Functions of Several Variables Partial Derivatives
description
Transcript of 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
Find the Find the domaindomain of the function of the function
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
Find the Find the domaindomain of the function of the function
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
Applied Example:Applied Example: Revenue Functions Revenue Functions
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
Applied Example 3, page 537
1000 2000
Applied Example:Applied Example: Revenue Functions Revenue Functions
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
Applied Example 3, page 537
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)
Graphs of Functions of Two VariablesGraphs of Functions of Two Variables
(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)
Graphs of Functions of Two VariablesGraphs of Functions of Two Variables
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
Graphs of Functions of Two VariablesGraphs of Functions of Two Variables
((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
Geometric Interpretation of the Partial DerivativeGeometric Interpretation of the Partial Derivative
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
Geometric Interpretation of the Partial DerivativeGeometric Interpretation of the Partial Derivative
ff((xx, , yy))
xx
yy
Geometric Interpretation of the Partial DerivativeGeometric Interpretation of the Partial Derivative
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
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/∂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
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 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
differentiatedifferentiate the resulting function of the resulting function of xx with respect towith respect to xx::
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
Find the Find the firstfirst partial derivativespartial derivatives ofof the the functionfunction
SolutionSolution To compute To compute ∂w/∂y∂w/∂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( , )
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
Find the Find the firstfirst partial derivativespartial derivatives ofof the the functionfunction
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
Find the Find the firstfirst partial derivativespartial derivatives ofof the the functionfunction
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
Find the Find the firstfirst partial derivativespartial derivatives ofof the the functionfunction
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
Find the Find the firstfirst partial derivativespartial derivatives ofof the the functionfunction
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
Find the Find the firstfirst partial derivativespartial derivatives ofof the the functionfunction
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
Find the Find the firstfirst partial derivativespartial derivatives ofof the the functionfunction
SolutionSolution To find To find ffxx, think of the , think of the variablesvariables yy and and zz as a as a constantconstant and and
differentiatedifferentiate the resulting function of the resulting function of xx with respect towith respect to xx::
( , , ) lnyzw f x y z xyz xe x y ( , , ) lnyzw f x y z xyz xe x y
( , 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
Find the Find the firstfirst partial derivativespartial derivatives ofof the the functionfunction
SolutionSolution To find To find ffyy, think of the , think of the variablesvariables xx and and zz as a as a constantconstant and and
differentiatedifferentiate the resulting function of the resulting function of yy with respect towith respect to yy::
( , , ) lnyzw f x y z xyz xe x y ( , , ) lnyzw f x y z xyz xe x y
( , , ) 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
Find the Find the firstfirst partial derivativespartial derivatives ofof the the functionfunction
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::
( , , ) lnyzw f x y z xyz xe x y ( , , ) lnyzw f x y z xyz xe x y
( , ) 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?
Applied Example 4, page 550
Applied Example:Applied Example: Marginal Productivity Marginal 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
Applied Example 4, page 550
Applied Example:Applied Example: Marginal Productivity Marginal Productivity
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
Applied Example 4, page 550
Applied Example:Applied Example: Marginal Productivity Marginal Productivity
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.
Applied Example 4, page 550
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..
Second Order Partial DerivativesSecond Order Partial Derivatives
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
Second Order Partial DerivativesSecond Order Partial Derivatives
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
Second Order Partial DerivativesSecond Order Partial Derivatives
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
Find the Find the second-order second-order partial derivativespartial derivatives of the function of the function
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
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::
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
Find the Find the second-order second-order partial derivativespartial derivatives of the function of the function
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))
Graphic ExampleGraphic Example
yyxx
zz
((ee, , f f ))
There is a There is a relative minimumrelative minimum at at ((ee, , f f ))..
Graphic ExampleGraphic Example
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))
Graphic ExampleGraphic Example
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::
Relative MinimaRelative Minima
( , ) 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::
Relative MinimaRelative Minima
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
Relative MaximaRelative Maxima
((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
Relative MaximaRelative Maxima
((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
Find the Find the relative extremarelative extrema of the function of the function
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
Find the Find the relative extremarelative extrema of the function of the function
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
Find the Find the relative extremarelative extrema of the function of the function
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
Find the Find the relative extremarelative extrema of the function of the function
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
Find the Find the relative extremarelative extrema of the function of the function
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
Find the Find the relative extremarelative extrema of the function of the function
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
Find the Find the relative extremarelative extrema of the function of the function
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
Applied Example 3, page 563
Applied Example:Applied Example: Maximizing Profit Maximizing Profit
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
Applied Example 3, page 563
Applied Example:Applied Example: Maximizing Profit Maximizing Profit
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
Applied Example 3, page 563
Applied Example:Applied Example: Maximizing Profit Maximizing Profit
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
Applied Example 3, page 563
Applied Example:Applied Example: Maximizing Profit Maximizing Profit
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
Applied Example 3, page 563
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
The Method of Least SquaresThe Method of Least Squares
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 Method of Least SquaresThe Method of Least Squares
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
The Method of Least SquaresThe Method of Least Squares
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
The Method of Least SquaresThe Method of Least Squares
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
( , ) ( ) ( ) ( )
( ) ( )
f m b mx b y mx b y mx b y
mx b y mx b y
The Method of Least SquaresThe Method of Least Squares
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[( ) ( )
( )]
x x x x x m x x x x x b
y x y x y x y x y x
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
( , ) ( ) ( ) ( )
( ) ( )
f m b mx b y mx b y mx b y
mx b y mx b y
The Method of Least SquaresThe Method of Least Squares
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
( , ) ( ) ( ) ( )
( ) ( )
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
( , ) ( ) ( ) ( )
( ) ( )
f m b mx b y mx b y mx b y
mx b y mx b y
The Method of Least SquaresThe Method of Least Squares
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
( ) ( )
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
( ) ( )
x x x x x m x x x x x b
y x y x y x y x y x
The Method of Least SquaresThe Method of Least Squares
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
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)
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
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)
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
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)
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
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)
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
Applied Example:Applied Example: Maximizing Profit Maximizing Profit
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
Applied Example 3, page 572
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
Applied Example:Applied Example: Maximizing Profit Maximizing Profit
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
Applied Example:Applied Example: Maximizing Profit Maximizing Profit
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
Applied Example 3, page 572
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
Applied Example:Applied Example: Maximizing Profit Maximizing Profit
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
Applied Example 3, page 572
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.
Applied Example:Applied Example: Maximizing Profit Maximizing Profit
( ) 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
Applied Example 3, page 572
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
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
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
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 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
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 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
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 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
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 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..
Applied Example 5, page 582
Applied Example:Applied Example: Designing a Cruise-Ship Pool Designing a Cruise-Ship Pool
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
Applied Example 5, page 582
Applied Example:Applied Example: Designing a Cruise-Ship Pool Designing a Cruise-Ship Pool
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
Applied Example 5, page 582
Applied Example:Applied Example: Designing a Cruise-Ship Pool Designing a Cruise-Ship Pool
SolutionSolution We haveWe have
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
Applied Example 5, page 582
Applied Example:Applied Example: Designing a Cruise-Ship Pool Designing a Cruise-Ship Pool
SolutionSolution We haveWe have
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
Applied Example 5, page 582
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
A Geometric Interpretation of the Double IntegralA Geometric Interpretation of the Double Integral
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
A Geometric Interpretation of the Double IntegralA Geometric Interpretation of the Double Integral
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
A Geometric Interpretation of the Double IntegralA Geometric Interpretation of the Double Integral
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))::
yy
xx
aa
bb
cc dd
xxyy
SSii
z = fz = f((xx,, y y))
yyaa
bb
cc dd
zz
xx
A Geometric Interpretation of the Double IntegralA Geometric Interpretation of the Double Integral
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
A Geometric Interpretation of the Double IntegralA Geometric Interpretation of the Double Integral
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