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Lagrange Multipliers


Date: 01/28/2001 at 04:56:22
From: li
Subject: Lagrange Multipliers

The temperature of a point(x,y,z) on the unit sphere is given by 
T(x,y,z)=xy+yz. Using Lagrange multipliers, find the temperature of 
the hottest point on the sphere.


Date: 01/28/2001 at 06:03:18
From: Doctor Mitteldorf
Subject: Re: Lagrange Multipliers

Dear Li,

The idea of Lagrange multipliers is this:

For globabl maximization, you would set all the partial derivatives 
= 0 (dT/dx = 0, dT/dy = 0, dT/dz = 0).

For Lagrange optimization, you set the vector of partial derivatives 
(dT/dx, dT/dy, dT/dz) equal to a constant multiple lambda times a 
vector that is orthogonal to your surface at every point. In this 
case, the vector (x,y,z) will do just fine.  

So you have the equation of the sphere plus three more equations in 
the four variables x,y,z, lambda.  The three equations are

dT/dx = lambda x
dT/dy = lambda y
dT/dz = lambda z

- Doctor Mitteldorf, The Math Forum
  http://mathforum.org/dr.math/   


Date: 01/28/2001 at 08:14:51
From: Doctor Anthony
Subject: Re: Lagrange Multipliers

The general problem is to find stationary points of f(x,y,z) subject 
to constraint g(x,y,z) = 0  [Note that the constraint must be written 
in this form]. So for our problem g(x,y,z) = x^2 + y^2 + z^2 - 1

and the required equation is

      phi(x,y) = f(x,y) - kg(x,y)

where k is the Lagrange multiplier and phi is the auxiliary function.

Then f(x,y,z) will have a stationary point subject to constraint 
g(x,y,z) = 0 when part(d(phi)/dx) = 0, part(d(phi)/dy) = 0, 
part(d(phi)/dz) = 0 and  g(x,y) = 0

This gives four equations to find x, y, z and k.

Applying these ideas to our problem, we have

f(x,y,z) = xy + yz  and  g(x,y,z) = x^2 + y^2 + z^2 - 1

The auxiliary function is

phi(x,y,z) = f(x,y,z) - kg(x,y,z)
           = xy + yz - k[x^2 + y^2 + z^2 - 1]

Then:

part(d(phi)/dx) = y - 2kx     = 0          (1)

part(d(phi)/dy) = x + z - 2ky = 0          (2)

part(d(phi)/dz) = y - 2kz     = 0          (3) 

      g(x,y,z) = x^2 + y^2 + z^2 - 1 = 0   (4)

From (1) and (3) z = x and then from (2)

         2x - 2ky = 0     so  ky = x  and returning to (1)

       y - 2k(ky) = 0     ->  1 - 2k^2 = 0

              k^2 = 1/2   ->   k = +-sqrt(1/2)

We have  x = z = ky and substituting into (4)

     k^2.y^2 + y^2 + k^2.y^2 - 1 = 0

          y^2(2k^2 + 1) - 1 = 0

             y^2(1 + 1) - 1 = 0

                       2y^2 = 1    y^2 = 1/2

Then x^2 = 1/4 and  z^2 = 1/4

So the stationary points will be at  [+-1/2, +-1/sqrt(2), +-1/2]

It is obvious that the temperature, xy+yz, will be maximum if all 
signs are positive or all signs are negative.

  So max temp = (1/2)(1/sqrt(2)) + (1/sqrt(2))(1/2)

              =  1/sqrt(2)

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
College Calculus
High School Calculus

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