MAT495_CHAPTER_9

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Part 1 PARTIAL DERIVATIVES MAT 295

121

Chapter 9

Lagrange

MultipliersAt the end of this chapter, students should be able to:

Define the Lagrange multipliers

Identify the procedure for solving extremum problems with a constraint

9.1 Introduction

In chapter 8, one of the most common problems in calculus which is finding

the extrema of a function has been explored. What if the given function is

subject to fix an outside conditions or constraints? For example, How to

minimize the wood needed to make a rectangular box which can hold 2

kilograms of detergent? In this chapter a powerful tool for solving this

particular problem will be introduced. The method is known as Lagrange

multiplier which is used to optimize functions of two or three variables with a

constraint.

9.2 Lagrange Multipliers

Lagrange Multipliers is the name given to a method of finding the extreme of

a function subject to a constraint.

How the Method Works

To see how Lagrange multipliers work, take a look at Figure 9.1. Draw the

function ),( yxf from above, along with a constraint cyxg ),( . In the

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drawing, the constraint is a plane that cuts through a lake. Lets draw some

level curves of ),( yxf .

Figure 9.1

Figure 9.2: At the constrained critical point the tangents to the level curves of f

and g are in the same direction

The aim is to move down to the bottom of the lake as low as possible but do

not move any lower than where the constraint cg cuts the valley. The

boundary where the constraint intersects the function is marked with a dotted

line. Along that line are the lowest points possible without crossing over the

constraint cg . Hence, this is where to start looking for a constrained

minimum.

While moving along the level curves, compare the slope of a tangent line of

),( yxf with the slope on the constraint ),( yxg . The lowest point which is

known as the constrained minimum is reached once the slope of the

constraint line just equals the slope of the level curve. In other words, the two

x3

x1

x2f=a1

f=a2

f(x1, x2)

g(x1, x2)=c

gf

Level curves of f Level curve g=0

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curves are parallel and point in the same direction. So the gradients of ),( yxf

and ),( yxg must also be parallel and differ at most by a scalar. Lets call that

scalar, .

gf

Solving for zero, we get

0 gf

This is the condition that must hold when the minimum of f subject to the

constraint cg is reached.

Lets consider minimizing ),( yxf subject to the constraint cyxg ),( . In this

case, ),( yxf is the objective function or the function to be minimized.

Geometrically, we want to find the extreme minimum of ),( yxf when thepoint ),( yx is restricted to lie on the level curve cyxg ),( . This situation

happens when these two curves intersect each other, that is, when the curves

have a common tangent. Based on the fact that the two curves are tangent at

a point if and only if their gradient vectors are parallel (or in the same

direction), hence for some scalar

),(),( yxgyxf

The scalar is called a Lagrange multiplier.

Definition

The Lagrangian associated with the constrained problem, defined as

),(),(),,( yxgyxfyxF

The following theorem provides the necessary conditions for the existence of

the Lagrange multiplier.

Theorem

Lagrange Mult ipl ier

Suppose ),( yxf and ),( yxg have continuous partial derivatives such

that ),( yxf has a local maximum or minimum at ),( 00 yx on the constraint

curve cyxg ),( and 0),( 00 yxg , then there exists such that

).,(),( 0000 yxgyxf

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To find the extremum points, in practice, consider the following equations

),(),( yxgyxf and cyxg ),(

These equations are solved for the unknowns x, y and . Then the local

extremum points are found among the solutions of these equations.

Method of Lagrange Multiplier

Let ),( yxf be the objective function such that f has a maximum or minimum

subject to the constraint cyxg ),( . To find the maximum or minimum

value(s) of f requires finding the solution(s) to the simultaneous equations

),(),( yxgyxf and cyxg ),( .

The largest and smallest point(s) are the maximum and minimum of f

respectively.

Example 1

Find the maximum and minimum values of yxyxf 2),( subject to the

constraint 122 yx .

Identify the objective function : ),( yxf

yxyxf 2),(

Identify the constraint : cyxg ),(

122 yx

Compute xf , yf , xg and yg

Solut ion

Steps : Lagrange multipliers



Find xf , yf , xg and yg

Set up equations

xx gf yy gf cyxg ),(

Solve the simultaneous equations for the unknownsx, yand

Determine the critical point(s) Determine the extreme of f

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1xf xgx 2

2yf ygy 2

Figure 9.3 Set up equations

xx gf x21 (1)

yy gf y22 (2)

1),( yxg 122 yx (3)

Solve the simultaneous equations

)1()2( :

x

y2 xy 2 (4)

Substitute (4) into (3) yields

15

14

1)2(

2

22

22

x

xx

xx

51

)5(5

12

x

x

Then substitute (5) into (4) yields

xy 2

When5

1x

5

2y

When5

1x

5

2y

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Thus there are two possible points on the constraint set where ),( yxf can

take an extremum.

5

2,

5

1

5

2,

5

1

Note that

55

2,

5

1

f 5

5

2,

5

1

f

Comparing the values, it can be concluded that yxyxf 2),( subject to

the constraint 122 yx has

(i) an absolute maximum of 5 at

5

2,

5

1

(ii) an absolute minimum of 5 at

5

2,

5

1

Example 2

Find the maximum of 22 21 yx subject to the constraint that 1 yx .


22 21),( yxyxf


1 yx


xfx 2 1xg

yfy 4 1yg

Set up equations

xx gf x2 (1)

yy gf y4 (2)

1),( yxg 1 yx (3)

Solve the simultaneous equations forxand y

)1()2( : 12 xy

2xy (4)

Solut ion

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Substitute (4) into (3)

12

xx

12

3

x

3

2x then

3

1y

Determine the critical point(s)

3

1,

3

2

Determine the extreme (maximum) of f

3

1

3

1,

3

2f

Example 3

Find the maximum and minimum values of 22),( xyyxf on the ellipse

44 22 yx .


22),( xyyxf


44 22 yx


xfx 2 xgx 2

yfy 2 ygy 8

Set up equations

xx gf xx 22 (1)

yy gf yy 82 (2)

4),( yxg 44 22 yx (3)

Solve the simultaneous equations forxand y

)2()1( :yx

yx

4

Solut ion

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)4(0@0

05

4

yx

xy

xyxy

Substitute (4) into (3)

When 0x :

1

1

44

2

2

y

y

y

When 0y :

2

42

x

x

Determine the critical point(s)

1,0 , 1,0 , 0,2 and 0,2

Determine the extreme of f

11,0 f 11,0 f

4,0,2 f 40,2 f

Hence, the maximum value of f is 1 and the minimum value is -4.

Example 4

Find the maximum and minimum values of 22),( xyyxf on the

circle 122 yx .

22),( xyyxf

1),( 22 yxyxg

xyxfx 2),( yyxfy 2),(

xyxgx 2),( yyxgy 2),(

Then

)2(22

)1(22

yx

yx

Solut ion

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122 yx (3)

Solving the equations yield

xyxy

y

x

y

x

)4(0@0

02

yx

xy

Substitute (4) into (3):

When 0x :

1

12

y

y

When 0y :

1

12

x

x

Thus, x = -1 and x = 1. Finally

11,0 f 11,0 f

10,1 f 10,1 f

Hence, the maximum value of f is 1 and the minimum value of f is -1.

Example 5

Find the volume of the largest rectangular box with a divider but no top that

can be constructed from 288 square cm of material.

Figure 9.4: Rectangular box with divider with no top

Solut ion

x

z

y

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Identify the objective and constraint functions:

28832),,(

),,(

yzxzxyzyxg

xyzzyxV

Determine the partial derivatives

yxgzxgzyg

xyVxzVyzV

zyx

zyx

3232

The Lagrange condition means

)4(28832

)3()32(

)2()3(

)1()2(

yzxzxy

yxxy

zxxz

zyyz

Solve the simultaneous equations:

(1) = (2):

yzxyxzxy

x

zx

y

zy

32

)3()2(

3

2

32

xy

yx

Replacing3

2xy in (3):

xx

xxx

yxxy

122

223

2

)32(

2

2

6@0

06

062

xx

xx

xx

Then

43

2xy

Substitute 6x in (2):

2@0

0)2(3

)36(6

z

z

zz

Replacingx, y, and z with 24,6 and , respectively, in the constraint

yields

2@2

028872

288242424

28832

2

222

yzxzxy

Since the dimension should be a positive then 2 then

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4

8

12

z

y

x

and

V(12, 8, 4) = 384 cubic cm

Consider

xyyxf ),( subject to 49),( 22 yxyxg

(i) Find yxyx gandgff ,, .

(ii) Set gf

(iii) Solve the simultaneous equations obtained in (ii)(iv) Determine the critical points and extrema.

Exercises 9

1. Use the method of Lagrange Multipliers to find the extrema of the following

functions subject to the given constraints.

Function Constraint

a) yxyxf 23),( 122 yx

b) yxyxf 2),( 122 yx

c) yxyxf 2),( 2522 yx

d) yxyxf ),( 12 22 yx

e) 22 2),( yxyxf 122 yx

f) yxyxf 2),( 122 yx

g) 22 2),( yxyxf 122 yx

h) 42),( yxyxf 122 yx

i) )(sin),( xyyxf 122 yx

j) xyeyxf ),( 122 yx

Warm up exercise

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2. A farmer has 400 meters of fence with which to enclose a rectangular field

bordering a river. What dimensions of the field maximize the area if the field is to

be fenced on only three sides (see picture below)?

3. Maximize the product of two positive numbers whose sum is 36.

4. Minimize the sum of two positive numbers whose product is 36.

5. A farmer has 400 meters of fence with which to fence in a rectangular field

adjoining two existing fences which meet at a right angle. What dimensions

maximize the area of the field?

6. A rectangular box with a square bottom and an open top is to have a volume of

1000 m3. What dimensions for the box yield the smallest surface area?

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7. A certain cylindrical can is to have a volume of 0.25 cubic meters. Find the

height hand radius rof the can that will minimize surface area of the can. What

is the relationship between the resulting rand h?

8. A nuclear reactor is to have the shape of a right circular cylinder with radius x

and height y. Neutron diffusion theory says that the reactor must have the

following constraint.

14048.2

22

y

p

x

What dimensionsxand yminimize the volume of the reactor?

9. Find the maximum of22

),( yxeyxf subject to the constraint

a) 1 yx

b) 122 yx

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MAT495_CHAPTER_9