INTRODUCTORY MATHEMATICAL INTRODUCTORY MATHEMATICAL ANALYSISANALYSISFor Business, Economics, and the Life and Social Sciences
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Chapter 17 Chapter 17 Multivariable CalculusMultivariable Calculus
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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
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9. Additional Topics in Probability10. Limits and Continuity11. Differentiation12. Additional Differentiation Topics13. Curve Sketching14. Integration15. Methods and Applications of Integration16. Continuous Random Variables
17. Multivariable Calculus
INTRODUCTORY MATHEMATICAL ANALYSIS
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• 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
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• 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
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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)
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Lagrange Multipliers
Lines of Regression
Multiple Integrals
17.8)
17.9)
17.10)
Chapter 17: Multivariable Calculus
Chapter OutlineChapter Outline
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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,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.
23,
yxyxf
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Chapter 17: Multivariable Calculus17.1 Functions of Several Variables
Example 3 – Graphing a PlaneSketch the planeSolution:The plane intersects the x-axis when y = 0 and z = 0.
.632 zyx
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Chapter 17: Multivariable Calculus17.1 Functions of Several Variables
Example 5 – Sketching a SurfaceSketch the surfaceSolution: z = x2 is a parabola.
.2xz
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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
yxfyhxfyxfhx
,,lim,0
h
yxfhyxfyxfhy
,,lim,0
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Chapter 17: Multivariable Calculus17.2 Partial Derivatives
Example 1 – Finding Partial DerivativesIf 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
ff
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Chapter 17: Multivariable Calculus17.2 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
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Chapter 17: Multivariable Calculus17.2 Partial Derivatives
Example 4 – Finding Partial Derivatives
If find
Solution: By partial differentiating, we get
,,,, 22 tsrtrsuutsrgp
. and ,1,1,1,0t
ptp
sp
22
2
222
22 2
srttsrtru
tsrtstrsurutsrt
sp
222
22222 22
tsrtsrtrsu
spsrttsrtrsu
tp
0
1,1,1,0
tp
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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
xyzyz
yxzxz
6000857507.0, 2 yxxyxfc
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Chapter 17: Multivariable Calculus17.3 Applications of Partial DerivativesExample 1 – Marginal Costs
Solution: The marginal costs are
Thus,
and
85$
89$7510014.0
50,100
50,100
yc
xc
85 and 7514.0
ycx
xc
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Chapter 17: Multivariable Calculus17.3 Applications of Partial Derivatives
Example 3 – Marginal ProductivityA 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.
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Chapter 17: Multivariable Calculus17.3 Applications of Partial DerivativesExample 3 – Marginal Productivity
Solution: Since ,
Thus, lk
lkP
lkkklk
lP
2 and
221 2/1
2/1lkP
25 and
101
16,40016,400
klkl kP
lP
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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
xzxz
0 2
02
0
22
22
zyxx
yzxz
xzyxzxzyxxz
xy
xyxxz
x
2
2,2,1
xz
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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
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Chapter 17: Multivariable Calculus17.5 Higher-Order Partial Derivatives
Example 3 – Second-Order Partial Derivative of an Implicit FunctionDetermine ifSolution: By implicit differentiation,
Differentiating both sides with respect to x, we obtain
Substituting ,
2
2
xz
.2 xyz
0 2
2
z
zy
xzxy
xz
x
xzyz
xzyz
xxz
x
2
2
21
21
21
zy
xz
2
0 422
13
22
2
2
z
zy
zyyz
xz
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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
sy
yz
sx
xz
sz
ry
yz
rx
xz
rz
and
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Chapter 17: Multivariable Calculus 17.6 Chain Rule
Example 1 – Rate of Change of CostFor 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
FCFF
ppqppc
qc 4002000 and 9000
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Chapter 17: Multivariable Calculus17.6 Chain RuleExample 1 – Chain Rule
Example 3 – Chain Rule
Solution: By the chain rule,
Thus,
11015.09000015.0302
CFC
FC
F
FC
C
CC
qpp
qpq
pc
pq
qc
pc
2.1232,50
FC ppCpc
a. Determine ∂y/∂r ifSolution: By the chain rule,
.3 and 6ln 642 srxxxy
6ln
62312 44
45 x
xxsrx
rx
xy
ry
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Chapter 17: Multivariable Calculus17.6 Chain RuleExample 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,
xyeyxry
yz
rx
xz
rz
22 45
452 srsr
sryxrx
esrrz
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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 1Find relative maximum or minimum when
RULE 2 Second-Derivative Test for Functions of Two VariablesLet 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,
yxfyxf
y
x
.,,,, 2yxfyxfyxfyxD xyyyxx
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Chapter 17: Multivariable Calculus17.7 Maxima and Minima for Functions of Two Variables
Example 1 – Finding Critical PointsFind the critical points of the following functions.a.Solution: Sincewe solve the system and get
b.Solution: Sincewe solve the system and get and
13522, 22 yxxyyxyxf
,0322, and 0524, yxyxfyxyxf yx
21
1yx
lkklklf 33,
,03, and 03, 22 lkklfklklf kl
00
kl .
3131
kl
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Chapter 17: Multivariable Calculus17.7 Maxima and Minima for Functions of Two VariablesExample 1 – Finding Critical Points
c.
Since
we solve the system and get
1001002,, 22 yxzyxyxzyxf 0100,,
02,,04,,
yxzyxf
zyxzyxfzyxzyxf
z
y
x
1757525
zyx
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Chapter 17: Multivariable Calculus17.7 Maxima and Minima for Functions of Two Variables
Example 3 – Applying the Second-Derivative TestExamine 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
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Chapter 17: Multivariable Calculus17.7 Maxima and Minima for Functions of Two Variables
Example 5 – Finding Relative ExtremaExamine 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
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Chapter 17: Multivariable Calculus17.7 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
ppqppq
29400400
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Chapter 17: Multivariable Calculus17.7 Maxima and Minima for Functions of Two VariablesExample 7 – Profit Maximization
Solution: The total profit is given byThe 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
pppPpp
pP
qpqpP
800 1600 8002
2
2
2
2
ABBA ppP
pP
pP
06 ,5.5 D
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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
410 and
21,
23
λλ
yλ
x
04021023
22 yxλyλx
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Chapter 17: Multivariable Calculus17.8 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
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Chapter 17: Multivariable Calculus17.8 Lagrange MultipliersExample 3 – Minimizing Costs
Solution: We minimize c = f(q1, q2), given the constraint q1 + q2 = 200.
2002002,, 212221
2121 qqλqqqqλqqF
0200
02
04
21
212
211
qqλF
λqqqF
λqqqF
150 ,50 21 qq
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Chapter 17: Multivariable Calculus17.8 Lagrange Multipliers
Example 5 – Least-Cost Input CombinationFind 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
0808
002
02
2
1
222
21
1
yzFyxF
λyyFλzλyzxF
λxyF
λ
λ
z
y
x
yz
yxλ
λzλyzx
λx
y
88
102
2
222
21
1
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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
yxxyxa
2
11
2
111
n
ii
n
ii
n
ii
n
iii
n
ii
xxn
yxyxnb
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Chapter 17: Multivariable Calculus17.9 Lines of Regression
Example 1 – Determining a Linear-Regression LineBy 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).
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Chapter 17: Multivariable Calculus17.9 Lines of RegressionExample 1 – Determining a Linear-Regression Line
Solution: Perform the arithmetic
83.921916
7362127486
3.8821916
27482173691
2
2
b
a
xy 83.93.88ˆ
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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
32
2322
12
1
1
231
1
10
1
1
1
0
xxxdxyxy
dxdyx
x
x
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Chapter 17: Multivariable Calculus17.10 Lines of Regression
Example 3 – Evaluating a Triple IntegralFind
Solution:
. 1
0 0 0
dxdydzxx yx
81
82
1
0
41
0 0
22
1
0 00
1
0 0 0
xdxxyyx
dxdyxzdxdydzx
x
xyx
x yx
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