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Lagrange Multipliers

Academic Resource Center

In This Presentation..

ͻWe will giǀe a definition

ͻDiscuss some of the lagrange multipliers

ͻLearn how to use it

ͻDo edžample problems

Definition

Lagrange method is used for maximizing or minimizing a general function f(x,y,z) subject to a constraint (or side condition) of the form g(x,y,z) =k.

Assumptions made: the extreme values exist

Then there is a number ʄ such that

and ʄ is called the Lagrange multiplier.

Finding all values of x,y,z and ʄ such that

and g(x,y,z) =k And then evaluating f at all the points, the values obtained are studied. The largest of these values is the maximum value of f; the smallest is the minimum value of f. components, give It is a system of four equations in the four unknowns, however it is not necessary to find explicit values for ʄ. A similar analysis is used for functions of two variables.

Examples

Example 1:

A rectangular box without a lid is to be made from 12 m2 of cardboard. Find the maximum volume of such a box.

Solution:

let x,y and z are the length, width and height, respectively, of the box in meters. and V= xyz

Constraint: g(x, y, z)= 2xz+ 2yz+ xy=12

Using Lagrange multipliers,

Vx= ʄgx Vy= ʄgy Vz= ʄgz 2xz+ 2yz+ xy=12 which become

Continued..

yz= ʄ(2zнy) (1) xz= ʄ(2zнdž) (2) xy= ʄ(2džн2y) (3)

2xz+ 2yz+ xy=12 (4)

Solving these equations;

Let's multiply (2) by dž, (3) by y and (4) by z, making the left hand sides identical.

Therefore,

x yz= ʄ(2džzнdžy) (6) x yz= ʄ(2yzнdžy) (7) x yz= ʄ(2džzн2yz) (8) continued

2xz+xy=2yz+xy

which gives xz = yz. But z т 0, so dž с y. From (7) and (8) we haǀe

2yz+xy=2xz+2yz

which gives 2xz = xy and so (since x т0) y=2z. If we now put x=y=2z in (5), we get

4z2+4z2+4z2=12

Since x, y, and z are all positive, we therefore have z=1 and so x=2 and y = 2.

More Examples

Example 2:

Find the extreme values of the function f(x,y)=x2+2y2 on the circle x2+y2=1.

Solution:

multipliers

Constraint: g(x, y)= x2+y2=1

Using Lagrange multipliers,

fx= ʄgx fy= ʄgy g(x,y) = 1 which become

2x= 2xʄ (9)

4y= 2yʄ (10)

x2+y2= 1 (11) From (9) we haǀe džс0 or ʄс1. If džс0, then (11) giǀes yс±1. If ʄс1, then yс0 from (10), so then (11) giǀes džс±1. Therefore f has possible extreme values at the points (0,1), (0,-1), (1,0), (1,0). Evaluating f at these four points, we find that f(0,1)=2 f(0,-1)=2 f(1,0)=1 f(-1,0)=1

Therefore the maximum value of f on the circle

x2+y2=1 is f(0,±1) =2 and the minimum value is f(±1,0) =1.

More Examples.

Example 3

Find the extreme values of f(x,y)=x2+2y2 on the disk x2+y2ч1.

Solution:

Compare the values of f at the critical points with values at the points on the boundary. Since fx=2x and fy=4y, the only critical point is (0,0). We compare the value of f at that point with the extreme values on the boundary from Example 2: f(0,0)=0 f(±1,0)=1 f(0,±1)=2 Therefore the maximum value of f on the disk x2+y2ч1 is f(0,±1)=2 and the minimum value is f(0,0)=0.

Example 4

Find the points on the sphere x2+y2+z2=4 that are closest to and farthest from the point (3,1,-1).

Solution:

The distance from a point (x,y,z) to the point (3,1,-1) is d=(džо3)2н(yо1)2н(zн1)2 But the algebra is simple if we instead maximize and minimize the square of the distance: d2=f(x,y,z)=(x-3)2+(y-1)2+(z+1)2

Constraint: g(x,y,z)= x2+y2+z2=4

This gives

Continued

2(x-3)=2xʄ (12)

2(y-1)с2yʄ (13)

2(zн1)с2zʄ (14)

x2+y2+z2=4 (15) The simplest way to solve these equations is to solve for x, y, and z in terms of ʄ from (12), (13), and (14), and then substitute these values into (15). From12 we have x-3=džʄ or x(1- ʄ)=3 or x = ଷ

5?ʄ

Continued

Similarly (13) and (14) give

Lଵ

5?ʄ

zൌ

Fଵ

5?ʄ

Therefore, from (15), we have

Which gives (1- ʄ)2=ଵଵ

8 , 1- ʄ=±ଵଵ

6, so

ɉ=1±ଵଵ

6 These values of ɉ then give the corresponding points (x,y,z):

55ǡ

F଺

55ǡ

f has a smaller value at the first of these points, so the closest

55ǡ

55ǡ

Two constraints

Say there is a new constraint, h(x,y,z)=c.

So there are numbers ʄ and quotesdbs_dbs14.pdfusesText_20