a,

We say the

b) to

that /(a, b) is a local

domain of/ with

(a, b)

be a local

maximum

as the center,

if

maximum or a local minimum? there exists a circular region in

such that

f(a,b]^/(x,y) for all (x, y) in the region. Similarly,

we

say that /(a, b)

is

a local

minimum

domain

of /with (a, b) as the center,

for all (x, y) in the region. In Section 16-1,

Figure 2 illustrates a local

if

there exists a circular region in the

such that /(a,

b)^/(x,y)

minimum. Figure 3 illustrates a local maximum, and Figure 4 illustrates a saddle point, which is neither. What happens to /x (a, b) and/y(a, b) if/(a, b) is a local minimum or a local

maximum and

the partials of/ exist in a circular region containing

Figure 7 suggests that /^(a,

b)

=

indicated curves are horizontal.

reasoning

Theorem

1

Let /(a,

and fy(a,

b)

Theorem

1

=

0,

indicates that our intuitive

correct.

is

b)

be an extreme

the function

/. If

both

/,

(a local

and f^.

maximum

exist at (a,

(a, b)?

since the tangents to the

b),

or a local

then

16-4

/x(a, b)

/(a, b)

=

is

and

a local

Maxima and Minima

961

and test these further to determine whether fy(a, b) = extreme or a saddle point. Points (a, b) such that (1) holds

are called critical points.

The next theorem, using second-derivative

tests,

gives us sufficient conditions for a local point to produce a local extreme or a saddle point.

without proof.

Theorem

2

Se If:

1.

2. 3.

4.

As was the case with Theorem

1,

we

state this

theorem

962

Multivariable Calculus

Step

3.

Evaluate

AC — B^

Theorem

2.

AC - B2 =

and

(- 2)(-2)

Therefore, case

in

1

try to classify the critical point

-

(OP

=

Theorem

4

>

That

is,

using

A = -20 A = -4 ^ «x«

30 and

1

«y^

3.

Population distribution. In order to study the population distribution of a certain species of insects, a biologist has constructed

habitat in the shape of a rectangle 16 feet long

and 12

only food available to the insects in this habitat

is

The

biologist has

square foot

at a

an

artificial

feet wide.

located at

its

The

center.

determined that the concentration C of insects per

point d units from the food supply (see the figure)

is

given approximately by

C = 10-^d2 What

is

the average concentration of insects throughout the habitat?

Express C as a function of x and

y, set

up

a

double integral, and

evaluate.

y

.(X, y)

Food supply

^^

-6

1002

Multivariable Calculus

34.

Population distribution. Repeat Problem 33 for a square habitat that

measures 12

feet

on each

side,

where the

insect concentration

is

given

by

35.

Pollution.

A

town emits

heavy industrial plant located

in the center of a small

particulate matter into the atmosphere. Suppose the con-

centration of particulate matter in parts per million at a point d miles

from the plant

is

given by

C = 100- ISd^ If

the boundaries of the town form a rectangle 4 miles long and 2

miles wide, what

is

the average concentration of particulate matter

throughout the city? Express

C

as a function of x

and

y, set

up

a

dou-

ble integral, and evaluate.

36.

Problem 35

Pollution. Repeat

if

the boundaries of the

town form

a

rectangle 8 miles long and 4 miles wide and the concentration of particulate matter

is

given by

C = 100-3d2 Social Sciences

37.

Safety research.

Under

ideal conditions,

on the brakes and skids is given by the formula L

=

if

a person driving a car slams

to a stop, the length of the skid

marks

(in feet)

0.000 013 3xy2

where x

is

the weight of the car in pounds and y

in miles per hour.

is

the speed of the car

What is the average length of the skid marks for cars

weighing between 2,000 and 3,000 pounds and traveling at speeds between 50 and 60 miles per hour? Set up a double integral and evaluate. 38.

Safety research. Repeat Problem 37 for cars weighing between 2,000

and 2,500 pounds and traveling at speeds between 40 and 50 miles per hour.

Double Integrals over More General Regions

16-8

39.

Psychology.

The

intelligence quotient

age X and chronological age y Q(x, y)

is

1003

Q for an individual with mental

given by

= 100y

In a group of sixth graders, the

mental age varies between

8

and 16

years and the chronological age varies between 10 and 12 years.

integral 40.

and evaluate.

Psychology. Repeat Problem 39 for a group with mental ages between 6 and 14 years and chronological ages

16-8

What

the average intelligence quotient for this group? Set up a double

is

between

8

and 10

years.

Double Integrals over More General Regions Regular Regions

Double Integrals over Regular Regions Reversing the Order of Integration

Volume and Double In this section

we

tangular regions.

Integrals

will extend the concept of double integration to nonrec-

We

begin with an example and some

new

terminology.

Regular Regions Let

R be the

region graphed in Figure 14.

following inequalities:

R=

{(x,

y)|x«y«6x-x^

f(x)

«

= 6x -

X


-l).

2).

the associated cumulative probability distribution

function. 20.

Use the associated cumulative probability distribution function find the value of x that satisfies P(0 « X ^ x) = |.

21.

Find the associated cumulative probability distribution function for

/(x)

=

{'

0«x«2

x-fx^

otherwise

Graph both functions (on separate Repeat Problem 21

22.

In

sets of axes).

for

fe-"

x^O

\

otherwise

.,,

to

Problems 23-26 jind the associated cumulative probability /unction, and

use

to find the

it

Find P(l

23.

/((x) ^ '

=

«X« f

In X

e

fxe"'

x^O otherwise is

= 3x/(8vT+x) 0«x=S3 otherwise

Find P(X ff

Problems 27-30 F(xJ

1

=

&

e) for

J(^" ^'/^^ \

^^^ other erwise

the cumulative probability distribution function

random variable

associated with each F(x).

F(x)

,

2) for

'^

26.

for

«X«

Find P(l

/(x)'

.

\

/or a continuous

27.

«

X

,

otherwise

[

fix

f(x]

24.

2) for 1 =s

1

Find P(X^l)

25.

In

indicated probability.

X. Find the probability density /unction

1034

Additional Probability Topics

In

Problems 31-34, find the associated cumulative distribution /unction

(X 2

-

X

0«x«

i

1< X «

i

32.

2

r

-1 ^

ixi

X

«

1

/(x)

i

otherwise

33.

0«x« b

f(x)

1 -

b-a _L

V

..

F(x)

-^x

Mean;

M--[a +

b]

= -(a +

b)

x„

Median:

Standard deviation:

(T

= —=(b —

a)

V12

Example 24

Standard electrical current

is

uniformly distributed between 110 and 120

Electrical Current

volts.

What is the probability that the current is between

Solution

Since

we

are told that the current

[110, 120].

/(x)

we

is

113 and 118 volts?

uniformly distributed on the interval

choose the uniform probability density function

^ 110^x^120 otherwise

1068

Additional Probability Topics

Then X — dx = — 10

r"°

P(113^X«118)=

1

Problem 24

in

Example

24,

what

is

"°_218 _

113

_

1

10

Jii3

the probability that the current

is

at least

116 volts?

Beta Distribution In

many

applications, the

outcomes of an experiment are expressed in

terms of fractions or percentages. For example,

^

or

90%

of the students

entering a certain college successfully complete the freshman year, ^ or

20%

show a profit during their first year of work with outcomes expressed as fractions necessary to use a probability density function whose

of fast-food restaurants

fail to

operation, and so on. In order to or percentages,

values

which

is

it is

in the interval

lie

[0, 1].

One

often used in this situation

A continuous random variable random variable

as a beta

is

special probability density function

the beta probability density /unction.

has a beta distribution* and

if its

is

probability density function

referred to is

the beta

probability density function

r{/?+l)(/?+2 l)(y?+2)x^(l -x) /(x)

"

,

0^x«l otherwise

where y? is a constant, fi^O. The value of fi is usually determined by examining the results of a particular experiment. The values of a beta random variable can be expressed as fractions or percentages; however, percentages should be converted tions involving a beta First,

we show

random

to fractions before

performing calcula-

variable.

that /satisfies the requirements for a probability density

function: /(x)

=

f[x]dx=

(y?+l)(/J-f 2)x^(l

I

-x)^0

(;?+l)(/3+2)x^(l

0«x«l

-x)dx

I

r (yJ+l)(/?+2)(x'«-x^+^)dx

= {/]+2]-{fi+l] = Thus, /is *

There

here.

is

l

a probability density function. a

more general

definition of a beta distribution, but

we

will not consider

it

Uniform, Beta, and Exponential Distributions

17-5

F(x)

If

then

for

the associated cumulative probability distribution function,

is




given by

0.

Find the mean. What restrictions must you place on p? Find the variance and standard deviation. What restrictions

must you place on p? (C)

is

Find the median.

1075

Uniform, Beta, and Exponential Distributions

17-5

Applications Business

& Economics

29.

Waiting time. The time

them

officer to give

the interval

40].

[0,

(in

a driver's

What

is

minutes) applicants must wait for an

examination

is

uniformly distributed on

the probability that an applicant must

wail more than 25 minutes? 30.

Business failures.

during the (A) (B)

of

is

computer hobby

stores that fail

random variable with fi=

a beta

4.

What is the expected percentage of failures? What is the probability that over 50% of the stores fail during the year?

first

31.

The percentage

year of operation

first

Absenteeism. The percentage of assembly line workers that are absent

one Monday each month centage is 50%. (A) (B)

What What

a beta

is

random

is

the appropriate value of y??

is

the probability that no

variable.

more than 75%

The mean

per-

of the workers

on one Monday each month? Waiting time. The waiting time (in minutes) for customers at a drivein bank is an exponential random variable. The average (mean) time a customer waits is 4 minutes. What is the probability that a customer will be absent

32.

33.

waits more than 5 minutes? Communication. The length of time

minutes)

is

conversation lasts less

34.

Life Sciences

35.

is

than

minutes.

3

2

What

is

telephone conversations

for

exponentially distributed.

The average (mean)

(in

length of a

the probability that a conversation

minutes?

(in years) of a component in a random variable. Half the compomicrocomputer is an exponential company that manufactures the nents fail in the first 3 years. The What is the probability that a component offers a 1 year warranty. warranty period? component will fail during the

Component failure. The

life

The percentage

Nutrition.

expectancy

of the daily requirement of vitamin

present in an 8 ounce serving of milk

is

a beta

random

D

variable with

yS=.2. (A) (B)

What What

D

is

the expected percentage of vitamin

is

the probability that a serving contains at least

per serving?

50% of the

daily requirement? 36.

Medicine.

A scientist is measuring the percentage of a drug present in The results indicate random variable with

the bloodstream 10 minutes after an injection. that the percentage of the

mean fi = (A) (B)

drug present

is

a beta

.75.

What What

is is

the value of yS? the probability that no

preseilt 10

minutes

after

more than 25%

an injection?

of the

drug

is

1076

Additional Probability Topics

37.

The time

Survival time.

tracted a certain disease

of death (in years) after patients is

that a patient dies within

What What

(A) (B)

exponentially distributed.

year

1

The

have con-

probability

is .3.

is

the expected time of death?

is

the probability that a patient survives longer than the

expected time of death? 38.

Survival time. Repeat Problem 37 virithin 1

Social Sciences

year

Education.

39.

The percentage

year of college

first

What

(A)

first

is

Psychology.

17-6

first

The time

30 seconds.

to find a

of entering

a beta

random

the probability that

through a maze is

is

freshmen that complete the

variable with

y?

=

17.

the expected percentage of students that complete the

is

complete the 40.

the probability that a patient dies

year?

What

(B)

if

is .5.

is

more than 95%

(in

seconds)

it

is

way The average (mean) time

takes rats to find their

exponentially distributed.

What

of the students

year?

the probability that

it

takes a rat over

1

minute

path through the maze?

Normal Distributions Normal Probability Density Functions The Standard Normal Curve Areas under Arbitrary Normal Curves Normal Distribution Approximation of a Binomial

We

Distribution

now

consider the most important of all probability density funcnormal probability density function. This function is at the heart of a great deal of statistical theory, and it is also a useful tool in its own right for solving problems. We will see that the normal probability density function can also be used to provide a good approximation to the binomial will

tions, the

distribution.

Normal Probability Density Functions

A

continuous random variable

to as a

X has a normal distribution

normal random variable

if its

normal probability density function -(x-/i)2/2cr!

/(x)

TV27r

and

is

referred

probability density function

is

the

17-6

where /I

is

any constant and

cris

Normal Distributions

any positive constant.

It

1077

can be shown, but

not easily, that

fix]

dx

=

1

/: E(X)

=/:

x/(x)

dx= fi

and V(X)^

Thus,

// is

\ythe

Hfj(x]

mean

dx

of the

=

a^

normal probability density function and a is the is always a bell-shaped curve called a

standard deviation. The graph of /(x)

normal curve. Figure 9 oifi and "

65. 5

66. 50

(C)

T

(C)

(B)

9

>"

-2

7.

90

(B)

45

^

82.-4 83.1

"

84.^

85.

-72m'2n>=

1

(D)

Commutative JD) Associative

-33

Commutative

40.

56.-4«x«3;^

-5/x

19. 27

3.

-4>/7

^ + 10V2

+

3V3

H'-4

21.

5.

23.

Ts

5\/2

7.

^ V5 +

2

9. 5-3\/5 11. u + 3V^ 4-\/6 3V2-2\/3

25.

27.

2zV3

41.

39.

m-n

29.

_ + 2Vx-15

47. 14

-7 + 3V5

31

n

57.

61. 5

59.

-

2

13xyV2xy 69.

71.

12

15.

+

5

V3

lOv'3-3^ ^

,

31.

49. 6a

372x

33.

^- Va

51.

-4

35.

10xy2x

,

9^2

15

17.

-6 +

^

11 V6

3

Vm^n^ - Vmn m + 3r^-^ ,

55.

45. x

43. 5

3

53.

loVemn

9V5

W\f3

,r-

37.

+ 5v'z z-25 ^

z

2

2

7n/x-x

13.

x-3Vx'y' + 2Vx2y2-6y

73.

20u-27Vuv + 9v 16u - 9v

X

r-

2V6

63.

4-V6

+ 4VX + 4

x-4

65.

2V6 67.

5

3

A50

Answers

Chapter Review

Exercise 3-6 1.

1

f

2.

15. (A) 5.3

125

3.

X

-J

4.

X

10>» (B) 4.9

Not

5.

10""

real

81

6.

7.

— m^

8.

1

10.



(2xVF5

16. (A) 38,000,000 (B) 0.000 057

{m'-n^y/^

(B)

10xy=

19.

17. (A)

21.

13.

iVv?

\/(7zp (B)

VlSuv —

'

22.

23.

26.

v^

— 16

1

34.

- 3 Vs -

27.

—«—

1

-

35. 2

36.

8

37.

73

47. t>/'2



48.

5.24X10'

r-

-

31. 9

n/B

m'^

52.

3x

4

i/s

43.

-v

1

42.

-

lOx'^^v'^'

8.32X10'

(C)

(D)

5.29X10"^

2X10"^;

54.

32. 3

-

7TT

1

40^^706

+

372-2v/3

+

9

— ab

33.

256v5

1

44. v'^

125v'

8n'

6

+

45. u^

3y

v2/9

5.83X10-"

(B)

15x

-

41.

7x^

51.

50.

^2

30. 3

5y= ^-

40.

10« u*

—r

49.

V?

3

1

39. v'2

2

„l/30

-

29. >/35

27u« 38.

1

46.

28. Vs

49b^

3m2

53. (A)

2

,

24.

5v

7y 25. x/x^

uV

14.

y

nl/7 a'"

Vl4xv

ix^y*42\



1

12.

y

,

6x2V2x

20.



11.

u^

,

18. (A)

— 1

9.

0.002

55. (A)

-sVp

Va

(B)

Ta 56. (A) -6x(2xy2)3/'' (B) iw'^'^



7xV3x

65.

66.

13-5n/7

4mnV3mn"

81.

-

6

68. 6

-

TTo

74.

75.

6

u

+

82.

83. (A)

:

9a

-4b

Practice Test;

9.

25y' — —

3a' b^

llaVso^

10.

(p + — —^ q)^

3.

2.

36x"

lOx

69.

+

4.

3X10-^

60.

13>/xy

62.

61.

-

V

2x

(B)

3w-5

w

63. 11 Vs

3y

70. x

-

VxV + Vxy - y

78.

2V4p +

2x

-1 —1

8.

5bV5a^

v'4x^

,

77. 6

l

79.

-8x

Chapter

(A) -5y(3x^y)'/' (B)

3

—^z^ >/(x

11.

3m'V25mn^

2x

3

5.

pq

-2V2

2 76.

Oa-Sx/ob-eb

35

1.

2xvV2v

uv

Vo

73.

37^35

,-

80.

+ 2Vxy

9x-y

3

59.

2

6x 72.

2xv'V4x2

3v'

7zV2z

67.

14 71.

58.

3^2^

9V14

,

64.

5xV

57.

V4X

5u*

5xy"z5V2x^z



mVl5n

,,

6.

7.

,,

5n

yf

4X-11

12.

X

2

Chapter 4

Exercise 4-1 1.

3

7.

No

-1

31.

f

solution

49.

3.

27.

-f

47.

No

5.

29.

2

solution 33.

800 pounds

f

9.

11. All real

solution

(B)

second painting $9,000

1,400 pounds

numbers 13. -6 15. 24 17. 76 19. 70 21. 30 numbers 37. -J 39. -3 41. 25.000 43. 120

35. All real

-5 51.3 53.-2 55.-2

61. First painting $6,000; 65. (A)

No

57. f 59. $6,500 at 11%; $3,500 at 63. $7,200 at 10%; $4,800 at 15%

67. 24,000 gallons

69. 5.000 trout

18%

71. 12.6 years

23.

4

45. 5

25. 3

A51

Answers

Exercise 4-2 1.

17.

xs=3

xS=-2

3.

-4''

1

p)"



_1

>

"J

"^

[3, =^1

.3

^'^

\

-4] U

^-^—

= $4

1^

2]

35.



>"

1

\

Nosolution

A52

Answers

Exercise 4-5 d

1.

=-

r

=

3. r

t

3x 19.

+

C — 2n

C

V — ac

=

b

7.

n

-Ax - C y= z

15

y

=—

d

5.

21.

= |C +

F

32

R^

35.

47. x

2A

,

h=

25.

,

d

51.

r-

1-dt

/? 33l

=

9.

13.

2.

-i

Sx« 12;

2

No

23.

-3

-^

i,

+3 2y-4 y

x

p

27.

W

=-

28.

=—

fl

x>4;^

-l±\/7 44.

43.

51.

« -7

x

or x

^

30. 5

-)-^

< 4;

40. z

±

45.

. •

/ ^

-o

-3«xx j2.

10

2±>/22

±2^3

19. 2

4;

T"

"''^

-45«C«65

58.

^

>

"

64. $40,000 or 69. 12



x>-12;

-^^

-• -4

10%; $12,000

54. „^.

5

-3„,„„„.,. < x < or x >

-7 ±715 60.

59.

s

MA «

more

-5±V29 46.

3

->x

1
/t

61. x^

-

+ x"' + C

6

5,0006-°°^'

+

3,000

Exercise 14-2 I.

II.

A = l,000e'"'»' 3. A = 8,0006°°^' 5. Q = 3e"'""'; Q(10) = 2.01 milliliters

353 people

(B)

p(x)

=

lOOe"""^"

7.

N = L(l - e-°°=")

13. 24,200 years (approximately)

15.

9. I = loe-"™*'^''; x 104 times; 67 times

=

74 feet

17. (A) 7 people;

400

Exercise 14-3 ,.(£!::i)! 6

-

13.

+ C 3.^(2x-l)- + C 5^^^^^^ + C 7.t^^it^ + c 9.(5^^1±^+c ii.(^^il±^ + c 24

3

-

(x^

2x

-

S)'/^

+C

15.

^

29.

p

(1

+

+C

In .x)V2

23. i2[x''

=

-(e"

-

e"-)-'

+C

+

31. fi(x)

^

19.

=

"^^

+

25. x

=

ir

Vx^

+9-

3; fl(4)

^

(t^

+

= $2,000

5)'/^

=

2x)^

-

y

=

3(t^

-

10,000

12,000

+C

33. E(t)

(e" -

+C

4

3

2x^

4

9

--{4-x^Y/' + C

17.

18

3

21.

16

—-(3t'+1]-^ + C

27.

4)>''^

:

Vt

+

+C

E(15)

=

9,500 students

l

Exercise 14-4 1.

5

3.

5

5.

2

48

7.

9.

- -7

1

11. 2

-(e'-l)

13.

3

25.

-[(e^-2)=l-l 6^

2

-

-

P(2)

C(2)

=

29.

^

^

(IS^/^

15. 2 In 3.5

I

=1

1

[R'(x)

dx

-

= $2

-

S'-'^)

= $6

-2

31.

-

(22/'

-

3^^^)

4

thousand per day

C'(x)]dx

17.

19. 14

21.

5^=15,625

23. 12

'

6

'

35. (A) C(4)

(C) P(4)

-3 -In

27.

2

(B) R(4)

thousand per day

-

'

R(2)

=1

(10

= ^^ 33.—;^ e->-e 1-e", -

2x)dx

= $8

thousand per day

A102

Answers

("10

fS 37.

500(t-

I

= -$23,750;

12)dt

-

500((

I

fw

= -$11,250

12)dt

Js

Jo

r™ 41.

(12

+

-10.5 beats per minute

J49



f" 25

=

0.006(")dt

~

-295x"^''^dx

39.

134 billion cubic feet

43.

Jio

=

dt

100 items

Vt

Ji

Exercise 14-5 7

1.

16

7

3.

5.

-

9

7.

5

e'-e"'

9.

-In

11.

17. 36

15. 32

13. 15

0.5

19. 9

-

21.

4

27.-

29.

+ ln2-2e°=

23. 2e

25.

23 —

2

3

Consumers'

3

4

surplus =

1;

=-

producers' surplus

31. 5 years;

33. (A) Solve

$25,000

lOOe"'"'^''

=

lOe""^*

r2303

(1006-°°="

(B)

-31.62)dx

=

639.47

Exercise 14-6 1.

(Al 120 (B) 124

time

11. (A)

1,

(A)

5.

at this

29.6.54

27.0.791

25. 32;r/3

23. 64;r

(A) 123 (B) 124

3.

59(B) Not possible

-4

time

(B) -5.33 7. (A) -5 (B) -5.33 9. (A) 1.63 (B) Not possible at this 13.250 15.2 17.45/28 = 1.61 19. 2(1 - e'^) == 1.73 21. 26;t

31. (A)

=

I

-200t

+

600

(B)

f'

-1

(-2001

+

=

600)dl

300

33. $16,000

3 Jo

35. $7.18

37.

10°C

Chapter Review

Exercise 14-7 1.

('-(2

9.

2(5

+

+C 17)

2.

=

44

12 10.

3.

i

-3t-'-3t + C

-

(6x

15/2

4.

+ C 11.2

5)"/^

5.

-2e-/T5

/(t)d(

=

+C

x^

21. 4

-

8 In 1.4

thousand barrels

/(t)dl

I

+^

10.2; producers'

barrels; 10.000

I

x

=

1/3

30. 45 thousand;

xMn

x

.37 .3

I

Chapter 15

Practice Test:

1.

-l/6(x2

8.

ln[(x

-

+

3)'

l)/x]

+C +C

-e-'

2.

9.

+ C 3.-6""* -

x(ln x)^

2x In x

+

(x

4.

2x

+C

+

2)6"

+C

10. 7.5

+

5.

(In

8 In 2

x)V8

=

+C

13.05

6. -^

-—

-

x"

+C

7.



11. 2,642 barrels; 10,000 barrels

Chapter 16

Exercise 16-1 1.

10

3.

25. 2y-

1

33. R(p, q)

T(60, 27)

5.

7.

27. E(0, 0,

=

1

3);

= -5p2 +

9.

6pq

33 minutes

6

-

150

11.

F(2, 0, 3)

iq^

13.

16n

15. 791

+

200p

37. C(6. 8)

=

+

300q;

fi(2, 3)

75; C(8.1. 9)

=

90

19. 118 21. lOOe"' = 222.55 23. 2x + Ax $142 thousand; $150 thousand; $8 thousand loss

17. 0.192

29, $4,400; $6,000; $7,100

31.

= $1,280;

R{3, 2)

39. Q(12, 10)

=

= $1,175

35. T(70, 47)

120; Q(10, 12)

=

83

=

29 minutes;

A105

Answers

Exercise 16-2 11. 60 13. 2x - 2y + 6 15. 6 17. -2 19. 2 21. 26""+^'' 23. Ge^'-*-^" - y')^; /^(x, y) = -9y2(x^ - y^f 31. /,(x, y) = 24xy(3x^v - 1)'; 35. /,(x, y) = y^e^^^ /,(x, V) = 12x^(3xV - 1)' 33. /Jx, y) = 2x/(x^ + y']: /,(x, y) = 2y/(x^ + y') 39. /,Jx, y) = 2y^ + 6x; /y(x, y) = 2xy'e'"" + 2ye'"" 37. /,(x, y) = 4xyV(x^ + yT: f,{x. V) = -4x V(x' + yT /,,{x, y) = 4xy = /,„(x, y); ^x, y) = 2x^ 41. /,,(x, y) = -2y/x^; /,,(x, y) = -1/y^ + \/x' = /„(x, y); /,^(x. y) = 2x/y' 43. /«,(x, y) = (2y + xy^Je'"'; /^^(x, y) = (2x + x'y]e''y = /^Jx, y); /yy(x, y) = x^e''>' 45. x = 2 and y = 4 49. (A) 2x (B) 4y 51. C,(x, y) = 70: 47. /,x(x, y) + fyy[x, y) = {2y^ - 2y}]/[x} + y'^f + (2x2 _ 2y2)/(x2 + y^)^ = Increasing x by one unit and holding y fixed will increase costs by $70 at any production level; Cy(x, y) = 100: Increasing y by one unit and holding x fixed will increase costs by $100 at any production level 53. P^(l, 2) = 4: Profit will increase approximately $4 thousand per 1.000 increase in production of type A calculator at the (1. 2) output level; P,.(l, 2) = — 2: Profit will decrease approximately $2 thousand per 1,000 increase in production of type B calculator at the (1, 2) output level 55. dx/dp = — 5: A $1 increase in the price of brand A will decrease the demand for brand A by 5 pounds at any price level (p. q); dy/Sp = 2: A $1 increase in the price of brand A will increase the demand for brand B by 2 pounds at any price level (p, q) 57. Tv(70, 47) = 0.41 minute per unit increase in volume of air when V = 70 cubic feet and x = 47 feet; T^(70, 47) = —0.36 minute per unit increase in depth when V = 70 cubic feet and x = 47 feet 59. CvX"[(y + 2x^y) dx + (x + 2xy^) dy] xz dy + xy dz) 29. $10,460 31. $5.40 33.1.1k

*

364

47.12 cubic inches

35.14.896

Exercise 16-4 1.

/(— 2,

7. /(3, 2)

0)

=

=

saddle point

/(- 3, -18) type

A and

advertising;

10

33

is

is

maximum 3. /(—I, 3) = 4 is a local minimum 5. /has a saddle point at (3, —2) maximum 9. /(2, 2) = 8 is a local minimum 11. /has a saddle point at (0, 0) 13. 15. /has a saddle point at (0, 0); /(3, 18) = —162 and 0); /(I, 1) = — 1 is a local minimum a local

a local

at (0,

= -162

are local

minima

17.

The

test fails at (0. 0);

/has saddle points

at (2, 2)

and

(2,

-2)

/has

a

19. 2.000

Max P = P(2, 4) = $15 million 21. $3 million on research and development and $2 million on Max P = P(3, 2) = $30 million 23. 8 inches by 4 inches by 2 inches 25. 20 inches by 20 inches by 40 inches

4,000 type B;

Exercise 16-5 Max/(x, y)=/(3, 3) = 18 3. Min /(x, y) = /(3, 4) = 25 5. Max /(x, y) =/(3, 3) =/(-3, -3) = 18; min/(x, y) =/(3, —3) =/(— 3, 3) = —18 7. Maximum product is 25 when each number is 5 = -6 13. 60 of 9. Min/(x, y. z)=/(-4, 2. -6] = 56 11. Max/(x, y, z) = /(2, 2, 2) = 6; min/(x, y, z) = /(-2, -2, -2) model A and 30 of model B will yield a minimum cost of $32,400 per week 15. A maximum volume of 16,000 cubic inches occurs for a box 40 inches by 20 inches by 20 inches. 17. 8 inches by 8 inches by 8/3 inches 19. x = 50 feet and y = 200 feet; maximum area is 10,000 square feet 1.

A106

Answers

Exercise 16-6 y

1.

y

=

V

5.

+

0.7x

1

5

y ..I

11

y 13. y 11.

u. y y

17. (A) 19. (A)

= =

+

X

2

-)x

= 2.12X + 10.8; y = 63.8 when x = 25 9. = - 1.53X + 26.67; y = 14.4 when x = 8 = 0.75x2 - 3.45X + 4.75 y

y

7.

=

-^x

0.7x

y

= - 1.2x +

15. (A)

12.6;

=

y

= 0.382x +

y

when

10.2

1.265

=

x

2

$10,815

(B)

-^x

+ 112 (B) Demand, 140,000 units; + 69.2 (B) 69.2 parts per million

revenue, $56,000

y

21. (A)

11.9X

=

lO.lx

+

weeks

10.7 (B) 9

Exercise 16-7 1.

(A) 3x^y*

9.

330

C[x] (B) 3x^

(56-20V5)/3

11.

19.^1

3.

=i

r2o

r2

;



33.^1

[10

I

per million

=

15.

6xy

+

49

17.

0.626 4 1

ln2

- tL(x2 +

37. .^oms

W

21. j

f°'

0.4 Joe

,15(2'"

y2)] dj,

+ E(y)

5x

f'

dx

-

1)(20'

=^

Js

"-

(B)

f

35

+

+ y)^

(x

I

23.

29.

-

xo^y"" dy dx =

ji,

16

In J

2v + 3xy2 ^=— ;^dvdx = 1 + x^

f'

+

(A) 2x^

13.

J-,

f'

Jo

dy dx

(x/y)

I

27.

31.

+

25.

30y

dy dx

I

10' ")

=

30 In 2

=

+

E(y) (B)

dy dx

= =

((x, y)|0 =s

y

«

X

{(x, y)|0

100.86 feet

35.

= $20.8

M

39. ^^

Je

J50

^ 4 - x', « V4-y, R

is

« x « 2) « y « 4}

both a

regular x-region

and

A*

^

"'

''

I '

'

)

X

a regular

y-region

3.

R

= {(x,

y)|x^

12

V^

7.

9

billion

I

[100

-

I

100

- dy dx =

Jio

«y«

12

-

«x«

2x,

y y = lo|B»v

+4-

= | + ^e^ - e

y

Exercise 16-8 1. fl

Vy

8.375 or 8,375 items

insects per square foot

0.0000133xy2 dy dx

x^

=f

V

-^!— dy dx = 1 - X

+

Vy

(A)

xe"*'

j

\

J2.000

5.

-

2x



H

.

IS

,

a regular

x-region

2)

ISfx^

+ y^)] dy dx = 600 In

1.2

=

75 parts

109.4

Answers

5.

R

=

((x,

y)|iy'«xSy + R

"=1^

^^x = y ^ 10

17.

R =

("x

+

is

f

11.

9. |§

-|

13.

15.

ie*-|

a regular

^x

«

l,

x

«

R

19.

1}

=

{(x, y)|0

(68

«

-24V2)/15

Vy Jo

Jo

y

« 4x - x^

r4x-x!

("4

l

VTTTTydy dx = Jo

l.\

y-region

«ysx +

{(x. y)|0

("l

4

-2SyS4}

4,

+

x^

Ay dx

=

Jo

y y

=

+

X

1

y

=

4x



x'

-)x

21.

R

=

fl + Vi.

f4

Jo

->/x«y« I +

((x, y)|l

y-l)'dy dx =

Jl-^

^

n/x,

«

x

23

4)

IT'''

^ y

=

3

-

+

2y)

X

or

-y kx = -y 3

y

=

1

+

^> i-L— 5^

\/x

'

i

Ol

I

) )

5

-!-^x y

=

1

-

\/x

x^dxdy = fj

25.

Jo

27.

=

y

y

x dx dy

I

={

Jo J4y2

Jo

-

1

x'

or

)x

12

(4-x-y) dydx = f

29.

4

31.

Jo

Jo Jo

3

dy dx

4

=

j

Jo

1=1- x' y

=

4

-

X

-^

X

^dxdy = lnl7

33.

Jo

^x 1

5

ol

Jo

1

+ y'

35.

4ye'''

Jo Jo

"

dy dx

=e-

1

X

dx dy

=f

sx«

4}

A107

A108

Answers

Chapter Review

Exercise 16-9

2. d^z/dx' = 6xy^ d^z/dx dy = 10) = 2,900; /,(x, y) = 40; fy{x, y] = 70 = 4xV^ dx + 3x''y2 dy 5. 2xy' + 2y^ + C(x) 6. 3x"y^ + 4xy + E(y) 7. 9. /(2. 3) = 7:/y(x, y) = -2x + 2y + 3;/y(2, 3) = 5 10. (-8)(-6) - 4^ = 32 II. Az = /(l.l, 2.2) -/(I. 2) - 7.8897; dz = 4(1)^(0.1) + 4(2)^(0.2) = 6.8 I. /(5,

dz

4.

12.

y

= -1.5x +

15.5;

y=

0.5

whenx =

13.18

10

/T

44.

[x

+ yf

6x'y 1

dx dy

dz

3.

=

2

dx

+

3

dy

i

8.

= 408

Jo Jo

15. /,(x, y) 16. /,(x, y) 18. /(2, 3)

= =

= 2e'"=+2y. ^^^(x, y) = 4xe'"+2y + y')*:f^y{x. y) = 80xy(x2 + y^)^ 17. 25.076

2xe'"+^>'; fy[x. y)

10x(x2

= -25

a local

is

-IT'' + A v)

increase in product units of B. P(2. profit in sixth

3)

dv dx if

is

=f

a saddle point at (-2, 3)

y dx dy

22,

Jo Jo production of product

= $100

year

minimum; /has

thousand

is

I

Jio

minute per foot increase y = 68 when x = 40

in

depth

B

is

=^

1.

(A) 7 (B) 8

Az =

0.85;

9. /(O, 2)

11.

y

=

= —4

1.08x

/Jx, y)

2.

=

dz is

+

0,8

a local

-4x

minimum

18.3; cost of

Chapter 17

6.

x°V^ dy dx =

70 cubic feet and x

- 2); fy[x. y) = + 2x6'"'' 7.72

2x)^(2xy' 2x='ye'">'

10.

I

=

8; profit will

(1

by 6 inches by

7.764

or

20.

.

3) (B)

f increase $8,000 for 100 units

For 200 units of

2 inches

7,764 units

25.

y

=

0.63x

A and 300 + 1.33; 0.924

27. T,(70, 17)

Ji

when V =

= 4[x^y^ -

5.

= iJfx + -™

y

23. (A) P,(l. 3)

24. 8 inches

Practice Test:

4.

19.

held fixed at an output level of

maximum

a local

26.^1

$5.11 million

inches

I

producing 40 units

(1

is



x

=

28. 65.6k

29. 50,000

30.

y

= Jx + 48;

Chapter 16 12x'y^[x'y^ 8.

-

R,(30. 20)

— y) dy dx

$61,500

17 feet

=

2x)=

3.

dz

= $1,100;

=

3x^y''

Ry(20, 30)

+ 4x'y' dy = -$600

dx

Answers

13. x,

A109

Alio

Answers

19. (A) p(

(B)

X

-(:)' OSn.QSf p(x)

(C)

p(x)

21. .035

23. (A) p( X)

(B)

X

=

g)(.6)> (.4)"

p(x)

(C)

p(x)

Answers

x2

1

y

=

f(x)

>"

i

x
-

2 In 2

1— xe" — e'

e

=

29. fix) " '

otherwise

x>

1

'

^e-"^'" dx

35. (A)

=1-

=

2e-'

.7358

.3863

x/T5/4 =

E(X)=Jf

2.75; V(X)

.9682

4.

-$0.25

4

(B)

/^

=

1.2,

(T=.85

6.

(1

-ix) dx

=J=

fM

.75

I

Ul

7.

/i

=

y

ix-

ix^)

(0 x-ix" 1

dx

=f«

.6667; V(X1

= T

(x^

- jx^)

dx

-

(f)^

=f=

.2222;

x2

9.

2

-

>/2

=

.5858

10. .4938

11. .4641

a = ^2/3

=

.4714

4x 1

2

Answers

12. (A)

A113

A114

Answers

8889

F(x]= \f[l-{l/x)]

4.

1

I

T x„ =

5.

(A)

=

//

60

-f

10. E(X)

(T=6.48

(B)

(6x

Appendix

- x^)

.8764

6.

dx

=

5

~

TT

x«l l«x«10 x>10

10

1-8182

(A) .5762 (B) .0668

7.

^, expected

demand

8.

f

333 dozen; x„

is

ie""^'

=6-

dx

= e"' = .3679

2\/2,

9.

^

median demand

is

approx. 317 dozen

A Exercise A-1

= 3;14, 17 (C) Not an arithmetic progression Not an arithmetic progression (B) = -10; -22, -32 3.02 = 11,03 = 15 5.021 = 82,531 = 1,922 7. S20 = 930 9.2,400 11.1,120 13. Use Qi = 1 and d = 2 in S„ = (n/2)[2a, + (n - l)d] 15. Firm A: $280,500; firm B: $278,500 + $4 + $2 = $600 17. $48 + $46 + 1.

(A)

d

(D)

d







Exercise A-2 1.

3.

= — 2;

= — 8,

=

16

(A)

r

(D)

Not a geometric progression -6, 03 = 12, a, = -24 5. S^

02

=

S.

13. (B)

a,

= 1=1.6

Q;

15.0.999

Not

(B)

17.

=

547

a

7.

geometric progression

Qio

=

199.90

About $11,670,000

9. r

=

(C)

1.09

r

= |,

11. S,o

=

a,

= {,

1,242, S„

05

=

=

^

1,250

19. $31,027; $251,600

Exercise A-3 1.

21.

720

3.

10

5.

1,320

r) a* + r\ a^b +

23. x«

-

6x=

+

29. 5,005p^q« '^ ^

15x*

-

7.

(

"*

20x^

31. 264x^y»