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English Pages 10 [1240] Year 1986
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 AlbrightKnox 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.
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Inc.
Library of Congress CataloginginPublication 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.
8513322
510
ISBN 0023055901 Printing:
4 5 6 7 8
ISBN 0053055101
Year:
7 8
9
J.
Preface
PART
Algebra
I
CHAPTER
CHAPTER
1
2
Preliminaries
CHAPTER
3
4
4
11
Sets
12
Real
13
Inequality Statements and Line Graphs
14
Basic Operations on Signed
15
Positive Integer Exponents
41
16
Chapter Review
48
Numbers and
the Rules of Algebra
Numbers
Polynomials and Fractional Forms
12 21
28
55
21
Basic Operations on Polynomials
56
22
Factoring Polynomials
67
23
Multiplying and Dividing Fractions
79
24
Adding and Subtracting Fractions Chapter Review
88
25
CHAPTER
IX
Exponents and Radicals
96
101
31
Integer Exponents
32
Scientific Notation
111
33
Rational Exponents
117
34
Radicals
125
35
Basic Operations on Radicals
135
36
Chapter Review
141
Equations and Inequalities
102
147
41
Linear Equations
148
42
Linear Inequalities
160
111
iv
Contents
CHAPTER
PART
5
CHAPTER
Quadratic Equations
168
44
Nonlinear Inequalities
182
45
Literal Equations
190
46
Chapter Review
194
6
7
51
Cartesian Coordinate System and Straight Lines
200
52
Relations and Functions
213
53
Graphing Functions
226
54
Chapter Review
242
249
61
Simple Interest and Simple Discount
250
62
Compound
257
63
Future Value of an Annuity; Sinking Funds
64
Present Value of an Annuity; Amortization
273
65
Chapter Review
283
Interest
267
Systems of Linear Equations; Matrices 72
289
Review: Systems of Linear Equations Systems of Linear Equations and Augmented Matrices
—
290
Introduction
308 317
76
Gauss Jordan Elimination Matrices Addition and Multiplication by a Number Matrix Multiplication Inverse of a Square Matrix; Matrix Equations
77
Leontief InputOutput Analysis (Optional)
361
78
Chapter Review
367
73
74 75
8
Mathematics
Mathematics of Finance
71
CHAPTER
199
Graphs and Functions
Finite
II
CHAPTER
43
—
Linear Inequalities and Linear Programming Two
81
Linear Inequalities in
82
Systems of Linear Inequalities
83
Linear Programming in
Approach
Two
337 347
373 374
Variables in
330
Variables
Two Dimensions — A
379
Geometric 389
Contents
84 85
PART
9
Constraints
412 432
87
Constraints (Optional)
448
88
Chapter Review
470
477
Probability
478
91
Introduction
92
The Fundamental
93
Permutations, Combinations, and Set Partitioning
485
94
Experiments, Sample Spaces, and Probability of an Event
497
95
Empirical Probability
515
96
Union, Intersection, and Complement of Events
524
97
Chapter Review
541
10
Principle of Counting
The Derivative
479
Introduction
550
102
Limits and Continuity
551
103
Increments, Tangent Lines, and Rates of Change
569
104
The
582
105
Derivatives of Constants,
107 108
11
549
101
106
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
86
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
111
Implicit Differentiation
112
Related Rates
637
113
HigherOrder Derivatives
645
114
The
649
115
Marginal Analysis in Business and Economics
658
116
Chapter Review
663
Differential
vi
Contents
CHAPTER
CHAPTER
CHAPTER
CHAPTER
CHAPTER
12
13
14
15
16
Graphing and Optimization 121
Asymptotes; Limits
122
First Derivative
123
124
Second Derivative and Graphs Curve Sketching
125
Optimization; Absolute
126
Elasticity of
127
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
131
Exponential Functions
132
Logarithmic Functions
133
The Constant
134
Derivatives of Logarithmic Functions
798
135
Derivatives of Exponential Functions
804
136
Chapter Review
811
e
and Continuous Compound
772 778 Interest
791
815
Integration 141
Antiderivatives and Indefinite Integrals
142
Differential Equations
143
General Power Rule
839
144
Definite Integral
846
145
Area and the Definite Integral
146
Definite Integral as a Limit of a
147
Chapter Review
— Growth and Decay
816 829
854
Sum
866 879
Additional Integration Topics by Substitution
885 886
151
Integration
152
Integration by Parts
900
153
Integration Using Tables
907
154
Improper Integrals
916
155
Chapter Review
925
Multivariable Calculus
931
161
Functions of Several Variables
932
162
Partial Derivatives
942
163
Total Differentials
164
Maxima and Minima
and Their Applications
951
959
Contents
CHAPTER
17
APPENDIX A
APPENDIX B
vii
165
Maxima and Minima Using Lagrange
166
Method
167 168
Double Integrals over Rectangular Regions Double Integrals over More General Regions
1003
169
Chapter Review
1014
Multipliers
968 977
of Least Squares
990
1021
Additional Probability Topics 171
Random
172
Binomial Distributions
1033
173
1044
174
Continuous Random Variables Expected Value, Standard Deviation, and Median of Continuous Random Variables Uniform, Beta, and Exponential Distributions
1056
175
176
Normal
1077
177
Chapter Review
Variable, Probability Distribution,
and Expectation
Distributions
Al
Al
Arithmetic Progressions
A2
Geometric Progressions
A3
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 81 and Section
Appendix
A. Part
II
131),
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 1013 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 86 or 87 (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 87 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 1013 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 onesided
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 153 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 belowaverage 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 realworld 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 12).
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
14).
key steps
introduce
(see Section 14).
new terms and
highlight important
comments.
Answers
to
oddnumbered 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 oddnumbered 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 evennumbered 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
13
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 15
A
Replace each question mark with an appropriate expression. 1.
47
48
Preliminaries
13.
16
12
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 13
Inequality statements
and
line graphs, inequality symbols, less than,
greater than, less than or equal set, line
bol,
a
14
22
^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.
34
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
527.
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 35.
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
34
_
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
= V125 = 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)
67T253745
(B)
—12r2727
(C)
yty^ + 2vV32y
73 (D)
V62510^
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
35
Example 26
137
Multiply and simplify whenever possible.
V5(VT54)
(A)
l_= Vs
n/i5 Vs •4j
•
= V754V5 = V25 34V5 = 5\/34^ •
+
(V3
(B)
4)(V35)
=%/3
I
•
V3
+ 4V35V32OJ
= 3V320 = 17V3 (3 ^/5
(C)

+
2 73){>/5
V3)
j=3V5n/52VT5 + 3>/T52V3V3
l=35 + VT523
= 15 + Vl56 = 9 + +
(Va5)(Va
(D)
3)
r=7a
•

I
7l5
VasVa + sValS
I
= a2Va15 [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(VT57)
(C)
(2V33V2)(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 34
+
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^ = 53 =
rationalize the
2
denominator of
4
V5V3 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 •
^/HV3
_
4V5
V5
+ V3 — + V3
.
.
is
,^. , A and Multiply numerator ,
denominator by V5
+
V3.
+ 4v^
53 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)
^—= ^— ;^
31
•
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
35
Vs
2V5
2V53V3
V5
(B)
+ 3v^~"2V5 +
Multiply
2V53V3
3x/3
numerator and denominator by
10 3VT5 2027 10 3715 ^7
2n/53>/3.
10 +
IO3V15
(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
+ 3VXI8 V5
+
(B)
1+2I7
(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 35 In the following
problems
all \'ar]ables
represent positive real numbers.
Simplify by writing in simplest radical form and combining terms whenever possible.
+ 3V35V7
1.
3Vx8Vx
2.
5V73V7
3.
V7
4.
VmsVn + sVm
5.
n/i2

6.
V50
7.
VT8
8.
727^3
+
2V2
V^
4^/2
Multiply and simplify whenever possible. 9.
V5(^/53)

10.
12.
^(^fa
15.
^/5(2^/T5 3>/2)
17.
(n/5

13.
5)
3)(V5
+
3)
VTl(VTT2) ^fx[7

Vx)
11.
Vu(Vu
14.
Vz(5
16.
V7(3V34Vl4)
18.
(V3
+ 4)(V34)
+
+
3)
Vz)
140
Exponents and Radicals
+
19.
(V2
21.
(n/w2)(Vvv
5)2
+
2)
20.
(V53)2
22.
(Va
+
5)(Va5)
Rationalize each denominator and simplify. 1
V52
5
3
.. 24.
23.
^f7

26.
25.
4
1
+
3
^6
V3
72 27.
28.
V6
B
+
29.
30.
vTo2
2
yfa
+
3
Simplify by writing in simplest radical form and combining terms whenever possible. 31.
V5OXV8X
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 4556 multiply and simplify the product whenever possible.

47.
+ 5) (2V3 1)13^3 + 4)
49.
(3Va5)(2N/a
+
3)
51.
(5V23n/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.
(3V55)(2V53)
50.
(5Vv
52.
(4n/33V5)(2\/32V5)
54.
(2Vx3Vy)(2N/x
5 5
+ 
V5
62.
65.
+
2)(3>/v
+
5)
+ 3Vy) (v^ 2W](W^ 3Vb
56.
= =
+ 4)(Vz7)
46.
V2T. V2T.
Rationalize each denominator and simplify.
59.
3t/32x*
Chapter Review
36
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
36
+ 3Vy
74.
73.
+
3^lv
Chapter Review Important Terms
31
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""
32
Scientijic notation, scientific notation,
33
Rational exponents, root, square root, cube root, nth root, principal
nth 34
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)"'
35
Rasic operations on radicals, simplifying radical expressions, addition, subtraction, multiplication, rationalizing
Exercise 36
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.
8V3
142
Exponents and Radicals
In
Problems 714 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)
jj7
"
sIm'

to rational
exponent form
n'
Write each expression in simplest radical form. (AJJ variables represent positive real numbers.)
19.
36
46.
(27m^n^)'/3
Chapter Review
a_^^^
(i/4(i/3
47_
143
n4/5
Problems 51 and 52 multiply and express each answer using positive
In
exponents. 51.
3xV4(5xV42x3/4)
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)
01/3
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 6373 and write each answer in simplest radical form. 63.
5V252V^ +
65.
V^+Vi
67.
2z>/T6^
+
69.
[5^ 
^f^][2^
3745
3V^ + 3s/^]
64.
3xV27^2^/^
66.
V8U^,/H
68.
(2^53721(3^5
70.
(^V^V^)(y^ +
n/73
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^
32 77.
^3
+ vT' + 3^ _ q4
76.
(w'/5
+
2w^/=)(w'/=

3^"/=)
Write each expression in Problems 7882 in simplest radical /orm. (All variables represent positive real numbers.)
78.
716x^
2x
+4
79.
+1 1
V4x2 80.
2nV3m^mV3mn^+ mnV243mi?
81.
VVf +
83.
Simplify st/x*
^
(A)
Practice Test:
Chapter
r
For x
2^fa3^/b
/
V3^

82.
^_^
3Vx^:
5^
For x
(B)
'^..*'
»
CHAPTER
4
Contents 41 Linear Equations
42 Linear Inequalities 43 Quadratic Equations
44 Nonlinear Inequalities 45 Literal Equations
46 Chapter Review
41
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:
3x5 =
2x3y=12
5(2xl)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 firstdegree and zerodegree terms
is
called a linear or firstdegree equation in
one variable. For
example,
=

=
3x
+
5
are linear equations.
A
solution or root of an equation
2(3x5)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
41
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
3x9 = 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
— 14—4 r4x = —
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 99121 + 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
41
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
5X3X +
_ 111' X
which
Example
3
4
The LCD
is 6.
6
n
/ x + 2 _ x\ ^^^^ = 65 \
/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(x30)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
41
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 15X21 = 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.
41
Linear Equations
155
Applications The methods discussed
in this section
can be utiUzed
to solve a large
variety of practical problems.
Example
7
BreakEven 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
onetime 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 breakeven 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 41
1.
final
is,
cocoa butter
Problem 9
157
Linear Equations
41
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)
Breakeven 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
Rreake\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
41
[Note:
°F
=f
°C
A
+
32]
and game department number of rainbow trout in a certain lake using the popular capturemarkrecapture 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
9yearold 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
42
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 41,
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
42
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
64
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
64
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 longterm 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
64
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 wordprocessing 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 twopart 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
65
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)
65
Chapter Review Important Terms and
61
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
62
= Prt, A = P(l + rt), D = Mdf,
Compound nominal P(l
63
+
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
64
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 65
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
65
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?
65
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
71 Review:
72 Systems of Linear Equations
and Augmented Matrices
— Introduction
73 Gauss Jordan Elimination
— Addition and Multiplication by
74 Matrices
a
Number
75 Matrix Multiplication 76 Inverse of a Square Matrix; Matrix Equations 77 Leontief
78 Chapter
InputOutput 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 largescale systems. suitable for systems involving
suitable for systems involving larger
71
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
71
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)
x2y =
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.
71
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 = 45(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
71
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
realworld problems are readily solved by applying twoequation
twounknown methods. We Example jj.
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
71
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
ki
kj
(1)
kj
encountered frequently. In fact, systems of equations are so important in solving realworld 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 higherorder 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 threedimensional 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
71
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 + 3y5z = 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 + 4y8z = 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
x5z = 5
(5)
+ 2z=
(6)
13x [It
(4) (5)
9x6y + 12z= 18 4x + 6y10z = 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)
13x65z = 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)
x5(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
71
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)
18
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 twoequationtwounknown case. There, when we obtained = 0, we knew that there were infinitely many solutions. We shall have more to say about this in Section 73. solutions or none.
Applications
Now
let
us consider a realworld 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 laborhours 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
71
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 laborhours 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 71 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
71
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 Tshirts 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
Breakeven 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 threeequation
threeunknown 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 breakeven point. Interpret the regions between the lines to the left and to the right of the breakeven point.
Breakeven 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: oneperson, twoperson, and fourperson
models. Each boat requires the services of three departments as listed in the table.
The
cutting, assembly,
and packaging departments have
i
i
71
available a
Review: Systems of Linear Equations
maximum of 380, How many boats
respectively.
week
330, of
and 120 laborhours per week,
each type must be produced each
for the plant to operate at full capacity?
OnePerson
TwoPerson
Boat
Boat
0.6 hr
1.0
Cutting
department
Assembly department
0.6 hr
Packaging department
0.2 hr
hr
307
FourPerson 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
1948,41:450465.)
72
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 fiveequation fiveunknown 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 largescale 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 76, 77, 85. 86, and 87. 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
72
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 71. 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 rowequivalent, denoted by the symbol matrices, tions.
if
(Think about
this.)
rowequivalent 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
72
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
32(2)J=7
98^1
3+4^7
The above process
is
written more compactly as follows:
R1R2
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,
+^2RjRi '
Systems of Linear Equations and Augmented Matrices
72
— 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 twoequation twounknown 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 largerscale
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
72
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 72 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
73
GaussJordan Elimination
317
Solve using augmented matrix methods. 29.
3Xi 2x,
31.
3Xi 2Xi
=7 + 3X2 = 1 + 2X2 = 4 — X2 = 5 
0.2X10.5X2
33.
0.8X1
73
X2

0.3X2
30.
2Xi3x2 = 8
= 1 3X2 = 26 11X2 = — 0.3X10.6X2 = 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
73
(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
Gausslordan 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
= 2x33
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
73
Gauss Jordan Elimination
321
322
Systems of Linear Equations: Matrices
Steps 14 outlined in the solution of Example 15 are referred to as
GaussJordan 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
13) 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 GaussJordan 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
73
GaussJordan 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,
= tf = 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 GaussJordan elimination: 2x,
324
Systems of Linear Equations: Matrices
Solud'or.
2
73
Solution
First,
we summarize
the relevant manufacturing data in a table:
LaborHours per Sculpture
A Casting department Finishing department
GaussJordan 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 laborhours per week and a finishing capacity of 200 laborhours 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
allzero
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,
73
1
B
Gauss Jordan Elimination
327
328
Systems of Linear Equations; Matrices
25.
73
44.
45.
46.
47.
48.
Life Sciences
49.
GaussJordan 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.
74
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
74
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:
AB = 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:
[23
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.
74
(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
BA
(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 74 ProbJems 118
2
1
refer to the following matrices: 2
B =
3
D=
E=[4
3
1
2
3
C=
2]
74
1.3
2.5
8.3
1.4
Matrices
— Addition and Multiplication by a Number
ii
r4.1
1.
7J
0.7
2.
21.
2.6 22.
[
335
336
Systems of Linear Equations; Matrices
Life Sciences
32.
Heredity. Gregor nist,
made
Mendel (18221884), 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
75
75
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 Hajhs
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
= 106 + = 16
Problem 23
n row
1
a^
which we
[23
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 laborhours in the
fabricating department
and
1
laborhour 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
laborhours in the fabricating department and 1.5 laborhours 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.
75
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
75
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
Lefthand distributive property
Righthand 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.
75
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
75
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
76
76
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
76
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
76
Inverse of a Square Matrix; Matrix Equations
351
B
I
31
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
31
3
Mi
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
76
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
76
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.
76
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.
77
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, largescale 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.
TwoIndustry 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.
77
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
IM^
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:
77
We summarize
Solution to a
these results for convenient reference.
TwoCompany 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 712 pertain io the following inputoutput 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
78
78
Chapter Review
367
Chapter Review Important Terms and
71
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 72
—
Systems of linear equations and augmented matrices introduction, augmented matrix, column, row, equivalent sys
matrix, element,
tems, rowequivalent matrices,
+
Ri
73
row
operations. R, *»R. kR,
—»R,.
kR, ^ R;
GaussJordan elimination, reduced matrix, basic variables, nonbasic Gauss Jordan elimination
variables, submatrix,
74
Matrices
— addition and multiplication
sion of a matrix,
mXn
matrix, equal matrices, a matrix
matrix 75
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.
76
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~^ 77
Leontief inputoutput analysis fopfional). inputoutput analysis,
technology matrix,
Exercise 78
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 311 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 GaussIordan elimination: 2x2
2x,
+ +
x,
+
2X2
(A)
24.
78
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
56
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 GaussJordan 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
78
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,
+
Xi
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 laborhour and wage requirements for the manufacturing of two inexpensive hand calculators as given in matrices and N
M
below:
LaborHours per Calculator Fabricating
M^
N
8
Linear Inequalities and Linear
Programming
.i
CHAPTER
8
Contents 81 Linear Inequalities in
Two
Variables
Two
82 Systems of Linear Inequalities in
Variables
—
A Geometric Approach Simplex Method 85 The Simplex Method: Maximization with « Problem Constraints 86 The Dual; Minimization with ^ Problem Constraints 83 Linear
A
84
Programming
in
Two
Geometric Introduction
87 Maximization
Dimensions
to the
and Minimization with Mixed Problem Constraints
(Optional)
88 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
81
Linear Inequalities in
in
Chapter
Two
7 will
The row operations on matrices
be particularly useful in Sections
y
=
intro
and
87.
Variables
Having graphed linear equations such 2x3
we now
85, 86,
2x3y =
and
as
12
turn to graphing linear inequalities in two variables such as
y«2xf3
2x3y>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 halfpIanes; a nonvertiupper and lower halfplanes (Fig. 1).
plane into
cal line divides a plane into
above the
line
81
i
R
(4, y)
y
It is
X — 2 is the upper halfplane determined by the graph of y = X — 2, and y < x — 2 is the lower halfplane. 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
81
true.
Linear Inequalities in
Hence, the graph of the inequality
is
1
Graph 6x

3y
>
18.
Variables
the upper halfplane.
^^^^
Problem
Two
377
378
Linear Inequalities and Linear Programming
— 2«0x + y ; I
5
I 1
5
H x
Exercise 81 Graph each 1.
inequality.
= 3
x
=
3
X
x
82
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.
82
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
82
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 83 through 87), solution regions are often referred to as feasible regions.
Systems of Linear Inequalities in
82
lOx
+
8x
Problem 6
A
+
2y
=
Two
385
Variables
84
4y = 120
Feasible region
manufacturing plant makes two types of inflatable boats,
a
twoperson
boat and a fourperson boat. Each twoperson boat requires 0.9 laborhour
department and 0.8 laborhour in the assembly department. Each fourperson boat requires 1.8 laborhours in the cutting department and 1.2 laborhours 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 twoperson boats and y fourperson 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)
82
Systems of Linear Inequalities in
Two
Variables
387
Exercise 82
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
xly=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 lowgrade
ore, 3 tons of
two
Summit mine can
mediumgrade
ore,
and
2
The North Ridge mine mediumgrade ore, and 2
tons of highgrade ore per hour of operation.
can produce
1
ton of lowgrade ore,
1
ton of
tons of highgrade 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 lowgrade ore, 45 tons of mediumgrade ore, and 31


The Dual; Minimization with ^ Problem Constraints
86
tons of highgrade 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 singlesided and
doublesided disk drives. The disk drives are supplied by two other
companies, Associated Electronics and Digital Drives. Associated Electronics charges $250 for a singlesided disk drive and $350 for a
doublesided disk drive. Digital Drives charges $290 for a singlesided disk drive and $320 for a doublesided disk drive. Each
month Asso
ciated Electronics can supply at most 1,000 disk drives in
nation
of
monthly drives.
singlesided
total
Acme
and doublesided
each type should the 28.
its
at least 1,200 singlesided 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 doublesided 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.
87
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
87
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
+X2Mai
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
2X1X2
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:
87
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.
87
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
+ X22x3)^(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:
87
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
87
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?
87
Xl
Maximization and Minimization with Mixed Problem Constraints (Optional)
459
460
Linear Inequalities and Linear Programming
Summary of Methods
87
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
= RC = 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
87
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
015
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 = $86atXi=^,
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 87 In
ProbJems 38:
(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
87
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 1724 are mixed. Some can besoJved by the methods presen(ed Sections 85 17.
and
86, 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.
87
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 highprotein, lowcarbohydrate 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 3339. 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
87
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
88
Cost per Student School I
School
$4
$8
6
4
3
9
II
Chapter Review Important Terms
81
Linear inequalities in two variables, graph of a linear inequality in two
82
Systems of linear inequalities
and Symbols
variables,
upper halfplane, lower halfplane
solution region,
bounded
in
regions,
two variables, graphical solution,
unbounded
regions, corner point,
feasible solution, feasible region
83
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, constantprofit line, isoprofit line,
optimal solution, minimization problem, linear function, multiple optimal solutions, no feasible region, unbounded objective function 84
A geometric introduction
to the
simplex method, slack variables, basic
solution, basic feasible solution, nonbasic variables, basic variables,
simplex method 85
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,
88
86
The dual: minimizad'on
tvith 3^
Chapter Review
471
problem constrainfs. dual problem,
solution of minimization problems, transportation problem 87
Exercise 88
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
88
473
Chapter Review
Applications Business & Economics
19.
—
Manufacturing resource allocation. Set up (but do not solve) Problem 19 in Exercise 83 with the added restrictions that the fabricating department must operate at least 60 laborhours per day and the finishing department at least 12 laborhours 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.
88
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
121
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,
linixc 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 pointbypoint 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 pointbypoint 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 pointbypoint 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 115). of producing x radios is
tor radio are $5,000
The
total cost
C(x)
=
5,000
I
2x
Asymptotes; Limits
121
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 righthand 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 121 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
121
7.
/(x)=(X

Problems 912
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) (34 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 pointbypoint
plotting in regions of
uncertainty.
27.
/(x)
=
2xh4
=
X
29.
/(x)
31.
Let/(x)
32.
v2
—
=
=
/(x)
28.
x4
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 3336
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.
122
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.
122
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
FirstDerivative 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^ + 9x6 [1,5]
[1,3]
(B)
= 2x33x212x + [3,4]
= (xl)(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
= x26x +
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
125
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(21^
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?
126

maximum
126
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):
126
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
132
— 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
132
— 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.
132
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
132
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(xl)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(x2)
first.
58.
Problems 59 and
789
logb 21
6)log,„(x3)
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 carbon14 dating. Cosmicray bombardment of the atmosphere produces neutrons, which in turn react with nitrogen to produce radioactive carbon14. Radioactive carbon14 enters all living tissues through carbon dioxide which is first absorbed by plants. As long as a plant or animal is alive, carbon14 is maintained at a constant level in its tissues. Once dead, however, it ceases taking in carbon and the carbon14 diminishes by radioactive decay according to the
equation
A = Aoe°'''"
133
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
133
e
A=
amount
of carbon14
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 (17071783), 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
133
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
133
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 38 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
133
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 twostep process discussed Step
1.
f (x
in Section 104.
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
134
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 107.
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
134
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(4xl)[ln(2x2x)]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, lnx
= D,
_
1_
X
In X
Since
Ixl
=
x for x
>
Antiderivatives and Indefinite Integrals
141
Indefinite Integral
Formulas
1
6.
dx
e""
\
I I
8.
I
— = lnx + C dx — = +C
r
2.
D^
,
x>0 x^O
lnx
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(My)
^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,
lce
•
Learning
•
Longterm 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 142
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
pricedemand 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
pricesupply equation.
Differential Equations
142
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
= klLN]
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{LN]
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,000N)
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 carbon14
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 kthranked person participated was given approximately by
Nlk]
= N,e
where N,
is
0.11kl)
the
number
in the discussion.
N,
=
1
180, estimate
If,
«k «
10
of times the firstranked person participated
in a particular discussion
how many
group of ten people,
times the sixthranked person partici
The tenthranked person. One of the oldest laws WeberFechner 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
143
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)
143
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 107), 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, = (xir2x then
 lY2x
X
/
=
dx
——^ + C
Since
2.
= (xxT'(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^ti
the r
du
,
u"idx 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
143
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"rdxforin 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"
=
(x310)'/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
{x22xn(xl)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
=
143
/
(x22xn(xl)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(x23p''2d> 2(x23p''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
143
845
Exercise 143 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
(xl)Vx22x3dx
13.
3f dx
I
I
11.
(jx
(x'S^x^dx
8.
I
9.
6x2
I
(x^
7.
+
(5x
6.
I
B
5
I
(3x2)'dx
5.
+
V2x ^
4.
I
1
(x^x)N/x^2x2
14.
I
^ dx

+ 7dx
I
^ 17.
dx
,
J
(52x^)5
J
(e"

2x)^(e''
r Vl
+
In x J
19.
dx
18.
V4x'

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(n28r dn
dt
dy
3t 28.
v't24 e^
+ e"
(e'e'j^
^

dx
dm _
^^ '
dt
5x2
(x^7)* ln(t
 5)
t5
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
144
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 146.
Suppose a manufacturing company's marginal product C'(x)
is
cost equation for a given
given by
= 20.2x
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
144
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 141).
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 146
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).
(3x22x)dx = (x3x2)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 141.
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
144
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) = 100.5t, 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
= 62x
is
given by
147
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(436
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
= 404t 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 151 Integration by Substitution
152 Integration by Parts 153 Integration Using Tables
154 Improper Integrals 155 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.
151
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 143
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^ti
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 114 that
du
887
Integration by Substitution
151
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
151
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:
151
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
'
u2

2 lnu
+
c
lnx
Substitute u
+
2
+
c
If
c
is
= x2lnx +
2[
1
2
x
+
2
•
2
+C
=x+
2.
an arbitrary
constant, so
D,(x2lnx +
solve
= X + 2 for x: u = x + 2 x = u 2
du
^x + 22
Check
in the
we
Check by
is
C=
c
+
2.
differentiating.
Integration by Substitution
151
893
Thus,
x
=
dx J
+
Vx
2
= u222u du
'Vx
"
J
t
dx
= 2u
+
=
2
r
du
u
4)du
(2u^
J
= u34u + 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
151
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 +
lnu
+C
2
=  (x 1
r=
dx
1)2
c^i
=

2(u
+ 2(x 
5)
1)
+ lnx 
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(u5)du
u
U
X X
=
= 5 + Vx = 1 implies = 9 implies
(2u10lnu:
= (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)
ln8
(A)
(5
2x
+
+
x^
4)'
+C
(B)
6"'+=
+C
+C
+ eT' + C
(B)
[\nx]'/^
+C
u u
= =
6 8
Integration by Substitution
151
3.
(A)
4.

+
x
+ l+C
lnx
1
8 In 4
5.
=
(B)
(x
+
1^==
+
2(x
+
1)'/^
+C
x^lx^
+
gj'dx
11.1
5
Exercise 151
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.
Ji
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 x2
22.
J 21
I
J
^\ (x2)
dx
V4X
f^i— dx
J (x2)2
J 23.
f^^dx J
3
24.
f^^dx (X2F
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,
'^

,
,
VlOx
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 3134, find each indefinite integral two ways: = Vx — 1 and then use the substitution u = x —
substitution u
31.
I
J 33.
I
, ,
Vxl
dx
use the
dx
1
J
—^^ dx
34.
,
J
xVx
32.
first 3.
Vx1
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
Ja
that
pdx 2n/x
;
if/(x)=/(x)], then
Then show
dx
,
+ Vx2
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
151
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 pricede
equation.
Consumers' surplus. (Refer
to Section
145.)
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 coinoperated 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
+ xVl2x
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
0«
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
152
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 141
/
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
152
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
lC 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
152
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^fe"")
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
152
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'"

2ln6ln 3
1
=
1.4849
Exercise 152 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 518 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''
,
rrrdx 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
153
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
153
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 / VuMa^ du and J u^vuMo^ 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
  ln3x + + 5 5ln3x
+
ln3x
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
ln2x
You must
lncu
e
4
ln2x 2x
Notice that the constant of integration table.

7V2 14
formulas in the
b
your antiderivatives.
to find
153
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 810 and
select formula 8
f^=du = iln
5
^ dx = J
xV25
x2
a
+
with a^ n/q^
—
=
+ V25x=
In
Thus.
J3
x\/25
X
dx 2
5
=
In 5 In
25 and a
u^
5
+ V25x2
—
Formula
+C
we examine
x^,
=
5.
8
for
910
Additional Integration Topics
SoJufion
In order to relate this integral to
we
(formulas 1924),
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
Vu225
J
4
= ^f^;=dt
X
=—
u
4
n/u225
64 J
dx
=—
du
4
This
can be evaluated by using formula 23 with a
last integral
=
du
/
^lu^25
(uVu^
— Q' +
a^ lnu
+
Ju^
—
=
5:
Formula 23
a^)
2
Thus,
dx
/
V16x225
=
—
Vu225
1
(un/u^
128 1
Use formula 23 with
nil I
64 J
a
25 +
+
25 lnu
(4xV16x225 + 25
Vu^
ln4x
+

25)
+C
^116x^25]]
128
Problem 12
Find/V9x216dx.
Example 13
Find
/ VxMM Solution
None u
1
=
r
1
Vxi
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 1118. 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
153
We
recognize the
=
du
+
Vu^
formula 15 with a
last integral as
+
lnu
+
Vu
=
911
1:
Formula 15
a'l
a
Thus,
= if^=du f^dx Vu^ + +
J
Jx'
2 J
l
+
lnu
= 1 Problem 13
Find /xVx"
Example 14
Evaluate ^'
+
Vu^
+
l
+C
=
Substitute u
1.
x^.
+ VF + Tl + C
lnx^
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
—+—" u2 u^
I
J3 •"
2
^
Use formula
du
;
u
„ = b = 2, = A 4.
'
1,
= 2lu^Yl"l" + 2ylnu2j =
2(5
=4+

ln7
2 In
+
ln3)

— = 5.5243 7
3
with a
=
d
= 2,
and
,
2(3

ln5
+
lnl)
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 142
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(My)
J
equation by dt/[y[M
J
=
kt
+C
(1)
To evaluate dy f J
we
y(My)
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
My
and simplifying yields
(2)
153
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
xVx^ 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
xV4x''
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
xVQx
,
X
J
dx
,
dx
22.
^
24.
J
dx
x(l +n/x)=
—
Problems 2532 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
xVx

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.
,
.,
JAT^dx (xip
fx+l, ——— dx
+X X + 1 ———dx x + 3x
J Xr 32.
J
,
915
Integration Using Tables
153
Applications Business
& Economics
33.
Consumers' surplus. (Refer
Find the consumers'
to Section 145.)
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'lg
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
154
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?
rdx 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
154
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
154
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
_
(50go.o5,
soeo ") 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 161
= 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
162
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
HigherOrder Partial Derivatives Just as there are higherorder ordinary derivatives, there are higherorder partials,
and we
discuss local
will be using some of these in Section 164 when we maxima and minima. The following secondorder partials will
be useful:
SecondOrder 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
162
15)
15. increasing
profit 9.
=
6y2
(A)
12xy
(B)
(C)
54
(D)
54
Exercise 162
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 4748, show that the /unction
= =
47.
/(x, y)
48.
/{x, y)
49.
For f[x,
For/{x,
x^
y)
find:
/(xfAxy)/(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. Ay0
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 161
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
162
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 tenspeed The weekly demand and cost functions are

I
is
the price of a tenspeed bicycle, $q
the weekly
equations.
the weekly
demand
is
the price of a
tenspeed bicycles, y for threespeed bicycles, and C(x, y) is the cost
is
demand
function. Find R,(10,
Demand
and
+
I
threespeed bicycle, x
55.
12x
= 230  9x y = 130 lx4y = 200 80x 30y
where $p is
2),
+
profit functions.
and threespeed 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)
=
^ xl33
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
163
951
where
Q = IQ M = MA C = CA Find Qm[12,
163
10)
and Qc(12,
10),
and
interpret.
Total Differentials and Their Applications The
Total Differential
Approximations Using Differentials
The
Total Differential
Recall (Section 114) 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
163
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,
ACdC = = =
C,(x, y]
dx
[105x^/2 _
+ Cy(x,
y]
dy
ioxV2yV2] 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
VxMy^ 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
163
=
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 163 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
= x22xy +
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)
(
3j,
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 1538 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
163
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,
164
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.
164
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 secondderivative
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 secondorder 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 maximaminima 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 161,
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
164
/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 secondderivative
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 = 1003d2 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
168
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
168
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«6xx^
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
xfx^
otherwise
Graph both functions (on separate Repeat Problem 21
22.
In
sets of axes).
for
fe"
x^O
\
otherwise
.,,
to
Problems 2326 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 2730 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 3134, 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 
ba _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 fastfood 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)(/Jf 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
175
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
175
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.
176
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
176
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 bellshaped 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. 53\/5 11. u + 3V^ 4\/6 3V22\/3
25.
27.
2zV3
41.
39.
mn
29.
_ + 2Vx15
47. 14
7 + 3V5
31
n
57.
61. 5
59.

2
13xyV2xy 69.
71.
12
15.
+
5
V3
lOv'33^ ^
,
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 z25 ^
z
2
2
7n/xx
13.
x3Vx'y' + 2Vx2y26y
73.
20u27Vuv + 9v 16u  9v
X
r
2V6
63.
4V6
+ 4VX + 4
x4
65.
2V6 67.
5
3
A50
Answers
Chapter Review
Exercise 36 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.
r»
(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
+
3722v/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.
135n/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)
3w5
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.
OaSx/obeb
35
1.
2xvV2v
uv
Vo
73.
37^35
,
80.
+ 2Vxy
9xy
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
4X11
12.
X
2
Chapter 4
Exercise 41 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 42 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 45 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
1dt
/? 33l
=
9.
13.
2.
i
Sx« 12;
2
No
23.
3
^
i,
+3 2y4 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 142 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 143 ,.(£!::i)! 6

13.
+ C 3.^(2xl) + 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
{4x^Y/' + C
17.
18
3
21.
16
—(3t'+1]^ + C
27.
4)>''^
:
Vt
+
+C
E(15)
=
9,500 students
l
Exercise 144 1.
5
3.
5
5.
2
48
7.
9.
 7
1
11. 2
(e'l)
13.
3
25.
[(e^2)=ll 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 1e", 
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 145 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.
+ ln22e°=
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 146 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 147 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 161 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 162 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 164 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 165 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 166 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 167 1.
(A) 3x^y*
9.
330
C[x] (B) 3x^
(5620V5)/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', « V4y, R
is
« x « 2) « y « 4}
both a
regular xregion
and
A*
^
"'
''
I '
'
)
X
a regular
yregion
3.
R
= {(x,
y)x^
12
V^
7.
9
billion
I
[100

I
100
 dy dx =
Jio
«y«
12

«x«
2x,
y y = loB»v
+4
=  + ^e^  e
y
Exercise 168 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
xregion
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^
r4xx!
("4
l
VTTTTydy dx = Jo
l.\
yregion
«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
yl)'dy dx =
Jl^
^
n/x,
«
x
23
4)
IT'''
^ y
=
3

+
2y)
X
or
y kx = y 3
y
=
1
+
^> iL— 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
(4xy) 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 169
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 xix" 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
18182
(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 A1
= 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 A2 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 A3 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»