INTRODUCTORY MATHEMATICAL INTRODUCTORY MATHEMATICAL ANALYSISANALYSISFor Business, Economics, and the Life and Social Sciences
2007 Pearson Education Asia
Chapter 17 Chapter 17 Multivariable CalculusMultivariable Calculus
2007 Pearson Education Asia
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
2007 Pearson Education Asia
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
2007 Pearson Education Asia
• 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
2007 Pearson Education Asia
• 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 17: Multivariable Calculus
Chapter ObjectivesChapter Objectives
2007 Pearson Education Asia
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 17: Multivariable Calculus
Chapter OutlineChapter Outline
17.4)
17.5)
17.6)
17.7)
2007 Pearson Education Asia
Lagrange Multipliers
Lines of Regression
Multiple Integrals
17.8)
17.9)
17.10)
Chapter 17: Multivariable Calculus
Chapter OutlineChapter Outline
2007 Pearson Education Asia
Chapter 17: Multivariable Calculus
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,
yxyxfz
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
xyxf
2007 Pearson Education Asia
Chapter 17: Multivariable Calculus
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
2007 Pearson Education Asia
Chapter 17: Multivariable Calculus
17.1 Functions of Several Variables
Example 5 – Sketching a SurfaceSketch the surfaceSolution: z = x2 is a parabola.
.2xz
2007 Pearson Education Asia
Chapter 17: Multivariable Calculus
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
yxfyhxfyxf
hx
,,lim,
0
h
yxfhyxfyxf
hy
,,lim,
0
2007 Pearson Education Asia
Chapter 17: Multivariable Calculus
17.2 Partial Derivatives
Example 1 – Finding Partial Derivatives
If f(x, y) = xy2 + x2y, 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
2007 Pearson Education Asia
Chapter 17: Multivariable Calculus
17.2 Partial Derivatives
Example 3 – Partial Derivatives of a Function of Three VariablesIf find
Solution:
By partial differentiating, we get
,,, 322 zzyxzyxf .,, and ,, ,,, zyxfzyxfzyxf zyx
xzyxfx 2,,
yzzyxfy 2,,
22 3,, zyzyxfz
2007 Pearson Education Asia
Chapter 17: Multivariable Calculus
17.2 Partial Derivatives
Example 4 – Finding Partial Derivatives
If find
Solution:
By partial differentiating, we get
,,,,22 tsrt
rsuutsrgp
. and ,
1,1,1,0t
p
t
p
s
p
22
2
222
22 2
srtt
srtru
tsrt
strsurutsrt
s
p
222
22222 2
2tsrt
srtrsu
s
psrttsrtrsu
t
p
0
1,1,1,0
t
p
2007 Pearson Education Asia
Chapter 17: Multivariable Calculus
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. held is when to respect with of change of rate the is
fixed. held is when to respect with of change of rate the is
xyzy
z
yxzx
z
6000857507.0, 2 yxxyxfc
2007 Pearson Education Asia
Chapter 17: Multivariable Calculus
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
85 and 7514.0
y
cx
x
c
2007 Pearson Education Asia
Chapter 17: Multivariable Calculus
17.3 Applications of Partial Derivatives
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.
2007 Pearson Education Asia
Chapter 17: Multivariable Calculus
17.3 Applications of Partial Derivatives
Example 3 – Marginal Productivity
Solution: Since ,
Thus, lk
l
k
P
lk
kklk
l
P
2 and
22
1 2/1
2/1lkP
2
5 and
10
1
16,40016,400
klkl k
P
l
P
2007 Pearson Education Asia
Chapter 17: Multivariable Calculus
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
022
yyx
xzx
z
0 2
02
0
22
22
zyxx
yz
x
z
xzyxzx
zyxxz
xy
xyx
xz
x
2
2,2,1
x
z
2007 Pearson Education Asia
Chapter 17: Multivariable Calculus
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
f fff
f fff
)(means and )( means
)(means and )( means
., 222 yxyxyxf
xyxyxfyyyxf
xyxyyxf
xyxx
x
42, and 22,
22,2
2
xyxyxfxyxf
yxxyxf
yxyy
y
42, and 2,
2,2
22
2007 Pearson Education Asia
Chapter 17: Multivariable Calculus
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
zz
y
x
zxy
xz
x
x
zyz
x
zyz
xx
z
x
2
2
21
2
1
2
1
z
y
x
z
2
0 422
13
22
2
2
z
z
y
z
yyz
x
z
2007 Pearson Education Asia
Chapter 17: Multivariable Calculus
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
2007 Pearson Education Asia
Chapter 17: Multivariable Calculus
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
ppqppc
qc 4002000 and 9000
2007 Pearson Education Asia
Chapter 17: Multivariable Calculus
17.6 Chain Rule
Example 1 – Chain Rule
Example 3 – Chain Rule
Solution: By the chain rule,
Thus,
11015.09000
015.0302
C
FC
FC
F
FC
C
CC
qpp
qp
q
p
c
p
q
q
c
p
c
2.1232,50
FC ppCp
c
a. Determine ∂y/∂r ifSolution: By the chain rule,
.3 and 6ln 642 srxxxy
6ln6
2312 4
4
45 x
x
xsrx
r
x
x
y
r
y
2007 Pearson Education Asia
Chapter 17: Multivariable Calculus
17.6 Chain Rule
Example 3 – Chain Rule
b. Given that z = exy, x = r − 4s, and y = r − s, find ∂z/∂r in terms of r and s.Solution: By the chain rule,
Since x = r − 4s and y = r − s,
xyeyxr
y
y
z
r
x
x
z
r
z
22 45
452 srsr
sryxrx
esrr
z
2007 Pearson Education Asia
Chapter 17: Multivariable Calculus
17.7 Maxima and Minima for Functions 17.7 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
.,,,, 2yxfyxfyxfyxD xyyyxx
2007 Pearson Education Asia
Chapter 17: Multivariable Calculus
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: Sincewe solve the system and get and
13522, 22 yxxyyxyxf
,0322, and 0524, yxyxfyxyxf yx
21
1
y
x
lkklklf 33,
,03, and 03, 22 lkklfklklf kl
0
0
k
l . 31
31
k
l
2007 Pearson Education Asia
Chapter 17: Multivariable Calculus
17.7 Maxima and Minima for Functions of Two Variables
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
2007 Pearson Education Asia
Chapter 17: Multivariable Calculus
17.7 Maxima and Minima for Functions of Two Variables
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, and 03, 22 xyyxfyxyxf yx
1, 6, 6, yxfyyxfxyxf xyyyxx
136166, 2 xyyxyxD
271
31
313
313
31
31
31 , f
2007 Pearson Education Asia
Chapter 17: Multivariable Calculus
17.7 Maxima and Minima for Functions of Two Variables
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
yxfyxyxf yxxyxf
yxyxfyxxyxf
xyyyxx
yx
2007 Pearson Education Asia
Chapter 17: Multivariable Calculus
17.7 Maxima and Minima for Functions of Two Variables
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
2007 Pearson Education Asia
Chapter 17: Multivariable Calculus
17.7 Maxima and Minima for Functions of Two Variables
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 ∂2P/∂p2A< 0, we indeed have a maximum.
01342 and 0122
32
BAB
BAA
BBAA
ppp
Ppp
p
P
qpqpP
800 1600 8002
2
2
2
2
ABBA pp
P
p
P
p
P
06 ,5.5 D
2007 Pearson Education Asia
Chapter 17: Multivariable Calculus
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: ConstraintConstruct 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
λy
λx
04
021
023
22 yx
λy
λx
2007 Pearson Education Asia
Chapter 17: Multivariable Calculus
17.8 Lagrange Multipliers
Example 3 – Minimizing CostsSuppose 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,, 2221
2121 qqqqλqqfc
2007 Pearson Education Asia
Chapter 17: Multivariable Calculus
17.8 Lagrange Multipliers
Example 3 – Minimizing Costs
Solution: We minimize c = f(q1, q2), given the constraint
q1 + q2 = 200. 2002002,, 21
2221
2121 qqλqqqqλqqF
0200
02
04
21
212
211
qqλ
F
λqqq
F
λqqq
F
150 ,50 21 qq
2007 Pearson Education Asia
Chapter 17: Multivariable Calculus
17.8 Lagrange Multipliers
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,,,, 222
121 yzλyxλyzxyλλzyxF
08
08
0
02
02
2
1
222
21
1
yzF
yxF
λyyF
λzλyzxF
λxyF
λ
λ
z
y
x
yz
yx
λ
λzλyzx
λx
y
88
1
022
222
21
1
2007 Pearson Education Asia
Chapter 17: Multivariable Calculus
17.9 Lines of Regression17.9 Lines of Regression
2
11
2
1111
2
n
ii
n
ii
n
iii
n
ii
n
ii
n
ii
xxn
yxxyx
a
2
11
2
111
n
ii
n
ii
n
ii
n
iii
n
ii
xxn
yxyxn
b
2007 Pearson Education Asia
Chapter 17: Multivariable Calculus
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).
2007 Pearson Education Asia
Chapter 17: Multivariable Calculus
17.9 Lines of Regression
Example 1 – Determining a Linear-Regression Line
Solution: Perform the arithmetic
83.921916
7362127486
3.8821916
27482173691
2
2
b
a
xy 83.93.88ˆ
2007 Pearson Education Asia
Chapter 17: Multivariable Calculus
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:
. 121
1
1
0
dxdyxx
3
2
23
22
12
1
1
231
1
10
1
1
1
0
xxx
dxyxy
dxdyx
x
x
2007 Pearson Education Asia
Chapter 17: Multivariable Calculus
17.10 Lines of Regression
Example 3 – Evaluating a Triple Integral
Find
Solution:
. 1
0 0 0
dxdydzxx yx
8
1
82
1
0
41
0 0
22
1
0 0
0
1
0 0 0
xdx
xyyx
dxdyxzdxdydzx
x
xyx
x yx