Functions of Several Variables Partial Derivatives

88

Functions of Several VariablesFunctions of Several Variables Partial DerivativesPartial Derivatives Maxima and Minima of Functions of Maxima and Minima of Functions of

Several VariablesSeveral Variables The Method of Least SquaresThe Method of Least Squares Constrained Maxima and Minima and Constrained Maxima and Minima and

the Method of Lagrange Multipliersthe Method of Lagrange Multipliers Double IntegralsDouble Integrals

Calculus of Several VariablesCalculus of Several Variables

8.18.1Functions of Several VariablesFunctions of Several Variables

z

y

x

f(x, y) = x2 + y2

x2 + y2 = 0

x2 + y2 = 1x2 + y2 = 4

x2 + y2 = 9

x2 + y2 = 16

Functions of Two VariablesFunctions of Two Variables

A real-valuedA real-valued function of two variables function of two variables ff, consists of, consists of

1.1. A set A set AA of of ordered pairsordered pairs of real numbers of real numbers ((xx, , yy)) called the called the domaindomain of the function. of the function.

2.2. A A rulerule that associates with each ordered pair in that associates with each ordered pair in the domain of the domain of ff one and onlyone and only oneone real number, real number, denoted by denoted by zz = = ff((xx, , yy))..

ExamplesExamples

Let Let ff be the function defined by be the function defined by

Compute Compute ff(0, 0)(0, 0), , ff(1, 2)(1, 2), and , and ff(2, 1)(2, 1)..

SolutionSolution

The The domaindomain of a function of of a function of two variablestwo variables ff((xx, , yy)), is a set of , is a set of ordered pairsordered pairs of real numbers and may therefore be of real numbers and may therefore be viewed as a viewed as a subset of thesubset of the xyxy-plane-plane..

2( , ) 2f x y x xy y 2( , ) 2f x y x xy y

2(0,0) 0 (0)(0) 0 2 2f 2(0,0) 0 (0)(0) 0 2 2f

2(1,2) 1 (1)(2) 2 2 9f 2(1,2) 1 (1)(2) 2 2 9f

2(2,1) 2 (2)(1) 1 2 7f 2(2,1) 2 (2)(1) 1 2 7f

Example 1, page 536

ExamplesExamples

Find the Find the domaindomain of the function of the function

SolutionSolution ff((xx, , yy)) is is defined for all real valuesdefined for all real values of of xx and and yy, so the , so the

domaindomain of the function of the function ff is the is the set of all points set of all points ((xx, , yy)) in thein the xyxy-plane-plane..

2 2( , )f x y x y 2 2( , )f x y x y

Example 2, page 536

ExamplesExamples


SolutionSolution gg((xx, , yy)) is is defined for all defined for all xx ≠≠ yy, so the , so the domaindomain of the function of the function

gg is the is the set of all points set of all points ((xx, , yy)) in thein the xyxy-plane -plane exceptexcept those those lying onlying on the the y y == x x line line..

2( , )g x y

x y

2

( , )g x yx y

x

y

y = x

Example 2, page 536

ExamplesExamples


SolutionSolution We require that We require that 1 – 1 – xx22 – – yy22 0 0 or or xx22 + + yy22 1 1 which is the which is the

set of all pointsset of all points ((xx, , yy)) lying lying on and inside on and inside thethe circle circle of of radiusradius 11 with with center center at the at the originorigin::

2 2( , ) 1h x y x y 2 2( , ) 1h x y x y

x

y

xx22 + + yy22 = 1 = 1

–1 1

1

–1Example 2, page 536

Applied Example:Applied Example: Revenue Functions Revenue Functions

Acrosonic manufactures a bookshelf Acrosonic manufactures a bookshelf loudspeaker systemloudspeaker system that may be bought that may be bought fully assembledfully assembled or or in a kitin a kit..

The The demand equationsdemand equations that relate the unit price, that relate the unit price, pp and and qq, , to the quantities demanded weekly, to the quantities demanded weekly, xx and and yy, of the , of the assembledassembled and and kit versionskit versions of the loudspeaker systems are of the loudspeaker systems are given bygiven by

a.a. What is the weekly What is the weekly total revenuetotal revenue function function RR((xx, , yy))??

b.b. What is the What is the domaindomain of the function of the function RR??

1 1 1 3300 240

4 8 8 8an d p x y q x y

1 1 1 3300 240

4 8 8 8an d p x y q x y

Applied Example 3, page 537


SolutionSolutiona.a. The weekly The weekly revenuerevenue from selling from selling xx units units assembled speaker assembled speaker

systemssystems at at pp dollars per unit dollars per unit isis given bygiven by xpxp dollars. dollars.Similarly, the weekly Similarly, the weekly revenuerevenue from selling from selling yy speaker kitsspeaker kits at at qq dollars per unit dollars per unit isis given bygiven by yqyq dollars. dollars.Therefore, the weekly Therefore, the weekly total revenuetotal revenue functionfunction RR is given by is given by

( , )R x y xp yq ( , )R x y xp yq

1 1 1 3300 240

4 8 8 8x x y y x y

1 1 1 3300 240

4 8 8 8x x y y x y

2 21 3 1300 240

4 8 4x y xy x y 2 21 3 1

300 2404 8 4

x y xy x y


1000 2000


SolutionSolutionb.b. To find the To find the domaindomain of the of the functionfunction RR, note that the , note that the

quantities quantities xx, , yy,, p p, and , and qq must be must be nonnegativenonnegative, which leads , which leads to the following to the following system of linear inequalitiessystem of linear inequalities::

1 1300 0

4 8x y

1 1300 0

4 8x y

1 3240 0

8 8x y

1 3240 0

8 8x y

0y 0y

0x 0x

Thus, the graph of the domain is:

x

y

2000

1000

1 1300 0

4 8x y

1 1300 0

4 8x y

1 3240 0

8 8x y

1 3240 0

8 8x y

D


PP(1, 2, 3)(1, 2, 3)

Graphs of Functions of Two VariablesGraphs of Functions of Two Variables

Consider the task of locating Consider the task of locating PP(1, 2, 3)(1, 2, 3) in in 3-space3-space:: One method to achieve this is to One method to achieve this is to start at thestart at the originorigin andand

measure out from theremeasure out from there, axis by axis:, axis by axis:

yy

xx

11

22

33

zz

PP(1, 2, 3)(1, 2, 3)


(1, 2)(1, 2)

yy

zz

xx

11

22

33

Consider the task of locating Consider the task of locating PP(1, 2, 3)(1, 2, 3) in in 3-space3-space:: Another common method is to Another common method is to find thefind the xyxy coordinate coordinate and and

from there elevatefrom there elevate to the level of the to the level of the z z value:value:

QQ(–1, 2, 3)(–1, 2, 3)


yy

xx

33

RR(1, 2, –2)(1, 2, –2)––22

SS(1, –1, 0)(1, –1, 0)

zz

LocateLocate the following the following pointspoints in in 3-space3-space::

QQ(–1, 2, 3)(–1, 2, 3),, RR(1, 2, –2)(1, 2, –2),, andand SS(1, –1, 0)(1, –1, 0). .

SolutionSolution


((xx, , yy))

((xx, , yy,, z z))

The The graphgraph of a function in of a function in 3-space 3-space is a is a surfacesurface. . For every For every ((xx, , yy)) in the in the domaindomain of of ff, there is a , there is a zz value value on the on the

surfacesurface..zz

yy

xx

z z == f f((xx, , yy))

Level CurvesLevel Curves

The graph of a The graph of a function of two variablesfunction of two variables is often is often difficult difficult to sketchto sketch..

It can therefore be It can therefore be usefuluseful to apply the to apply the methodmethod used to used to construct construct topographic mapstopographic maps..

This method is relatively This method is relatively easy to applyeasy to apply and conveys and conveys sufficient informationsufficient information to enable one to obtain a feel for the to enable one to obtain a feel for the graph of the function.graph of the function.

zz

yy

xx

In the In the 3-space 3-space graph we just saw, we can graph we just saw, we can delineate the delineate the contourcontour of the graph as it is of the graph as it is cutcut by a by a z z == c c plane: plane:

Level CurvesLevel Curves

z z == c c

ff((xx, , yy) = ) = cc

z z == f f((xx, , yy))

zz

y

x

ExamplesExamples Sketch a Sketch a contour mapcontour map of the function of the function ff((xx, , yy) = ) = xx22 + + yy22..SolutionSolution The function The function ff((xx, , yy) = ) = xx22 + + yy22 is a revolving parabola called is a revolving parabola called

a paraboloid.a paraboloid.

ff((xx, , yy) = ) = xx22 + + yy22

Example 5, page 540

– 4 – 2 2 4

4

2

– 2

– 4

zz

y

x

ExamplesExamples Sketch a Sketch a contour mapcontour map of the function of the function ff((xx, , yy) = ) = xx22 + + yy22..SolutionSolution A A level curvelevel curve is the is the graphgraph of the equation of the equation xx22 + + yy22 = = cc, which , which

describes a describes a circlecircle with with radiusradius . . Taking Taking different valuesdifferent values of of cc we obtain: we obtain:

y

x

ff((xx, , yy) = ) = xx22 + + yy22

xx22 + + yy22 = 0 = 0

xx22 + + yy22 = 1 = 1 xx22 + + yy22 = 4 = 4

xx22 + + yy22 = 9 = 9

xx22 + + yy22 = 16 = 16 xx22 + + yy22 = 0 = 0

xx22 + + yy22 = 1 = 1 xx22 + + yy22 = 4 = 4

xx22 + + yy22 = 9 = 9 xx22 + + yy22 = 16 = 16

c

Example 5, page 540

– 2 – 1 1 2

4

3

2

1

0

– 1

– 2

ExamplesExamples Sketch Sketch level curveslevel curves of the function of the function ff((xx, , yy) = 2) = 2xx22 – – yy

corresponding to corresponding to z z = –2= –2, , –1–1, , 00, , 11, and , and 22..SolutionSolution The The level curveslevel curves are the are the graphsgraphs of the equation of the equation 22xx22 – – yy = = kk

or for or for k k = –2= –2, , –1–1, , 00, , 11, and , and 22: :

y

x

22xx22 – – yy = –= – 22

22xx22 – – yy = –= – 11

22xx22 – – yy = 0= 0

22xx22 – – yy = 1= 1

22xx22 – – yy = 2= 2

Example 6, page 540

8.28.2Partial DerivativesPartial Derivatives

ff

x

x

y

y

f

x

f

x

f

y

f

y

x

x

y

y

2f f

x y x y

2f f

x y x y

2

2

f f

y y y

2

2

f f

y y y

x

x

y

y

2

2

f f

x x x

2

2

f f

x x x

2f f

y x y x

2f f

y x y x

2 2f f

y x x y

2 2f f

y x x y

When both are

continuous

First Partial DerivativesFirst Partial Derivatives

First Partial Derivatives of First Partial Derivatives of ff((xx, , yy) ) Suppose Suppose ff((xx, , yy) ) is a function of is a function of two variablestwo variables xx and and yy.. Then, the Then, the first partial derivativefirst partial derivative of of ff with respectwith respect to to xx

atat the pointthe point ((xx, , yy) ) isis

provided the limit exists.provided the limit exists. The The first partial derivativefirst partial derivative of of ff with respectwith respect to to yy atat the the

pointpoint ((xx, , yy) ) isis

provided the limit exists.provided the limit exists.

0

( , ) ( , )limh

f f x h y f x y

x h

0

( , ) ( , )limh

f f x h y f x y

x h

0

( , ) ( , )limk

f f x y k f x y

y k

0

( , ) ( , )limk

f f x y k f x y

y k

xx

yy

f

x

What does mean?f

x

What does mean?

Geometric Interpretation of the Partial DerivativeGeometric Interpretation of the Partial Derivative

zz

ff((xx, , yy))

zz


ff((xx, , yy))

y = by = b planeplane

aa

xx

yybb

((a, ba, b))

( , )f

f x bx

slope of ( , )

ff x b

x

slope of

ff((x, bx, b))ff((x, bx, b))

f

x

What does mean?f

x

What does mean?

zz

xx

yy

What does ean? mf

y

What does ean? mf

y


ff((xx, , yy))

xx

yy


ff((xx, , yy))

cc((cc, , dd))

x = c x = c planeplane

ff((c, yc, y)) ( , )slope of f

f c yy

( , )slope of

ff c y

y

What does ean? mf

y

What does ean? mf

y

zz

dd

ExamplesExamples

Find the Find the partial derivativespartial derivatives ∂f/∂x∂f/∂x and and ∂f/∂y∂f/∂y of the of the functionfunction

Use the partials to determine the Use the partials to determine the rate of changerate of change of of ff in the in the xx-direction-direction and in the and in the yy-direction-direction at the point at the point (1, 2)(1, 2) . .

SolutionSolution To compute To compute ∂f/∂x∂f/∂x, think of the , think of the variablevariable yy as a as a constantconstant and and

differentiatedifferentiate the resulting function of the resulting function of xx with respect towith respect to xx::

2 2 3( , )f x y x xy y 2 2 3( , )f x y x xy y

2 2 3( , )f yx y yx x 2 2 3( , )f yx y yx x

22f

xyx

22

f

xyx

Example 1, page 546

ExamplesExamples


Use the partials to determine the Use the partials to determine the rate of changerate of change of of ff in the in the xx-direction-direction and in the and in the yy-direction-direction at the point at the point (1, 2)(1, 2)..

SolutionSolution To compute To compute ∂f/∂y∂f/∂y, think of the , think of the variablevariable xx as a as a constantconstant and and

differentiatedifferentiate the resulting function of the resulting function of yy with respect towith respect to yy::

2 2 3( , )f x y x xy y 2 2 3( , )f x y x xy y

2 2 3( , )f y y yx x x 2 2 3( , )f y y yx x x

22 3xf

y yy

22 3x

fy y

y

Example 1, page 546

ExamplesExamples


Use the partials to determine the Use the partials to determine the rate of changerate of change of of ff in the in the xx-direction-direction and in the and in the yy-direction-direction at the point at the point (1, 2)(1, 2)..

SolutionSolution The The rate of changerate of change of of ff in the in the xx-direction-direction at the point at the point (1, 2)(1, 2)

is given byis given by

The The rate of changerate of change of of ff in the in the yy-direction-direction at the point at the point (1, 2)(1, 2) is given byis given by

2 2 3( , )f x y x xy y 2 2 3( , )f x y x xy y

2

(1,2)

2(1) 2 2f

x

2

(1,2)

2(1) 2 2f

x

2

(1,2)

2(1)(2) 3(2) 8f

y

2

(1,2)

2(1)(2) 3(2) 8f

y

Example 1, page 546

ExamplesExamples

Find the Find the firstfirst partial derivativespartial derivatives ofof the the functionfunction

SolutionSolution To compute To compute ∂w/∂x∂w/∂x, think of the , think of the variablevariable yy as a as a constantconstant and and


2 2( , )

xyw x y

x y

2 2( , )

xyw x y

x y

2 2( , )

x

yw

y

xyx

2 2( , )

x

yw

y

xyx

2

2 2

2

2

( ) (2 )

( )

w x x x

x x

y y y

y

2

2 2

2

2

( ) (2 )

( )

w x x x

x x

y y y

y

2 2

2 2 2

( )

( )

y y x

x y

2 2

2 2 2

( )

( )

y y x

x y

Example 2, page 547

ExamplesExamples


SolutionSolution To compute To compute ∂w/∂y∂w/∂y, think of the , think of the variablevariable xx as a as a constantconstant and and


2 2( , )

xyw x y

x y

2 2( , )

xyw x y

x y

2 2( , )

xyw

yxyx

2 2( , )

xyw

yxyx

2 2

2 22

( ) (2 )

( )

x x x

x

w y y y

y y

2 2

2 22

( ) (2 )

( )

x x x

x

w y y y

y y

2 2

2 2 2

( )

( )

x x y

x y

2 2

2 2 2

( )

( )

x x y

x y

Example 2, page 547

ExamplesExamples


SolutionSolution To compute To compute ∂g/∂s∂g/∂s, think of the , think of the variablevariable tt as a as a constantconstant and and

differentiatedifferentiate the resulting function of the resulting function of ss with respect towith respect to ss::

2 2 5( , ) ( )g s t s st t 2 2 5( , ) ( )g s t s st t

2 2 5( , ) ( )g s t sts t 2 2 5( , ) ( )g s t sts t

22 45( ) (2 )g

s st t tss

22 45( ) (2 )

gs st t ts

s

2 2 45(2 )( )s t s st t 2 2 45(2 )( )s t s st t

Example 2, page 547

ExamplesExamples


SolutionSolution To compute To compute ∂g/∂t∂g/∂t, think of the , think of the variablevariable ss as a as a constantconstant and and

differentiatedifferentiate the resulting function of the resulting function of tt with respect towith respect to tt::

2 2 5( , ) ( )g s t s st t 2 2 5( , ) ( )g s t s st t

2 2 5( , ) ( )g s t sts t 2 2 5( , ) ( )g s t sts t

42 25( ) ( 2 )s sg

t tstt

42 25( ) ( 2 )s s

gt tst

t

2 2 45(2 )( )t s s st t 2 2 45(2 )( )t s s st t

Example 2, page 547

ExamplesExamples


SolutionSolution To compute To compute ∂h/∂u∂h/∂u, think of the , think of the variablevariable vv as a as a constantconstant and and

differentiatedifferentiate the resulting function of the resulting function of uu with respect towith respect to uu::

2 2

( , ) u vh u v e 2 2

( , ) u vh u v e

2 2

( , ) u vh evu 2 2

( , ) u vh evu

2 2

2u vhe u

u

2 2

2u vhe u

u

2 2

2 u vue 2 2

2 u vue

Example 2, page 547

ExamplesExamples


SolutionSolution To compute To compute ∂h/∂v∂h/∂v, think of the , think of the variablevariable uu as a as a constantconstant and and

differentiatedifferentiate the resulting function of the resulting function of vv with respect towith respect to vv::

2 2

( , ) u vh u v e 2 2

( , ) u vh u v e

2 2

( , ) u vh v eu 2 2

( , ) u vh v eu

22

( 2 )vuhe v

u

22

( 2 )vuhe v

u

2 2

2 u vve 2 2

2 u vve

Example 2, page 547

ExamplesExamples


SolutionSolution Here we have Here we have a function ofa function of three variablesthree variables, , xx,, yy, and , and zz, and , and

we are required to computewe are required to compute

For short, we can For short, we can labellabel these these first partial derivativesfirst partial derivatives respectively respectively ffxx, , ffyy, and , and ffzz..

( , , ) lnyzw f x y z xyz xe x y ( , , ) lnyzw f x y z xyz xe x y

, , f f f

x y z

, , f f f

x y z

Example 3, page 549

ExamplesExamples


SolutionSolution To find To find ffxx, think of the , think of the variablesvariables yy and and zz as a as a constantconstant and and



( , ln, ) yzy z yz ew yf x x x x ( , ln, ) yzy z yz ew yf x x x x

lnyx

zyz ef y lnyx

zyz ef y

Example 3, page 549

ExamplesExamples


SolutionSolution To find To find ffyy, think of the , think of the variablesvariables xx and and zz as a as a constantconstant and and



( , , ) lnyzw f y yx z x xez yx ( , , ) lnyzw f y yx z x xez yx

yy

z xxz xzf e

y y

yz x

xz xzf ey

Example 3, page 549

ExamplesExamples


SolutionSolution To find To find ffzz, think of the , think of the variablesvariables xx and and yy as a as a constantconstant and and

differentiatedifferentiate the resulting function of the resulting function of zz with respect towith respect to zz::


( , ) ln, zyx y xyw f z yz ex x ( , ) ln, zyx y xyw f z yz ex x

zyzf xy xye zyzf xy xye

Example 3, page 549

The Cobb-Douglas Production FunctionThe Cobb-Douglas Production Function

The The Cobb-Douglass Production FunctionCobb-Douglass Production Function is of the form is of the form

ff((xx,, y y) =) = ax axbbyy11

–– bb (0 < (0 < b b < 1) < 1)

wherewhereaa and and bb are are positive constantspositive constants,, xx stands for the stands for the cost of laborcost of labor,,

yy stands for the stands for the cost of capital equipmentcost of capital equipment, and , and f f measures the measures the output of the finished productoutput of the finished product..

The Cobb-Douglas Production FunctionThe Cobb-Douglas Production Function

The The Cobb-Douglass Production FunctionCobb-Douglass Production Function is of the form is of the form

ff((xx,, y y) =) = ax axbbyy11

–– bb (0 < (0 < b b < 1) < 1)

The The first partial derivativefirst partial derivative ffxx is called the is called the marginal marginal

productivity of laborproductivity of labor. . ✦ It measures the It measures the rate of change of productionrate of change of production with respect with respect

to the amount of to the amount of money spent on labormoney spent on labor, with the level of , with the level of capital kept constantcapital kept constant..

The The first partial derivativefirst partial derivative ffyy is called the is called the marginal marginal productivity of capitalproductivity of capital. . ✦ It measures the It measures the rate of change of productionrate of change of production with respect with respect

to the amount of to the amount of money spent on capitalmoney spent on capital, with the level of , with the level of labor kept constantlabor kept constant..

Applied Example:Applied Example: Marginal Productivity Marginal Productivity

A certain country’s A certain country’s productionproduction in the early years following in the early years following World War II is described by the functionWorld War II is described by the function

ff((xx,, y y) =) = 3030xx2/32/3yy1/31/3

when when xx units of units of laborlabor and and yy units of units of capitalcapital were used. were used. Compute Compute ffxx and and ffyy..

Find the Find the marginal productivity of labormarginal productivity of labor and the and the marginal marginal productivity of capitalproductivity of capital when the when the amount expended on amount expended on labor and capitallabor and capital was was 125125 units and units and 2727 units, respectively. units, respectively.

Should the government have Should the government have encouraged capital encouraged capital investmentinvestment rather than rather than increase expenditure on laborincrease expenditure on labor to to increase the country’s productivity?increase the country’s productivity?



ff((xx,, y y) =) = 3030xx2/32/3yy1/31/3

SolutionSolution The The first partial derivativesfirst partial derivatives are are

1/31/3 1/32

30 203x

yf x y

x

1/31/3 1/32

30 203x

yf x y

x

2/3

2/3 2/3130 10

3y

xf x y

y

2/3

2/3 2/3130 10

3y

xf x y

y



ff((xx,, y y) =) = 3030xx2/32/3yy1/31/3

SolutionSolution The required The required marginal productivity of marginal productivity of laborlabor is given by is given by

or or 1212 units of output per unit units of output per unit increase inincrease in labor labor expenditure expenditure (keeping (keeping capital constantcapital constant).).

The required The required marginal productivity of marginal productivity of capitalcapital is given by is given by

or or 2727 7 7//99 units of output per unit units of output per unit increase inincrease in capitalcapital expenditure (keeping expenditure (keeping labor constantlabor constant).).

1/327 3

(125,27) 20 20 12125 5xf

1/327 3

(125,27) 20 20 12125 5xf

2/3

79

125 25(125,27) 10 10 27

27 9yf

2/3

79

125 25(125,27) 10 10 27

27 9yf



ff((xx,, y y) =) = 3030xx2/32/3yy1/31/3

SolutionSolution The government The government should definitely have should definitely have encouraged capital encouraged capital

investmentinvestment.. A A unitunit increase in capital expenditureincrease in capital expenditure resulted in a much resulted in a much

faster increase in productivityfaster increase in productivity than a than a unitunit increase in laborincrease in labor: : 2727 7 7//99 versus versus 1212 per unit of investment, respectively. per unit of investment, respectively.


Second Order Partial DerivativesSecond Order Partial Derivatives

The The first partial derivativesfirst partial derivatives ffxx((xx,, y y) ) andand ffyy((xx,, y y) ) of a function of a function

ff((xx,, y y) ) of two variablesof two variables x x and and y y are also functionsare also functions of of xx and and yy.. As such, As such, we may differentiatewe may differentiate each of the functions each of the functions ffxx and and ffyy

to obtain the to obtain the second-order partial derivativessecond-order partial derivatives of of ff..


DifferentiatingDifferentiating the function the function ffxx with respect towith respect to xx leads to the leads to the

second partial derivativesecond partial derivative

But the function But the function ffxx can also be can also be differentiateddifferentiated with respect with respect

toto yy leading to a different leading to a different second partial derivativesecond partial derivative

2

2( )xx x

ff f

x x

2

2( )xx x

ff f

x x

2

( )xy x

ff f

y x y

2

( )xy x

ff f

y x y


Similarly,Similarly, differentiating differentiating the function the function ffyy with respect towith respect to yy

leads to the leads to the second partial derivativesecond partial derivative

Finally, the function Finally, the function ffyy can also be can also be differentiateddifferentiated with with

respect torespect to xx leading to the leading to the second partial derivativesecond partial derivative

2

2( )yy y

ff f

y y

2

2( )yy y

ff f

y y

2

( )yx y

ff f

x y x

2

( )yx y

ff f

x y x


Thus, Thus, fourfour second-order partial derivativessecond-order partial derivatives can be can be obtained of a function of obtained of a function of two variablestwo variables::

ff

x

x

y

y

f

x

f

x

f

y

f

y

x

x

y

y

2f f

x y x y

2f f

x y x y

2

2

f f

y y y

2

2

f f

y y y

x

x

y

y

2

2

f f

x x x

2

2

f f

x x x

2f f

y x y x

2f f

y x y x

2 2f f

y x x y

2 2f f

y x x y

When both are

continuous

ExamplesExamples

Find the Find the second-order second-order partial derivativespartial derivatives of the function of the function

SolutionSolution First, calculate First, calculate ffxx and use it to find and use it to find ffxxxx and and ffxyxy::

3 2 2 2( , ) 3 3f x y x x y xy y 3 2 2 2( , ) 3 3f x y x x y xy y

2 23 6 3x xy y 2 23 6 3x xy y

3 2 2 2( 3 3 )xf x x y xy yx

3 2 2 2( 3 3 )xf x x y xy y

x

2 2(3 6 3 )xxf x xy yx

2 2(3 6 3 )xxf x xy y

x

6 6x y 6 6x y

2 2(3 6 3 )xyf x xy yy

2 2(3 6 3 )xyf x xy y

y

6 6x y 6 6x y

6( )x y 6( )x y 6( )y x 6( )y x Example 6, page 552

ExamplesExamples


SolutionSolution Then, calculate Then, calculate ffyy and use it to find and use it to find ffyxyx and and ffyyyy::

3 2 2 2( , ) 3 3f x y x x y xy y 3 2 2 2( , ) 3 3f x y x x y xy y

23 6 2x xy y 23 6 2x xy y

3 2 2 2( 3 3 )yf x x y xy yy

3 2 2 2( 3 3 )yf x x y xy y

y

2( 3 6 2 )yxf x xy yx

2( 3 6 2 )yxf x xy y

x

6 6x y 6 6x y

2( 3 6 2 )yyf x xy yy

2( 3 6 2 )yyf x xy y

y

6 2x 6 2x

6( )y x 6( )y x 2(3 1)x 2(3 1)x Example 6, page 552

ExamplesExamples


SolutionSolution First, calculate First, calculate ffxx and use it to find and use it to find ffxxxx and and ffxyxy::

2

( , ) xyf x y e2

( , ) xyf x y e

22 xyy e22 xyy e

2

( )xyxf e

x

2

( )xyxf e

x

22( )xyxxf y e

x

22( )xy

xxf y ex

24 xyy e24 xyy e

22( )xyxyf y e

y

22( )xy

xyf y ey

2 232 2xy xyye xy e 2 232 2xy xyye xy e 2 22 (1 )xyye xy 2 22 (1 )xyye xy

Example 7, page 553

ExamplesExamples


SolutionSolution Then, calculate Then, calculate ffyy and use it to find and use it to find ffyxyx and and ffyyyy::

2

( , ) xyf x y e2

( , ) xyf x y e

2

2 xyxye2

2 xyxye

2

( )xyyf e

y

2

( )xyyf e

y

2

(2 )xyyxf xye

x

2

(2 )xyyxf xye

x

2 232 2xy xyye xy e 2 232 2xy xyye xy e

2

(2 )xyyyf xye

y

2

(2 )xyyyf xye

y

2 2

2 (2 )(2 )xy xyxe xy xy e 2 2

2 (2 )(2 )xy xyxe xy xy e 2 22 (1 2 )xyxe xy 2 22 (1 2 )xyxe xy

2 22 (1 )xyye xy 2 22 (1 )xyye xy

Example 7, page 553

8.38.3Maxima and Minima Maxima and Minima of Functions of Several Variablesof Functions of Several Variables

yx

z

(a, b) (c, d)

(e, f )

(g, h)

Relative Extrema of a Function of Two VariablesRelative Extrema of a Function of Two Variables

Let Let ff be a be a functionfunction defined on a defined on a regionregion RR containing the containing the pointpoint ((aa, , bb))..

Then, Then, ff has a has a relative maximumrelative maximum at at ((aa, , bb)) if if ff((xx, , yy)) ff((aa, , bb)) for for all pointsall points ((xx, , yy)) that that are are sufficiently closesufficiently close to to ((aa, , bb))..✦ The number The number ff((aa, , bb)) is called a is called a relative relative

maximum valuemaximum value.. Similarly, Similarly, ff has a has a relative minimumrelative minimum at at ((aa, , bb))

if if ff((xx, , yy)) ff((aa, , bb)) for for all pointsall points ((xx, , yy)) that are that are sufficiently closesufficiently close to to ((aa, , bb))..✦ The number The number ff((aa, , bb)) is called a is called a relative relative

minimum valueminimum value..

yyxx

zz

((aa, , bb))

There is a There is a relative maximumrelative maximum at at ((aa, , bb))..

Graphic ExampleGraphic Example

yyxx

zz

((cc, , dd))

There is an There is an absolute maximumabsolute maximum at at ((cc, , dd))..(It is also a (It is also a relative maximumrelative maximum))


yyxx

zz

((ee, , f f ))

There is a There is a relative minimumrelative minimum at at ((ee, , f f ))..


yyxx

zz

((gg, , hh))

There is an There is an absolute minimumabsolute minimum at at ((gg, , hh))..(It is also a (It is also a relative minimumrelative minimum))


zz

yy

xx

At a At a minimum pointminimum point of the graph of a function of two of the graph of a function of two variables, such as point variables, such as point ((aa, , bb)) below, below, the plane tangent to the the plane tangent to the graph of the function is graph of the function is horizontal horizontal (assuming the surface of (assuming the surface of the graph is smooth):the graph is smooth):

Relative MinimaRelative Minima

((aa, , bb))

zz

yy

xx

Thus, at a Thus, at a minimum pointminimum point, the graph of the function has a , the graph of the function has a slope of zeroslope of zero along a direction along a direction parallelparallel to the to the xx-axis-axis::


( , ) 0f

a bx

( , ) 0

fa b

x

((aa, , bb))

zz

yy

xx

( , ) 0f

a by

( , ) 0

fa b

y

((aa, , bb))

Similarly, at a Similarly, at a minimum pointminimum point, the graph of the function , the graph of the function has a has a slope of zeroslope of zero along a direction along a direction parallelparallel to the to the yy-axis-axis::


At a At a maximum pointmaximum point of the graph of a function of two of the graph of a function of two variables, such as point variables, such as point ((aa, , bb)) below, below, the plane tangent to the the plane tangent to the graph of the function is graph of the function is horizontal horizontal (assuming the surface of (assuming the surface of the graph is smooth):the graph is smooth):

yy

xx

Relative MaximaRelative Maxima

zz

((aa, , bb))

zz

yy

xx


((aa, , bb))

Thus, at a Thus, at a maximum pointmaximum point, the graph of the function has a , the graph of the function has a slope of zeroslope of zero along a direction along a direction parallelparallel to the to the xx-axis-axis::

( , ) 0f

a bx

( , ) 0

fa b

x

zz

yy

xx


((aa, , bb))

Similarly, at a Similarly, at a maximum pointmaximum point, the graph of the function , the graph of the function has a has a slope of zeroslope of zero along a direction along a direction parallelparallel to the to the yy-axis-axis::

( , ) 0f

a by

( , ) 0

fa b

y

xx

Saddle PointSaddle Point In the case of a saddle point, In the case of a saddle point, bothboth partials are equal to zeropartials are equal to zero, ,

but the point is but the point is neither a maximum nor a minimumneither a maximum nor a minimum..

yy

zz

In the case of a saddle point, the function is at a In the case of a saddle point, the function is at a minimumminimum along one vertical planealong one vertical plane……

xx

Saddle PointSaddle Point

yy

zz

((aa, , bb))

( , ) 0f

a bx

( , ) 0

fa b

x

zz

… … but at a but at a maximummaximum along the perpendicular vertical planealong the perpendicular vertical plane..

xx

Saddle PointSaddle Point

yy

((aa, , bb))

( , ) 0f

a by

( , ) 0

fa b

y

((aa, , bb))

zz

yy

xx

((aa, , bb,, f f((aa,, b b))))

Extrema When Partial Derivatives are Not DefinedExtrema When Partial Derivatives are Not Defined A A maximummaximum (or (or minimumminimum) may also occur ) may also occur when both when both

partial derivatives are not definedpartial derivatives are not defined, such as point , such as point ((aa, , bb)) in the in the graph below:graph below:

Critical Point of a FunctionCritical Point of a Function

A A critical pointcritical point of of ff is a point is a point ((aa, , bb)) in the in the domaindomain of of f f such that bothsuch that both

or or at least oneat least one of the of the partial derivativespartial derivatives does not existdoes not exist..

( , ) 0 ( , ) 0an df f

a b a bx y

( , ) 0 ( , ) 0an d

f fa b a b

x y

Determining Relative ExtremaDetermining Relative Extrema

1.1. FindFind the the critical pointscritical points of of ff((xx, , yy)) by solving the system of by solving the system of simultaneous equationssimultaneous equations

ffxx = 0 = 0 ffyy = 0 = 0

2.2. The The second derivative testsecond derivative test: Let: Let

DD((xx, , yy) = ) = ffxx xx ffyyyy – f – f 22xy xy

3.3. Then,Then,a.a. DD((aa, , bb)) > 0> 0 and and ffxxxx((aa, , bb)) < 0< 0 impliesimplies that that ff((xx, , yy)) has a has a

relative maximumrelative maximum at the point at the point ((aa, , bb))..b.b. DD((aa, , bb)) > 0> 0 and and ffxxxx((aa, , bb)) > 0> 0 impliesimplies that that ff((xx, , yy)) has a has a

relative minimumrelative minimum at the point at the point ((aa, , bb))..c.c. DD((aa, , bb)) < 0< 0 impliesimplies that that ff((xx, , yy)) has has neitherneither a a relative relative

maximummaximum nornor a a relative minimumrelative minimum at the point at the point ((aa, , bb)), it , it has instead a has instead a saddle pointsaddle point..

d.d. DD((aa, , bb)) = 0= 0 impliesimplies that the that the test is inconclusivetest is inconclusive, so some , so some other techniqueother technique must be used to solve the problem. must be used to solve the problem.

ExamplesExamples

Find the Find the relative extremarelative extrema of the function of the function

SolutionSolution We have We have ffxx = 2 = 2xx and and ffyy = 2 = 2yy..

To To findfind the the critical pointscritical points, we , we setset ffxx = 0 = 0 and and ffyy = 0 = 0 and and solvesolve

the resulting system of the resulting system of simultaneous equationssimultaneous equations

22xx = 0 = 0 and and 22yy = 0 = 0

obtaining obtaining xx = 0 = 0, , yy = 0 = 0, or , or (0, 0)(0, 0), as the , as the sole critical pointsole critical point.. Next, apply the Next, apply the second derivative testsecond derivative test to determine the to determine the

naturenature of the of the critical pointcritical point (0, 0)(0, 0). .

We compute We compute ffxxxx = 2 = 2, , ffyyyy = 2 = 2, and , and ffxyxy = 0 = 0, ,

Thus,Thus, D D((xx, , yy) = ) = ffxxxx ffyyyy – f – f 22xyxy = (2)(2) – (0) = (2)(2) – (0)22 = 4 = 4..

2 2( , )f x y x y 2 2( , )f x y x y

Example 1, page 561

ExamplesExamples


SolutionSolution We haveWe have D D((xx, , yy) = 4) = 4, and , and in particularin particular, , DD(0, 0) = 4(0, 0) = 4.. Since Since DD(0, 0) > 0 (0, 0) > 0 and and ffxxxx = 2 = 2 > 0> 0, we conclude that , we conclude that ff has a has a

relative minimumrelative minimum at the point at the point (0, 0)(0, 0).. The The relative minimum valuerelative minimum value, , ff (0, 0) = 0(0, 0) = 0, also happens to , also happens to

be the be the absolute minimumabsolute minimum of of ff..

2 2( , )f x y x y 2 2( , )f x y x y

Example 1, page 561

ExamplesExamples


SolutionSolution2 2( , )f x y x y 2 2( , )f x y x y

The The relative minimum relative minimum valuevalue, , ff(0, 0) = 0(0, 0) = 0, also , also happens to be the happens to be the absolute minimumabsolute minimum of of ff::

z

y

x

f(x, y) = x2 + y2

Absolute Absolute minimumminimum atat (0, 0, 0) (0, 0, 0)..Example 1, page 561

ExamplesExamples


SolutionSolution We have We have

To To findfind the the critical pointscritical points, we , we setset ffxx = 0 = 0 and and ffyy = 0 = 0 and and solvesolve

the resulting system of the resulting system of simultaneous equationssimultaneous equations

66xx – 4 – 4yy – 4 = 0 – 4 = 0 and and – 4– 4xx + 8 + 8yy + 8 = 0 + 8 = 0

obtaining obtaining xx = 0 = 0, , yy = –1 = –1, or , or (0, –1)(0, –1), as the , as the sole critical pointsole critical point.. Next, apply the Next, apply the second derivative testsecond derivative test to determine the to determine the

naturenature of the of the critical pointcritical point (0, –1)(0, –1). .

We compute We compute ffxxxx = 6 = 6, , ffyyyy = 8 = 8, and , and ffxyxy = – = – 44, ,

Thus,Thus, D D((xx, , yy) = ) = ffxx xx · · ffyyyy – f – f 22xyxy = (6)(8) – (– = (6)(8) – (– 4)4)22 = 32 = 32..

2 2( , ) 3 4 4 4 8 4f x y x xy y x y 2 2( , ) 3 4 4 4 8 4f x y x xy y x y

6 4 4 4 8 8 and x yf x y f x y 6 4 4 4 8 8 and x yf x y f x y

Example 2, page 562

ExamplesExamples


SolutionSolution We haveWe have D D((xx, , yy) = 32) = 32, and , and in particularin particular, , DD(0, –1) = 32(0, –1) = 32.. Since Since DD(0, –1) > 0 (0, –1) > 0 and and ffxxxx = 6 = 6 > 0> 0, we conclude that , we conclude that ff has a has a

relative minimumrelative minimum at the point at the point (0, –1)(0, –1).. The The relative minimum valuerelative minimum value, , ff (0, –1) = 0(0, –1) = 0, also happens to , also happens to

be the be the absolute minimumabsolute minimum of of ff..

2 2( , ) 3 4 4 4 8 4f x y x xy y x y 2 2( , ) 3 4 4 4 8 4f x y x xy y x y

Example 2, page 562

ExamplesExamples


SolutionSolution We have We have

To To findfind the the critical pointscritical points, we , we setset ffxx = 0 = 0 and and ffyy = 0 = 0 and and solvesolve the resulting system of the resulting system of simultaneous equationssimultaneous equations

The The first equationfirst equation impliesimplies that that xx = 0 = 0, while the , while the second second equationequation implies implies that that yy = –1 = –1 or or yy = 3 = 3..

Thus, there are Thus, there are two critical pointstwo critical points of of ff : : (0, –1)(0, –1) and and (0, 3)(0, 3).. To apply the To apply the second derivative testsecond derivative test, we calculate, we calculate

ffxxxx = 2 = 2 ffyyyy = 24( = 24(yy – 1) – 1) ffxyxy = 0 = 0

DD((xx, , yy) = ) = ffxx xx · · ffyyyy – f – f 22xyxy = (2) = (2)· · 24(24(yy – 1) – (0) – 1) – (0)22 = 48( = 48(yy – 1) – 1)

3 2 2( , ) 4 12 36 2f x y y x y y 3 2 2( , ) 4 12 36 2f x y y x y y

22 12 24 36 andx yf x f y y 22 12 24 36 andx yf x f y y

22 0 12 24 36 0 nd ax y y 22 0 12 24 36 0 nd ax y y

Example 3, page 562

ExamplesExamples


SolutionSolution Apply the Apply the second derivative testsecond derivative test to the to the critical pointcritical point (0, –1)(0, –1): :

We haveWe have D D((xx, , yy) = 48() = 48(yy – 1) – 1).. In particularIn particular, , DD(0, –1) = 48[(–1) – 1] = –(0, –1) = 48[(–1) – 1] = – 9696.. Since Since DD(0, –1) = –(0, –1) = – 96 < 0 96 < 0 we conclude that we conclude that ff has a has a saddle saddle

pointpoint at at (0, –1)(0, –1).. The The saddle point valuesaddle point value is is ff (0, –1) = 22(0, –1) = 22, so there is a saddle , so there is a saddle

point at point at (0, –1, 22)(0, –1, 22)..

3 2 2( , ) 4 12 36 2f x y y x y y 3 2 2( , ) 4 12 36 2f x y y x y y

Example 3, page 562

ExamplesExamples


SolutionSolution Apply the Apply the second derivative testsecond derivative test to the to the critical pointcritical point (0, 3)(0, 3): :

We haveWe have D D((xx, , yy) = 48() = 48(yy – 1) – 1).. In particularIn particular, , DD(0, 3) = 48[(3) – 1] = 96(0, 3) = 48[(3) – 1] = 96.. Since Since DD(0, –1) = 96 > 0 (0, –1) = 96 > 0 and and ffxx xx (0, 3) = 2(0, 3) = 2 > 0> 0,, we conclude that we conclude that

ff has a has a relative minimumrelative minimum at the point at the point (0, 3)(0, 3). . The The relative minimum valuerelative minimum value, , ff (0, 3) = –106(0, 3) = –106, so there is a , so there is a

relative minimum at relative minimum at (0, 3, –106)(0, 3, –106)..

3 2 2( , ) 4 12 36 2f x y y x y y 3 2 2( , ) 4 12 36 2f x y y x y y

Example 3, page 562

Applied Example:Applied Example: Maximizing Profit Maximizing Profit

The total The total weekly revenueweekly revenue that Acrosonic realizes in that Acrosonic realizes in producing and selling its loudspeaker system is given byproducing and selling its loudspeaker system is given by

where where xx denotes the number of denotes the number of fully assembled unitsfully assembled units and and yy denotes the number ofdenotes the number of kits kits produced and sold produced and sold each weekeach week..

The total The total weekly costweekly cost attributable to the attributable to the productionproduction of of these loudspeakers isthese loudspeakers is

Determine Determine how manyhow many assembled unitsassembled units and and how manyhow many kitskits should be producedshould be produced per week per week to maximize profitsto maximize profits..

2 21 3 1( , ) 300 240

4 8 4R x y x y xy x y 2 21 3 1

( , ) 300 2404 8 4

R x y x y xy x y

( , ) 180 140 5000C x y x y ( , ) 180 140 5000C x y x y



SolutionSolution The The contributioncontribution to Acrosonic’s to Acrosonic’s weekly profitweekly profit stemming stemming

from the production and sale of the from the production and sale of the bookshelf loudspeaker bookshelf loudspeaker systemsystem is given by is given by

2 2

2 2

( , ) ( , ) ( , )

1 3 1300 240 (180 140 5000)

4 8 4

1 3 1120 100 5000

4 8 4

P x y R x y C x y

x y xy x y x y

x y xy x y

2 2

2 2

( , ) ( , ) ( , )

1 3 1300 240 (180 140 5000)

4 8 4

1 3 1120 100 5000

4 8 4

P x y R x y C x y

x y xy x y x y

x y xy x y



SolutionSolution We haveWe have

To find the To find the relative maximumrelative maximum of the of the profit functionprofit function PP, we , we first locate the first locate the critical pointscritical points of of PP..

Setting Setting PPxx and and PPyy equal to equal to zerozero, we obtain, we obtain

Solving the system of equations we get Solving the system of equations we get xx = 208 = 208 and and yy = 64 = 64.. Therefore, Therefore, PP has has only oneonly one critical pointcritical point at at (208, 64)(208, 64)..

2 21 3 1( , ) 120 100 5000

4 8 4P x y x y xy x y 2 21 3 1

( , ) 120 100 50004 8 4

P x y x y xy x y

1 1 3 1120 0 100 0

2 4 4 4 and x yP x y P y x

1 1 3 1120 0 100 0

2 4 4 4 and x yP x y P y x



SolutionSolution To To testtest if the point if the point (208, 64)(208, 64) is a is a solution to the problemsolution to the problem, we , we

use the use the second derivative testsecond derivative test.. We computeWe compute

So,So,

In particular, In particular, DD(208, 64) = 5/16 > 0(208, 64) = 5/16 > 0.. Since Since DD(208, 64) > 0(208, 64) > 0 and and PPxxxx(208, 64) < 0(208, 64) < 0, the point , the point (208, 64)(208, 64)

yields a yields a relative maximumrelative maximum of of PP..

21 3 1 3 1 5

( , )2 4 4 8 16 16

D x y

21 3 1 3 1 5

( , )2 4 4 8 16 16

D x y

1 3 1

2 4 4 xx yy xyP P P

1 3 1

2 4 4 xx yy xyP P P



SolutionSolution The The relative maximumrelative maximum at at (208, 64)(208, 64) is also the is also the absolute absolute

maximummaximum of of PP.. We conclude that Acrosonic can We conclude that Acrosonic can maximize its weeklymaximize its weekly profit profit

by manufacturing by manufacturing 208208 assembled units and assembled units and 6464 kits. kits. The The maximum weekly profitmaximum weekly profit realizable with this output is realizable with this output is

2 21 3 1(208,64) (208) (64) (208)(64)

4 8 4120(208) 100(64) 5000

$10,680

P

2 21 3 1(208,64) (208) (64) (208)(64)

4 8 4120(208) 100(64) 5000

$10,680

P

2 21 3 1( , ) 120 100 5000

4 8 4P x y x y xy x y 2 21 3 1

( , ) 120 100 50004 8 4

P x y x y xy x y


8.48.4The Method of Least SquaresThe Method of Least Squares

10

5

5 10x

y

d4

d5

d3

d2

d1

L

1010

55

55 1010

The Method of Least SquaresThe Method of Least Squares

Suppose we are given the Suppose we are given the data pointsdata points

PP11((xx11, , yy11)), , PP22((xx22, , yy22)), , PP33((xx33, , yy33)), , PP44((xx44, , yy44)), and , and PP55((xx55, , yy55))

that that describe describe thethe relationshiprelationship between between two variables two variables xx and and yy..

By By plotting these data pointsplotting these data points, we obtain a , we obtain a scatter diagramscatter diagram: :

xx

yy

PP11

PP22

PP33

PP44

PP55

1010

55

55 1010


Suppose we try to fit a Suppose we try to fit a straight linestraight line LL to the data points to the data points PP11, , PP22, , PP33, , PP44, and , and PP55..

The line will The line will miss these pointsmiss these points by the by the amountsamounts dd11, , dd22, , dd33, , dd44, and , and dd55 respectively. respectively.

xx

yy LL

d4

d5

d3

d2

d1

1010

55

55 1010


The principle of The principle of least squaresleast squares states that the states that the straight linestraight line L L that that fits the data points bestfits the data points best is the one chosen by requiring is the one chosen by requiring that the that the sum of the squaressum of the squares of of dd11, , dd22, , dd33, , dd44, and , and dd55, that is, that is

be made be made as small as possibleas small as possible..

xx

yy

2 2 2 2 21 2 3 4 5d d d d d 2 2 2 2 21 2 3 4 5d d d d d

LL

d4

d5

d3

d2

d1

1010

55

55 1010


Suppose the Suppose the regression lineregression line LL is is yy = = ff((xx) = ) = mxmx + + bb, where , where mm and and bb are are to be determinedto be determined..

The distances The distances dd11, , dd22, , dd33, , dd44, and , and dd55, represent the , represent the errorserrors the the line line LL is making in estimating these points, so that is making in estimating these points, so that

xx

yy

d4

d5

d3

d2

d1

LL

1 1 1 2 2 2 3 3 3( ) ( ) ( ), , , and so on.d f x y d f x y d f x y 1 1 1 2 2 2 3 3 3( ) ( ) ( ), , , and so on.d f x y d f x y d f x y


Observe that Observe that

This may be viewed as a This may be viewed as a function of two variablesfunction of two variables mm and and bb.. Thus, the Thus, the least-squares criterionleast-squares criterion is equivalent to is equivalent to minimizing minimizing

the functionthe function

2 2 2 2 21 2 3 4 5d d d d d 2 2 2 2 21 2 3 4 5d d d d d

2 2 21 1 2 2 3 3

2 24 4 5 5

[ ( ) ] [ ( ) ] [ ( ) ]

[ ( ) ] [ ( ) ]

f x y f x y f x y

f x y f x y

2 2 21 1 2 2 3 3

2 24 4 5 5

[ ( ) ] [ ( ) ] [ ( ) ]

[ ( ) ] [ ( ) ]

f x y f x y f x y

f x y f x y

2 2 2

1 1 2 2 3 3

2 24 4 5 5

[ ] [ ] [ ]

[ ] [ ]

mx b y mx b y mx b y

mx b y mx b y

2 2 21 1 2 2 3 3

2 24 4 5 5

[ ] [ ] [ ]

[ ] [ ]

mx b y mx b y mx b y

mx b y mx b y

2 2 21 1 2 2 3 3

2 24 4 5 5

( , ) ( ) ( ) ( )

( ) ( )

f m b mx b y mx b y mx b y

mx b y mx b y

2 2 21 1 2 2 3 3

2 24 4 5 5

( , ) ( ) ( ) ( )

( ) ( )


mx b y mx b y


We want to We want to minimizeminimize

We first find the We first find the partial derivativepartial derivative with respect towith respect to mm::

1 1 1 2 2 2 3 3 3

4 4 4 5 5 5

2( ) 2( ) 2( )

2( ) 2( )

fmx b y x mx b y x mx b y x

mmx b y x mx b y x

1 1 1 2 2 2 3 3 3

4 4 4 5 5 5

2( ) 2( ) 2( )

2( ) 2( )

fmx b y x mx b y x mx b y x

mmx b y x mx b y x

2 2 21 1 1 1 2 2 2 2 3 3 3 3

2 24 4 4 4 5 5 5 5

2[

]

mx bx y x mx bx y x mx bx y x

mx bx y x mx bx y x

2 2 21 1 1 1 2 2 2 2 3 3 3 3

2 24 4 4 4 5 5 5 5

2[

]

mx bx y x mx bx y x mx bx y x

mx bx y x mx bx y x

2 2 2 2 21 2 3 4 5 1 2 3 4 5

1 1 2 2 3 3 4 4 5 5

2[( ) ( )

( )]

x x x x x m x x x x x b

y x y x y x y x y x

2 2 2 2 21 2 3 4 5 1 2 3 4 5

1 1 2 2 3 3 4 4 5 5

2[( ) ( )

( )]


y x y x y x y x y x

2 2 21 1 2 2 3 3

2 24 4 5 5

( , ) ( ) ( ) ( )

( ) ( )


mx b y mx b y

2 2 21 1 2 2 3 3

2 24 4 5 5

( , ) ( ) ( ) ( )

( ) ( )


mx b y mx b y


We want to We want to minimizeminimize

We now find the We now find the partial derivativepartial derivative with respect towith respect to bb::

1 1 2 2 3 3

4 4 5 5

2( ) 2( ) 2( )

2( ) 2( )

fmx b y mx b y mx b y

bmx b y mx b y

1 1 2 2 3 3

4 4 5 5

2( ) 2( ) 2( )

2( ) 2( )

fmx b y mx b y mx b y

bmx b y mx b y

1 2 3 4 5 1 2 3 4 52[( ) 5 ( )]x x x x x m b y y y y y 1 2 3 4 5 1 2 3 4 52[( ) 5 ( )]x x x x x m b y y y y y

2 2 21 1 2 2 3 3

2 24 4 5 5

( , ) ( ) ( ) ( )

( ) ( )


mx b y mx b y

2 2 21 1 2 2 3 3

2 24 4 5 5

( , ) ( ) ( ) ( )

( ) ( )


mx b y mx b y


SettingSetting

givesgives

andand

SolvingSolving the two the two simultaneous equationssimultaneous equations for for mm and and bb then then leads to an equationleads to an equation yy = = mxmx + + bb..

This equation will be the This equation will be the ‘best fit’ line‘best fit’ line, or , or regression lineregression line for for the given data points.the given data points.

1 2 3 4 5 1 2 3 4 5( ) 5x x x x x m b y y y y y 1 2 3 4 5 1 2 3 4 5( ) 5x x x x x m b y y y y y

0 0an df f

m b

0 0an d

f f

m b

2 2 2 2 21 2 3 4 5 1 2 3 4 5

1 1 2 2 3 3 4 4 5 5

( ) ( )


y x y x y x y x y x

2 2 2 2 21 2 3 4 5 1 2 3 4 5

1 1 2 2 3 3 4 4 5 5

( ) ( )


y x y x y x y x y x


Suppose we are given Suppose we are given nn data pointsdata points::

PP11((xx11, , yy11)), , PP22((xx22, , yy22)), , PP33((xx33, , yy33)), , … … , , PPnn((xxnn, , yynn))

Then, the Then, the least-squares (regression) lineleast-squares (regression) line for the data for the data is given by the is given by the linear equationlinear equation

yy = = ff((xx) = ) = mxmx + + bb

where the where the constantsconstants mm and and bb satisfy the equations satisfy the equations

andand

simultaneously.simultaneously. These last two equations are called These last two equations are called normal equationsnormal equations..

1 2 3 1 2 3( ... ) ...n nx x x x m nb y y y y 1 2 3 1 2 3( ... ) ...n nx x x x m nb y y y y

2 2 2 21 2 3 1 2 3

1 1 2 2 3 3

( ... ) ( ... )

... n n

n n

x x x x m x x x x b

y x y x y x y x

2 2 2 21 2 3 1 2 3

1 1 2 2 3 3

( ... ) ( ... )

... n n

n n

x x x x m x x x x b

y x y x y x y x

ExampleExample

Find the equation of the Find the equation of the least-squares lineleast-squares line for the data for the data

PP11(1, 1)(1, 1), , PP22(2, 3)(2, 3), , PP33(3, 4)(3, 4), , PP44(4, 3)(4, 3), and , and PP55(5, 6)(5, 6)

66

55

44

33

22

11

11 22 33 44 55xx

yy

Example 1, page 570

ExampleExample



SolutionSolution Here, we have Here, we have nn = 5 = 5 and and

xx11 = 1 = 1 xx22 = 2 = 2 xx33 = 3 = 3 xx44 = 4 = 4 xx55 = 5 = 5

yy11 = 1 = 1 yy22 = 3 = 3 yy33 = 4 = 4 yy44 = 3 = 3 yy55 = 6 = 6

SubstitutingSubstituting in the in the first equationfirst equation we get we get2 2 2 21 2 3 1 2 3

1 1 2 2 3 3

( ... ) ( ... )

... n n

n n

x x x x m x x x x b

y x y x y x y x

2 2 2 21 2 3 1 2 3

1 1 2 2 3 3

( ... ) ( ... )

... n n

n n

x x x x m x x x x b

y x y x y x y x

2 2 2 2 2(1 2 3 4 5 ) (1 2 3 4 5)

(1)(1) (3)(2) (4)(3) (3)(4) (6)(5)

m b

2 2 2 2 2(1 2 3 4 5 ) (1 2 3 4 5)

(1)(1) (3)(2) (4)(3) (3)(4) (6)(5)

m b

55 15 61m b 55 15 61m b Example 1, page 570

ExampleExample



SolutionSolution Here, we have Here, we have nn = 5 = 5 and and

xx11 = 1 = 1 xx22 = 2 = 2 xx33 = 3 = 3 xx44 = 4 = 4 xx55 = 5 = 5

yy11 = 1 = 1 yy22 = 3 = 3 yy33 = 4 = 4 yy44 = 3 = 3 yy55 = 6 = 6

SubstitutingSubstituting in the in the second equationsecond equation we get we get

15 5 17m b 15 5 17m b

1 2 3 1 2 3( ... ) 5 ...n nx x x x m b y y y y 1 2 3 1 2 3( ... ) 5 ...n nx x x x m b y y y y

(1 2 3 4 5) 5 1 3 4 3 6m b (1 2 3 4 5) 5 1 3 4 3 6m b

Example 1, page 570

ExampleExample



SolutionSolution Solving the Solving the simultaneous equationssimultaneous equations

gives gives mm = 1 = 1 and and bb = 0.4 = 0.4.. Therefore, the Therefore, the required required least-squares lineleast-squares line is is

yy = = xx + 0.4 + 0.4

15 5 17m b 15 5 17m b 55 15 61m b 55 15 61m b

ExampleExample



SolutionSolution Below is the graph of the Below is the graph of the required required least-squares lineleast-squares line

yy = = xx + 0.4 + 0.4

66

55

44

33

22

11

11 22 33 44 55xx

yyLL

Example 1, page 570


A market research study provided the following data A market research study provided the following data based on the based on the projected monthly salesprojected monthly sales xx (in thousands) of (in thousands) of an adventure movie DVD.an adventure movie DVD.

FindFind the the demand equationdemand equation if the demand curve is the if the demand curve is the least-least-squares linesquares line for these data. for these data.

The The total monthly cost functiontotal monthly cost function associated with producing associated with producing and distributing the DVD is given byand distributing the DVD is given by

CC((xx) = 4) = 4xx + 25 + 25

where where xx denotes the denotes the number of discsnumber of discs (in thousands) (in thousands) produced and sold, and produced and sold, and CC((xx)) is in thousands of dollars. is in thousands of dollars.

DetermineDetermine the unit the unit wholesale pricewholesale price that will that will maximize maximize monthly profitsmonthly profits..

pp 3838 3636 34.534.5 3030 28.528.5

xx 2.22.2 5.45.4 7.07.0 11.511.5 14.614.6


SolutionSolution The calculations required for obtaining the The calculations required for obtaining the normal normal

equationsequations may be summarized as follows: may be summarized as follows:

Thus, the Thus, the nominal equationsnominal equations are are


x p x2 xp

2.2 38.0 4.84 83.6

5.4 36.0 29.16 194.4

7.0 34.5 49.00 241.5

11.5 30.0 132.25 345.0

14.6 28.5 213.16 416.1

40.7 167.0 428.41 1280.6

5 40.7 167 40.7 428.41 1280.6 an db m b m 5 40.7 167 40.7 428.41 1280.6 an db m b m Applied Example 3, page 572

SolutionSolution SolvingSolving the the system of linear equationssystem of linear equations simultaneously, we simultaneously, we

find thatfind that

Therefore, the required Therefore, the required demand equationdemand equation is given by is given by


0.81 39.99an dm b 0.81 39.99an dm b

( ) 0.81 39.99 (0 49.37) p f x x x ( ) 0.81 39.99 (0 49.37) p f x x x


SolutionSolution TheThe total revenue function total revenue function in this case is given byin this case is given by

Since the Since the total cost functiontotal cost function is is

CC((xx) = 4) = 4xx + 25 + 25

we see that the we see that the profit functionprofit function is is


2

( )

( 0.81 39.99)

0.81 39.99

R x xp

x x

x x

2

( )

( 0.81 39.99)

0.81 39.99

R x xp

x x

x x

2

2

( )

0.81 39.99 (4 25)

0.81 35.99 25

P x R C

x x x

x x

2

2

( )

0.81 39.99 (4 25)

0.81 35.99 25

P x R C

x x x

x x


SolutionSolution To find the To find the absolute maximumabsolute maximum of of PP((xx)) over the closed over the closed

interval interval [0, 49.37][0, 49.37], we compute, we compute

Since Since PP((xx) = 0) = 0 , we find that , we find that xx ≈ 22.22≈ 22.22 as the as the only critical only critical pointpoint of of PP..

Finally, from the Finally, from the tabletable

we see that the we see that the optimal wholesale priceoptimal wholesale price is is

or or $21.99$21.99 per disc. per disc.


( ) 1.62 35.99P x x ( ) 1.62 35.99P x x

xx 00 22.2222.22 49.3749.37

PP((xx)) ––2525 374.78374.78 – – 222.47222.47

0.81(22.22) 39.99 21.99p 0.81(22.22) 39.99 21.99p


8.58.5Constrained Maxima and Minima and Constrained Maxima and Minima and the Method of Lagrange Multipliersthe Method of Lagrange Multipliers

z

y

x

f(x, y) = 2x2 + y2

g(x, y) = 0

h(x) = 3x2 – 2x + 1

(a, b)

(a, b, f(a, b))

Constrained Maxima and MinimaConstrained Maxima and Minima

In many In many practical optimization problemspractical optimization problems, we must , we must maximize or minimizemaximize or minimize a function in which the a function in which the independent independent variablesvariables are are subjected to certain further constraintssubjected to certain further constraints..

We shall discuss a We shall discuss a powerful methodpowerful method for determining for determining relative extrema of a relative extrema of a functionfunction ff((xx, , yy)) whose whose independent independent variablesvariables xx and and yy are required to satisfy one or more are required to satisfy one or more constraintsconstraints of the form of the form gg((xx, , yy) = 0) = 0..

ExampleExample Find the Find the relative minimumrelative minimum of of ff((xx, , yy)) = 2= 2xx22 + +yy22 subject to subject to

the the constraint constraint gg((xx, , yy)) = = xx + + y y – 1 = 0– 1 = 0..

SolutionSolution SolvingSolving thethe constraint equation constraint equation forfor yy explicitly in terms explicitly in terms

of of xx, we obtain, we obtainyy = – = – xx + 1 + 1

SubstitutingSubstituting this value of this value of yy into into ff((xx, , yy)) results in a results in a

function of function of xx,,

2 2 2( ) 2 ( 1) 3 2 1h x x x x x 2 2 2( ) 2 ( 1) 3 2 1h x x x x x

Example 1, page 580



SolutionSolution

zz

y

x

ff((xx, , yy) = 2) = 2xx22 + + yy22

We have We have

hh((xx)) = 3= 3xx22 – 2 – 2xx + 1 + 1.. The function The function hh describes describes

the curve the curve lying on the lying on the graphgraph of of f f on which the on which the constrained relative constrained relative minimumminimum point point ((aa, , bb)) of of ff occurs: occurs:

gg((xx, , yy) = 0) = 0

hh((xx)) = 3= 3xx22 – 2 – 2xx + 1 + 1

((aa, , bb))

((aa, , bb, , ff((aa, , bb))))

Example 1, page 580



SolutionSolution To To find this pointfind this point ((aa, , bb)), we determine the , we determine the relative extremarelative extrema

of a of a function of one variablefunction of one variable::

Setting Setting hh′ ′ = 0= 0 gives gives xx = = as the as the sole critical pointsole critical point of of hh.. Next, we find Next, we find hh″″((xx) = 6) = 6 and, and, in particularin particular, , hh″″( ( ) = 6 > 0) = 6 > 0.. Therefore, by the Therefore, by the second derivative testsecond derivative test, the point gives , the point gives

rise to a rise to a relative minimumrelative minimum of of hh.. SubstituteSubstitute xx = = into the into the constraint equationconstraint equation xx + +y y – 1 = 0– 1 = 0

to get to get yy = = ..

( ) 6 2 2(3 1)h x x x ( ) 6 2 2(3 1)h x x x

1

31

3

1

31

3

Example 1, page 580



SolutionSolution Thus, the point Thus, the point gives rise to the required gives rise to the required constrained constrained

relative minimumrelative minimum of of ff..

Since Since , the required , the required constrainedconstrained

relative minimum valuerelative minimum value of of ff is at the point . is at the point .

It may be shown that It may be shown that is in fact a is in fact a constrained absolute constrained absolute

minimum valueminimum value of of ff..

2 21 2 1 2 2

, 23 3 3 3 3

f

2

31 2

,3 3

2

3

Example 1, page 580

1 2,

3 3

The Method of Lagrange MultipliersThe Method of Lagrange Multipliers

To find the To find the relative extremarelative extrema of the function of the function ff((xx, , yy)) subject to the constraintsubject to the constraint gg((xx, , yy)) = 0= 0 (assuming that (assuming that these extreme values exist),these extreme values exist),

1.1. Form an Form an auxiliary functionauxiliary function

called the called the Lagrangian functionLagrangian function (the variable (the variable λλ is is called the called the Lagrange multiplierLagrange multiplier).).

2.2. Solve the systemSolve the system that consists of the equations that consists of the equations

FFxx = 0 = 0 FFyy = 0 = 0 FFλλ = 0= 0

for all values of for all values of xx, , yy, and , and λλ..

3.3. The The solutionssolutions found in found in step 2step 2 are are candidates for candidates for the extremathe extrema of of ff..

( , , ) ( , ) ( , )F x y f x y g x y ( , , ) ( , ) ( , )F x y f x y g x y

ExampleExample Using the Using the method of Lagrange multipliersmethod of Lagrange multipliers, find the , find the

relative minimumrelative minimum of the function of the function ff((xx, , yy)) = 2= 2xx22 + +yy22 subject subject to the to the constraint constraint xx + + y y = 1= 1..

SolutionSolution Write the Write the constraint equationconstraint equation

xx + + y y = 1= 1 in the form in the form

gg((xx, , yy)) = = xx + + y y – 1 = 0– 1 = 0

Then, form the Then, form the Lagrangian functionLagrangian function

2 2

( , , ) ( , ) ( , )

(2 ) ( 1)

F x y f x y g x y

x y x y

2 2

( , , ) ( , ) ( , )

(2 ) ( 1)

F x y f x y g x y

x y x y

Example 2, page 582



SolutionSolution We haveWe have To find the To find the critical point(s)critical point(s) of the function of the function FF, solve the , solve the

systemsystem composed of the composed of the equationsequations

SolvingSolving the the firstfirst and and second equationssecond equations for for xx and and yy in terms in terms of of λλ, we obtain, we obtain

Which, upon Which, upon substitutionsubstitution into the into the third equationthird equation yields yields

2 22 ( 1)F x y x y 2 22 ( 1)F x y x y

4 0 2 0 1 0 x yF x F y F x y 4 0 2 0 1 0 x yF x F y F x y

1 1

4 2and x y

1 1

4 2and x y

1 1 41 0

4 2 3 or

1 1 41 0

4 2 3 or

Example 2, page 582



SolutionSolution SubstitutingSubstituting

Therefore, Therefore, and and , and , and results in a results in a

constrained minimumconstrained minimum of the function of the function ff..

1 4 1 1 4 2( ) ( )

4 3 3 2 3 3 yields and x y

1 4 1 1 4 2( ) ( )

4 3 3 2 3 3 yields and x y

4 1 1

3 4 2x y into and

4 1 1

3 4 2x y into and

1

3x

2

3y

1 2,

3 3

Example 2, page 582

Applied Example:Applied Example: Designing a Cruise-Ship Pool Designing a Cruise-Ship Pool

The operators of the Viking Princess, a luxury cruise liner, The operators of the Viking Princess, a luxury cruise liner, are contemplating the addition of another are contemplating the addition of another swimming poolswimming pool to the ship.to the ship.

The chief engineer has suggested that an area of the form The chief engineer has suggested that an area of the form of an of an ellipseellipse located in the rear of the promenade deck located in the rear of the promenade deck would be suitable for this purpose.would be suitable for this purpose.

It has been determined that the It has been determined that the shape of the ellipseshape of the ellipse may be may be described by the described by the equationequation

xx22 + 4 + 4yy22 = 3600 = 3600

where where xx and and yy are measured in feet. are measured in feet. Viking’s operators would like to know Viking’s operators would like to know the dimensions of the dimensions of

the rectangular poolthe rectangular pool with the with the largest possible arealargest possible area that that would would meet these requirementsmeet these requirements..



SolutionSolution We want to We want to maximizemaximize the areathe area of the of the rectanglerectangle that will that will fit the fit the

ellipseellipse::

Letting the sides of the rectangle be Letting the sides of the rectangle be 22xx and and 22yy feet, we see that feet, we see that the area of the rectangle is the area of the rectangle is AA = 4 = 4xyxy..

Furthermore, the point Furthermore, the point ((xx, , yy)) must be constrained to lie on the must be constrained to lie on the ellipse so that it satisfies the equation ellipse so that it satisfies the equation xx22 + 4 + 4yy22 = 3600 = 3600..

xx

yy

((xx, , yy))

xx22 + 4 + 4yy22 = 3600 = 3600



SolutionSolution Thus, the problem is Thus, the problem is equivalentequivalent to the problem of to the problem of maximizingmaximizing

the the functionfunction

subject to the subject to the constraintconstraint

gg((xx,, y y) = ) = xx22 + 4 + 4yy22 – 3600 = 0 – 3600 = 0 The The Lagrangian functionLagrangian function is is

To find the To find the critical pointscritical points of of FF, we , we solve the systemsolve the system of of equationsequations

2 2( , , ) ( , ) ( , ) 4 ( 4 3600)F x y f x y g x y xy x y 2 2( , , ) ( , ) ( , ) 4 ( 4 3600)F x y f x y g x y xy x y

2 2

4 2 0

4 8 0

4 3600 0

x

y

F y x

F x y

F x y

2 2

4 2 0

4 8 0

4 3600 0

x

y

F y x

F x y

F x y

ff((xx,, y y) = 4) = 4xyxy




SolvingSolving the the first equationfirst equation for for λλ, we obtain, we obtain

Which, Which, substitutingsubstituting into the into the second equationsecond equation, yields, yields

Solving for Solving for xx yields yields xx = ± 2 = ± 2yy. .

2 y

x

2 y

x

2 224 8 0 4 0 or

yx y x y

x

2 224 8 0 4 0 or

yx y x y

x

2 2

4 2 04 8 0

4 3600 0

x

y

F y xF x y

F x y

2 2

4 2 04 8 0

4 3600 0

x

y

F y xF x y

F x y




SubstitutingSubstituting xx = ± 2 = ± 2y y into the into the third equationthird equation, we have, we have

Which, upon Which, upon solvingsolving for for y y yields yields

The The correspondingcorresponding values of values of xx are are

Since both Since both xx and and yy must be nonnegativemust be nonnegative, we have, we have

or or approximatelyapproximately 42 42 ☓☓ 85 85 feet. feet.

2 24 4 3600 0y y 2 24 4 3600 0y y

450 15 2y 450 15 2y

2 2(15 2) 30 2x y 2 2(15 2) 30 2x y

30 2 15 2an dx y 30 2 15 2an dx y

2 2

4 2 04 8 0

4 3600 0

x

y

F y xF x y

F x y

2 2

4 2 04 8 0

4 3600 0

x

y

F y xF x y

F x y


8.68.6Double IntegralsDouble Integrals

z

x

y

21y x 21y x

z = f(x, y) = y

You may recall that we can do a You may recall that we can do a Riemann sumRiemann sum to to approximate approximate thethe areaarea under the graph under the graph of a function of of a function of one one variablevariable by adding the areas of the rectangles that form by adding the areas of the rectangles that form below the graph resulting from below the graph resulting from small incrementssmall increments of of x x ((xx)) within a given within a given intervalinterval [[aa,, b b]]::

xx

yy

A Geometric Interpretation of the Double IntegralA Geometric Interpretation of the Double Integral

y= fy= f((xx))

aa bb

xx

zz


Similarly, it is possible to obtain an Similarly, it is possible to obtain an approximationapproximation of the of the volumevolume of theof the solidsolid under the graph under the graph of a function of of a function of two two variablesvariables..

yy

xx

aa

bb

cc dd

y= fy= f((xx,, y y))

zz


yy

xx

aa

bb

cc dd

xxyy

To find the To find the volumevolume of the of the solid under the surfacesolid under the surface, we can , we can perform a perform a Riemann sumRiemann sum of the volume of the volume SSii of of parallelepipedsparallelepipeds

with with basebase RRii = = xx ☓☓ yy and and heightheight ff((xxii, , yyii))::

z = fz = f((xx,, y y))

zz




yy

xx

aa

bb

cc dd

xxyy

SSii

z = fz = f((xx,, y y))

yyaa

bb

cc dd

zz

xx




z = fz = f((xx,, y y))

zz


z = fz = f((xx,, y y))

yy

xx

aa

bb

cc dd

xxyy

The The limitlimit of the of the Riemann sumRiemann sum obtained when the obtained when the number of number of rectanglesrectangles mm along the along the xx-axis-axis, and the , and the number of subdivisions number of subdivisions nn along the along the yy-axis-axis tends to infinitytends to infinity is the value of the is the value of the double double integralintegral of of ff((xx,, y y)) over the region over the region RR and is and is denoteddenoted by by

( , )R

f x y dA ( , )R

f x y dA

yy ·n·n

xx ·m·m

Theorem 1Theorem 1Evaluating a Double Integral Over a Plane RegionEvaluating a Double Integral Over a Plane Region

a.a. Suppose Suppose gg11((xx)) and and gg22((xx)) are are continuous functionscontinuous functions on on [[aa, , bb]] and the and the

regionregion RR is is defineddefined by by RR = = {( {(xx, , yy)| )| gg11((xx) ) y y gg22((xx);); aa xx bb}}..

Then,Then,2

1

( )

( )( , ) ( , )

b g x

R a g xf x y dA f x y dy dx

2

1

( )

( )( , ) ( , )

b g x

R a g xf x y dA f x y dy dx

xx

yy

aa bb

y y == gg11((xx))

y y == gg22((xx))

RR

Theorem 1Theorem 1Evaluating a Double Integral Over a Plane RegionEvaluating a Double Integral Over a Plane Region

b.b. Suppose Suppose hh11((yy)) and and hh22((yy)) are are continuous functionscontinuous functions on on [[cc, , dd]] and the and the

regionregion RR is is defineddefined by by RR = = {( {(xx, , yy)| )| hh11((yy) ) x x hh22((yy);); cc yy dd}}..

Then,Then,2

1

( )

( )( , ) ( , )

d h y

R c h yf x y dA f x y dx dy

2

1

( )

( )( , ) ( , )

d h y

R c h yf x y dA f x y dx dy

xx

yy

cc

dd

x x == hh11((yy)) x x == hh22((yy))

RR

44

33

22

11

11 22 33 44

ExamplesExamples

Evaluate Evaluate ∫∫RR∫∫ff((xx, , yy))dAdA given that given that ff((xx, , yy)) == x x22 ++ y y22 and and RR is the is the

regionregion boundedbounded by the by the graphsgraphs of of gg11((xx)) = = xx and and gg22((xx)) = 2= 2xx

for for 0 0 xx 2 2..

SolutionSolution The The regionregion under consideration is: under consideration is:

xx

gg22((xx)) = 2= 2xx

RR

yy

gg11((xx)) = = xx

Example 2, page 593

ExamplesExamples

Evaluate Evaluate ∫∫RR∫∫ff((xx, , yy))dAdA given that given that ff((xx, , yy)) == x x22 ++ y y22 and and RR is the is the

regionregion boundedbounded by the by the graphsgraphs of of gg11((xx)) = = xx and and gg22((xx)) = 2= 2xx

for for 0 0 xx 2 2..

SolutionSolution Using Using Theorem 1Theorem 1, we find:, we find:

2 2 2 2

0( , ) ( )

x

R xf x y dA x y dy dx

2 2 2 2

0( , ) ( )

x

R xf x y dA x y dy dx

22 2 3

0

1

3

x

x

x y y dx

2

2 2 3

0

1

3

x

x

x y y dx

2 3 3 3 3

0

8 12

3 3x x x x dx

2 3 3 3 3

0

8 12

3 3x x x x dx

2 3

0

10

3x dx

2 3

0

10

3x dx

24

0

5

6x

24

0

5

6x 1

313 1313

Example 2, page 593

11

11

ExamplesExamples

Evaluate Evaluate ∫∫RR∫∫ff((xx, , yy))dAdA, where , where ff((xx, , yy)) == xe xeyy and and RR is the is the planeplane

regionregion boundedbounded by the by the graphsgraphs of of yy = = xx22 and and yy = = xx..

SolutionSolution The The regionregion under consideration is: under consideration is:

xx

gg11((xx)) = = xx22

RR

yy

gg22((xx)) = = xx

The The points of intersectionpoints of intersection of of the two curves are found by the two curves are found by solving the equation solving the equation xx22 = = xx, , giving giving xx = 0 = 0 and and x x = 1= 1..

Example 3, page 593

ExamplesExamples

Evaluate Evaluate ∫∫RR∫∫ff((xx, , yy))dAdA, where , where ff((xx, , yy)) == xe xeyy and and RR is the is the planeplane

regionregion boundedbounded by the by the graphsgraphs of of yy = = xx22 and and yy = = xx..

SolutionSolution Using Using Theorem 1Theorem 1, we find:, we find:

2

1

0( , )

x y

R xf x y dA xe dy dx 2

1

0( , )

x y

R xf x y dA xe dy dx 2

1

0

xy

xxe dx 2

1

0

xy

xxe dx

21

0( )x xxe xe dx

21

0( )x xxe xe dx

21 1

0 0

x xxe dx xe dx 21 1

0 0

x xxe dx xe dx

2

1

0

1( 1)

2x xx e e

2

1

0

1( 1)

2x xx e e

1 1 11 (3 )

2 2 2e e

1 1 11 (3 )

2 2 2e e

Integrating by parts on Integrating by parts on the right-hand sidethe right-hand side

Example 3, page 593

The Volume of a Solid Under a SurfaceThe Volume of a Solid Under a Surface

Let Let RR be a be a regionregion in the in the xyxy-plane and let -plane and let ff be be continuouscontinuous and and nonnegativenonnegative on on RR..

Then, the Then, the volumevolume of the of the solid solid under aunder a surface surface bounded abovebounded above by by zz = = ff((xx, , yy)) and and belowbelow by by RR is is given bygiven by

( , )R

V f x y dA ( , )R

V f x y dA

ExampleExample Find the Find the volumevolume of the of the solidsolid bounded abovebounded above by the plane by the plane

z z = = ff((xx, , yy) = ) = yy and and belowbelow by the plane region by the plane region RR defined by defined by

SolutionSolution The The graphgraph of the of the regionregion RR is: is:

11

11xx

yy

21y x 21y x

RR

Observe that Observe that ff((xx, , yy) = ) = y y > 0> 0 for for ((xx, , yy) ) ∈∈ RR..

21 (0 1) y x x 21 (0 1) y x x

Example 4, page 594

Find the Find the volumevolume of the of the solidsolid bounded abovebounded above by the plane by the plane z z = = ff((xx, , yy) = ) = yy and and belowbelow by the plane region by the plane region RR defined by defined by

SolutionSolution Therefore, the required volume is given byTherefore, the required volume is given by

21 1

0 0

x

RV ydA ydy dx

21 1

0 0

x

RV ydA ydy dx

211 2

00

1

2

x

y dx

21

1 2

00

1

2

x

y dx

1 2

0

1(1 )

2x dx

1 2

0

1(1 )

2x dx

1

3

0

1 1

2 3x x

1

3

0

1 1

2 3x x

1

3

1

3

ExampleExample

21 (0 1) y x x 21 (0 1) y x x

Example 4, page 594

Find the Find the volumevolume of the of the solidsolid bounded abovebounded above by the plane by the plane z z = = ff((xx, , yy) = ) = yy and and belowbelow by the plane region by the plane region RR defined by defined by

SolutionSolution

21 (0 1) y x x 21 (0 1) y x x

zz

xx

yy

ExampleExample

The The graphgraph of the solid of the solid in question is:in question is:

21y x 21y x

z z = = ff((xx, , yy) = ) = yy

Example 4, page 594

End of End of Chapter Chapter

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Functions of Several Variables Partial Derivatives