Chapter 17 - Multivariable Calculus

INTRODUCTORY MATHEMATICAL INTRODUCTORY MATHEMATICAL ANALYSISANALYSISFor Business, Economics, and the Life and Social Sciences

2007 Pearson Education Asia

Chapter 17 Chapter 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 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


• 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.

23,

yxyxf


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


Chapter 17: Multivariable Calculus17.1 Functions of Several Variables

Example 5 – Sketching a SurfaceSketch the surfaceSolution: z = x2 is a parabola.

.2xz



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


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



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



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



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


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


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.


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



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



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


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



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


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


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


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



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


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


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



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



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



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


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



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


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


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


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



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


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).


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



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


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

Chapter 17 - Multivariable Calculus

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Transcript of Chapter 17 - Multivariable Calculus