Chapter 17 - Multivariable Calculus

INTRODUCTORY MATHEMATICALINTRODUCTORY MATHEMATICAL
ANALYSISANALYSISFor Business, Economics, and the Life and Social Sciences
©2007 Pearson Education Asia
Chapter 17Chapter 17
Multivariable CalculusMultivariable Calculus

INTRODUCTORY MATHEMATICAL
ANALYSIS
0. Review of Algebra
1. Applications and More Algebra
2. Functions and Graphs
3. Lines, Parabolas, and Systems
4. Exponential and Logarithmic Functions
5. Mathematics of Finance
6. Matrix Algebra
7. Linear Programming
8. Introduction to Probability and Statistics

9. Additional Topics in Probability
10. Limits and Continuity
11. Differentiation
12. Additional Differentiation Topics
13. Curve Sketching
14. Integration
15. Methods and Applications of Integration
16. Continuous Random Variables
17. Multivariable Calculus
INTRODUCTORY MATHEMATICAL
ANALYSIS

• To discuss functions of several variables and to
compute function values.
• To compute partial derivatives.
• To develop the notions of partial marginal cost,
marginal productivity, and competitive and
complementary products.
• To find partial derivatives of a function defined
implicitly.
• To compute higher-order partial derivatives.
• To find the partial derivatives of a function by
using the chain rule.
Chapter 17: Multivariable Calculus
Chapter ObjectivesChapter Objectives

• To apply the second-derivative test for a
function of two variables.
• To find critical points for a function.
• To develop the method of least squares.
• To compute double and triple integrals.
Chapter ObjectivesChapter Objectives

Functions of Several Variables
Partial Derivatives
Applications of Partial Derivatives
Implicit Partial Differentiation
Higher-Order Partial Derivatives
Chain Rule
Maxima and Minima for Functions of Two Variables
17.1)
17.2)
17.3)
Chapter OutlineChapter Outline
17.4)
17.5)
17.6)
17.7)

Lagrange Multipliers
Lines of Regression
Multiple Integrals
17.8)
17.9)
17.10)
Chapter OutlineChapter Outline

17.1 Functions of Several Variables17.1 Functions of Several Variables
Example 1 – Functions of Two Variables
• A function can involve 2 or more variables, e.g.
( ) 22
2
,
yx
yxfz
−
==
a. is a function of two variables. Because the
denominator is zero when y = 2, the domain of f is all
(x, y) such that y ≠ 2.
b. h(x, y) = 4x defines h as a function of x and y. The
domain is all ordered pairs of real numbers.
c. z2
= x2
+ y2
does not define z as a function of x and y.
( )
2
3
,
−
+
=
y
x
yxf

17.1 Functions of Several Variables
Example 3 – Graphing a Plane
Sketch the plane
Solution:
The plane intersects the x-axis when y = 0 and z = 0.
.632 =++ zyx

17.1 Functions of Several Variables
Example 5 – Sketching a Surface
Sketch the surface
Solution:
z = x2
is a parabola.
.2
xz =

17.2 Partial Derivatives17.2 Partial Derivatives
• Partial derivative of f with respect to x is given
by
• Partial derivative of f with respect to y is given
by
( ) ( ) ( )
h
yxfyhxf
yxf
h
x
,,
lim,
0
−+
=
→
( ) ( ) ( )
h
yxfhyxf
yxf
h
y
,,
lim,
0
−+
=
→

17.2 Partial Derivatives
Example 1 – Finding Partial Derivatives
If f(x, y) = xy2
+ x2
y, find fx(x, y) and fy(x, y). Also, find
fx(3, 4) and fy(3, 4).
Solution: The partial derivatives are
Thus, the solutions are
( ) ( ) ( ) xyyyxyyxfx 221, 22
+=+=
( ) ( ) ( ) 22
212, xxyxyxyxfy +=+=
( )
( ) 334,3
404,3
=
=
xy
x
f
f

Example 3 – Partial Derivatives of a Function of
Three Variables
If find
Solution:
By partial differentiating, we get
( ) ,,, 322
zzyxzyxf ++= ( ) ( ) ( ).,,and,,,,, zyxfzyxfzyxf zyx
( ) xzyxfx 2,, =
( ) yzzyxfy 2,, =
( ) 22
3,, zyzyxfz +=

Example 4 – Finding Partial Derivatives
If find
Solution:
By partial differentiating, we get
( ) ,,,, 22
tsrt
rsu
utsrgp
+
==
( )
.and,
1,1,1,0t
p
t
p
s
p
∂
∂
∂
∂
∂
∂
( )( ) ( )( )
( )
( )
( )22
2
222
22
2
srtt
srtru
tsrt
strsurutsrt
s
p
+
−
=
+
−+
=
∂
∂
( ) ( ) ( )
( )222
2
2222 2
2
tsrt
srtrsu
s
p
srttsrtrsu
t
p
+
+
−=
∂
∂
⇒++−=
∂
∂ −
( )
0
1,1,1,0
=
∂
∂
t
p

17.3 Applications of Partial Derivatives17.3 Applications of Partial Derivatives
Example 1 – Marginal Costs
• Interpretation of rate of change:
A company manufactures two types of skis, the Lightning and the
Alpine models. Suppose the joint-cost function for producing x pairs
of the Lightning model and y pairs of the Alpine model per week is
where c is expressed in dollars. Determine the marginal costs ∂c/∂x
and ∂c/∂y when x = 100 and y = 50, and interpret the results.
fixed.heldiswhentorespectwithofchangeofratetheis
fixed.heldiswhentorespectwithofchangeofratetheis
xyz
y
z
yxz
x
z
∂
∂
∂
∂
( ) 6000857507.0, 2
+++== yxxyxfc

17.3 Applications of Partial Derivatives
Example 1 – Marginal Costs
Solution: The marginal costs are
Thus,
and
( )
( )
( )
85$
89$7510014.0
50,100
50,100
=
∂
∂
=+=
∂
∂
y
c
x
c
85and7514.0 =
∂
∂
+=
∂
∂
y
c
x
x
c

Example 3 – Marginal Productivity
A manufacturer of a popular toy has determined
that the production function is P = √(lk), where l is
the number of labor-hours per week and k is the
capital (expressed in hundreds of dollars per week)
required for a weekly production of P gross of the
toy. (One gross is 144 units.) Determine the
marginal productivity functions, and evaluate them
when l = 400 and k = 16. Interpret the results.

Example 3 – Marginal Productivity
Solution: Since ,
Thus, ( )
lk
l
k
P
lk
k
klk
l
P
2
and
22
1 2/1
=
∂
∂
==
∂
∂ −
( ) 2/1
lkP =
2
5
and
10
1
16,40016,400
=
∂
∂
=
∂
∂
==== klkl k
P
l
P

17.4 Implicit Partial Differentiation17.4 Implicit Partial Differentiation
Example 1 – Implicit Partial Differentiation
• We will look into how to find partial derivatives of a
function defined implicitly.
If , evaluate when x = −1, y = 2, and
z = 2.
Solution: Using partial differentiation, we get
02
2
=+
+
y
yx
xz
x
z
∂
∂
( ) ( )
( ) ( )
( )
0
2
02
0
22
2
2
≠
+
−=
∂
∂
=−++
∂
∂
+
∂
∂
=
∂
∂
+





+∂
∂
z
yxx
yz
x
z
xzyxz
x
z
yxxz
x
y
xyx
xz
x
( )
2
2,2,1
=
∂
∂
−x
z

17.5 Higher-Order Partial Derivatives17.5 Higher-Order Partial Derivatives
Example 1 – Second-Order Partial Derivatives
• We obtain second-order partial derivatives of f
as
Find the four second-order partial derivatives of
Solution:
yyyyxyyx
yxxyxxxx
ffff
ffff
)(meansand)(means
)(meansand)(means
( ) ., 222
yxyxyxf +=
( )
( ) ( ) xyxyxfyyyxf
xyxyyxf
xyxx
x
42,and22,
22,
2
2
+=+=
+=
( )
( ) ( ) xyxyxfxyxf
yxxyxf
yxyy
y
42,and2,
2,
2
22
+==
+=

17.5 Higher-Order Partial Derivatives
Example 3 – Second-Order Partial Derivative of an
Implicit Function
Determine if
Solution: By implicit differentiation,
Differentiating both sides with respect to x, we obtain
Substituting ,
2
2
x
z
∂
∂ .2
xyz =
( ) ( ) 0
2
2
≠=
∂
∂
⇒
∂
∂
=
∂
∂
z
z
y
x
z
xy
x
z
x
x
z
yz
x
z
yz
xx
z
x ∂
∂
−=
∂
∂
⇒





∂
∂
=





∂
∂
∂
∂ −− 2
2
2
1
2
1
2
1
z
y
x
z
2
=
∂
∂
0
422
1
3
2
2
2
2
≠−=





−=
∂
∂ −
z
z
y
z
y
yz
x
z

17.6 Chain Rule17.6 Chain Rule
• If f, x, and y have continuous partial derivatives,
then z is a function of r and s, and
s
y
y
z
s
x
x
z
s
z
r
y
y
z
r
x
x
z
r
z
∂
∂
∂
∂
+
∂
∂
∂
∂
=
∂
∂
∂
∂
∂
∂
+
∂
∂
∂
∂
=
∂
∂
and

17.6 Chain Rule
Example 1 – Rate of Change of Cost
For a manufacturer of cameras and film, the total
cost c of producing qC cameras and qF units of film is
given by
The demand functions for the cameras and film are
given by
where pC is the price per camera and pF is the price
per unit of film. Find the rate of change of total cost
with respect to the price of the camera when pC = 50
and pF = 2.
900015.030 +++= FFC qqqqcc
FCF
F
ppq
ppc
qc 4002000and
9000
−−==

17.6 Chain Rule
Example 1 – Chain Rule
Solution: By the chain rule,
Thus,
( ) ( )( )11015.0
9000
015.030 2
−++







−
+=
∂
∂
∂
∂
+
∂
∂
∂
∂
=
∂
∂
C
FC
F
C
F
FC
C
CC
q
pp
q
p
q
p
c
p
q
q
c
p
c
2.123
2,50
−≈
∂
∂
== FC ppCp
c
a. Determine ∂y/∂r if
( ) ( ) .3and6ln
642
srxxxy +=+=
( ) ( )





++
+
+=
∂
∂
∂
∂
=
∂
∂
6ln
6
2
312 4
4
4
5
x
x
x
srx
r
x
x
y
r
y

17.6 Chain Rule
b. Given that z = exy
, x = r − 4s, and y = r − s, find
∂z/∂r in terms of r and s.
Since x = r − 4s and y = r − s,
( ) xy
eyx
r
y
y
z
r
x
x
z
r
z
+=
∂
∂
∂
∂
+
∂
∂
∂
∂
=
∂
∂
( )
22
45
4
52 srsr
sry
xrx
esr
r
z +−
−=
−=
−=
∂
∂

17.7 Maxima and Minima for Functions17.7 Maxima and Minima for Functions
of Two Variablesof Two Variables
• Relative maximum at the point (a, b) is shown as
RULE 1
Find relative maximum or minimum when
RULE 2 Second-Derivative Test for Functions of Two Variables
Let D be the function defined by
1. If D(a, b) > 0 and fxx(a, b) < 0, relative maximum at (a, b);
2. If D(a, b) > 0 and fxx(a, b) > 0, relative minimum at (a, b);
3. If D(a, b) < 0, then f has a saddle point at (a, b);
4. If D(a, b) = 0, no conclusion.
( ) ( )yxfbaf ,, ≥
( )
( )


=
=
0,
0,
yxf
yxf
y
x
( ) ( ) ( ) ( )( ) .,,,,
2
yxfyxfyxfyxD xyyyxx −=

17.7 Maxima and Minima for Functions of Two Variables
Example 1 – Finding Critical Points
Find the critical points of the following functions.
a.
Solution:
Since
we solve the system and get
b.
Solution:
Since
we solve the system and get and
( ) 13522, 22
+−+−+= yxxyyxyxf
( ) ( ) ,0322,and0524, =−+−==+−= yxyxfyxyxf yx



=
−=
2
1
1
y
x
( ) lkklklf −+= 33
,
( ) ( ) ,03,and03, 22
=−==−= lkklfklklf kl



=
=
0
0
k
l
.
3
1
3
1



=
=
k
l

Example 1 – Finding Critical Points
c.
Since
we solve the system and get
( ) ( )1001002,, 22
−+−+++= yxzyxyxzyxf
( )
( )
( ) 0100,,
02,,
04,,
=+−−=
=−+=
=−+=
yxzyxf
zyxzyxf
zyxzyxf
z
y
x





=
=
=
175
75
25
z
y
x

Example 3 – Applying the Second-Derivative Test
Examine f(x,y) = x3
+ y3
− xy for relative maxima or
minima by using the second derivative test.
Solution: We find critical points,
which gives (0, 0) and (1/3, 1/3).
Now,
Thus,
D(0, 0) < 0  no relative extremum at (0, 0).
D(1/3,1/3)>0 and fxx(1/3,1/3)>0 relative minimum at (1/3,1/3)
Value of the function is
( ) ( ) 03,and03, 22
=−==−= xyyxfyxyxf yx
( ) ( ) ( ) 1,6,6, −=== yxfyyxfxyxf xyyyxx
( ) ( )( ) ( ) 136166,
2
−=−−= xyyxyxD
( ) ( ) ( ) ( )( ) 27
1
3
1
3
13
3
13
3
1
3
1
3
1
, −=−+=f

Example 5 – Finding Relative Extrema
Examine f(x, y) = x4
+ (x − y)4
for relative extrema.
Solution: We find critical points at (0,0) through
D(0, 0) = 0  no information.
f has a relative (and absolute) minimum at (0, 0).
( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) 0,12,02112,
04,044,
222
333
=−==−+=
=−−==−+=
yxfyxyxfyxxyxf
yxyxfyxxyxf
xyyyxx
yx

Example 7 – Profit Maximization
A candy company produces two types of candy, A and
B, for which the average costs of production are
constant at $2 and $3 per pound, respectively. The
quantities qA, qB (in pounds) of A and B that can be
sold each week are given by the joint-demand
functions
where pA and pB are the selling prices (in dollars per
pound) of A and B, respectively. Determine the selling
prices that will maximize the company’s profit P.
( )
( )BAB
ABA
ppq
ppq
29400
400
−+=
−=

Example 7 – Profit Maximization
Solution:
The total profit is given by
The profits per pound are pA − 2 and pB − 3,
The solution is pA = 5.5 and pB = 6.
Since ∂2
P/∂p2
A< 0, we indeed have a maximum.
( ) ( )
01342and0122
32
=+−=
∂
∂
=−+−=
∂
∂
−+−=
BA
B
BA
A
BBAA
pp
p
P
pp
p
P
qpqpP
8001600800
2
2
2
2
2
=
∂∂
∂
−=
∂
∂
−=
∂
∂
ABBA pp
P
p
P
p
P
( ) 06,5.5 >D

17.8 Lagrange Multipliers17.8 Lagrange Multipliers
Example 1 – Method of Lagrange Multipliers
• Lagrange multipliers allow us to obtain critical points.
• The number λ0 is called a Lagrange multiplier.
Find the critical points for z = f(x,y) = 3x − y + 6,
subject to the constraint x2
+ y2
= 4.
Solution: Constraint
Construct the function
Setting , we solve the equations to be
( ) 04, 22
=−+= yxyxg
( ) ( ) ( ) ( )463,,,, 22
−+−+−=−= yxλyxyxgλyxfλyxF
0=== λyx FFF
4
10
and
2
1
,
2
3
±=−== λ
λ
y
λ
x





=+−−
=−−
=−
04
021
023
22
yx
λy
λx

17.8 Lagrange Multipliers
Example 3 – Minimizing Costs
Suppose a firm has an order for 200 units of its
product and wishes to distribute its manufacture
between two of its plants, plant 1 and plant 2. Let q1
and q2 denote the outputs of plants 1 and 2,
respectively, and suppose the total-cost function is
given by
How should the output be distributed in order to
minimize costs?
( ) .2002,, 2
221
2
121 +++== qqqqλqqfc

Example 3 – Minimizing Costs
Solution: We minimize c = f(q1, q2), given the constraint
q1 + q2 = 200.
( ) ( )2002002,, 21
2
221
2
121 −+−+++= qqλqqqqλqqF









=+−−=
∂
∂
=−+=
∂
∂
=−+=
∂
∂
0200
02
04
21
21
2
21
1
qq
λ
F
λqq
q
F
λqq
q
F
150,50 21 == qq

Example 5 – Least-Cost Input Combination
Find critical points for f(x, y, z) = xy+ yz, subject to the
constraints x2
+ y2
= 8 and yz = 8.
Solution:
Set
We obtain 4 critical points:
(2, 2, 4) (2,−2,−4) (−2, 2, 4) (−2,−2,−4)
( ) ( ) ( )88,,,, 2
22
121 −−−+−+= yzλyxλyzxyλλzyxF









=+−=
=+−−=
=−=
=−−+=
=−=
08
08
0
02
02
2
1
22
2
21
1
yzF
yxF
λyyF
λzλyzxF
λxyF
λ
λ
z
y
x










=
=+
=
=−−+
=
y
z
yx
λ
λzλyzx
λ
x
y
8
8
1
02
2
22
2
21
1

17.9 Lines of Regression17.9 Lines of Regression
2
11
2
1111
2






−












−











=
∑∑
∑∑∑∑
==
====
n
i
i
n
i
i
n
i
ii
n
i
i
n
i
i
n
i
i
xxn
yxxyx
a
2
11
2
111






−












−
=
∑∑
∑∑∑
==
===
n
i
i
n
i
i
n
i
i
n
i
ii
n
i
i
xxn
yxyxn
b

17.9 Lines of Regression
Example 1 – Determining a Linear-Regression Line
By means of the linear-regression line, use the
following table to represent the trend for the index of
total U.S. government revenue from 1995 to 2000
(1995 = 100).

Example 1 – Determining a Linear-Regression Line
Solution: Perform the arithmetic
( ) ( )
( ) ( )
( ) ( )
( ) ( )
83.9
21916
7362127486
3.88
21916
27482173691
2
2
≈
−
−
=
≈
−
−
=
b
a
xy 83.93.88ˆ +=

17.10 Multiple Integrals17.10 Multiple Integrals
Example 1 – Evaluating a Double Integral
• Definite integrals of functions of two variables are
called (definite) double integrals, which involve
integration over a region in the plane.
Find
Solution:
( ) .12
1
1
1
0
dxdyx
x
∫ ∫−
−
+
( )
[ ]
3
2
23
2
2
12
1
1
231
1
1
0
1
1
1
0
=





++−=+=
+
−−
−
−
−
∫
∫ ∫
x
xx
dxyxy
dxdyx
x
x

Example 3 – Evaluating a Triple Integral
Find
Solution:
.
1
0 0 0
dxdydzx
x yx
∫∫ ∫
−
[ ]
8
1
82
1
0
41
0 0
2
2
1
0 0
0
1
0 0 0
=





=





−=
=
∫
∫∫∫∫ ∫
−
−
x
dx
xy
yx
dxdyxzdxdydzx
x
x
yx
x yx

Chapter 17 - Multivariable Calculus

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