Lagrange Multipliers

Date: 07/25/99 at 14:21:09
From: Robert Botnick
Subject: Lagrange Multipliers

My class was given this problem for extra credit. Now the class is 
over, and it's still driving me crazy. Can you help?

Let F(x,y) = xy + yz + xz. Find the extreme values of F. The 
constraint is x^2 + y^2 + z^2 = 1. Solve using Lagrange multipliers.

Date: 07/25/99 at 17:30:25
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

At stationary points of f(x,y) we have

     df = part(df/dx)*dx + part(df/dy)*dy + part(df/dz)*dz = 0

This implies that the vector [part(df/dx), part(df/dy), part(df/dz)] 
is perpendicular to the vector [dx, dy, dz]

Since g(x,y,z) = 0, we can write

     dg = part(dg/dx)*dx + part(dg/dy)*dy + part(dg/dz)*dz = 0

Thus the vector [part(dg/dx), part(dg/dy), part(dg/dz)] is also 
perpendicular to the vector [dx, dy, dz]. This implies that the vector 
[part(df/dx), part(df/dy), part(df/dz)] is parallel to the vector 
[part(dg/dx), part(dg/dy), part(dg/dz)] and that we can find a scalar 
k such that

     [part(df/dx), part(df/dy), part(df/dz)] - k[part(dg/dx), 
part(dg/dy), part(dg/dz)] = [0, 0, 0]
                                    
This can be summarized by writing

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

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,z) = 0.

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

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

Applying these ideas to our problem, we have

     f(x,y,z)= xy + yz + zx   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+zx - k(x^2 + y^2 + z^2 - 1)

Then:
     part(d(phi)/dx) = y + z - 2kx = 0   ...............(1)

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

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

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

Now we must solve (1), (2), (3) and (4) for k, x, y and z.

Adding (1), (2) and (3)

     2(x+y+z) - 2k(x+y+z) = 0

dividing out by 2(x+y+z) this gives 1-k = 0 and so k = 1.

Subtracting (2) from (1)

     (y-x) - 2(x-y) = 0
             3(x-y) = 0

This clearly requires x = y, similarly we can show x = z, so we must 
have: 

     x = y = z

and also

     x^2 + y^2 + z^2 = 1 
                3x^2 = 1
                 x^2 = 1/3
                   x = +-1/sqrt(3)

So the stationary points are 

     x = +-1/sqrt(3),   y = +-1/sqrt(3),   z = +-1/sqrt(3)

The combinations of signs must be such as to satisfy equations (1), 
(2) and (3) above. This means we can take all + signs or all - signs.

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

Date: 07/28/99 at 23:29:51
From: Robert Durant
Subject: Re: Lagrange Multipliers

Thanks for writing back so quickly. I have one more question though. 
In this problem you assume that (x+y+z) is not equal to zero. What if 
(x+y+z) is equal to zero? Would you then be able to calculate f min? 
Is there a min in that case?

Date: 07/29/99 at 06:54:03
From: Doctor Anthony
Subject: Re: Lagrange Multipliers

Yes. You are correct. I should have considered the alternative 
possibility that x+y+z = 0.

This means y+z = -x and equation (1) above becomes

      -x -2kx = 0
     -x(1+2k) = 0

so k = -1/2. Putting this into (1), (2) and (3) gives x+y+z = 0 in 
each case, so we must use equation (4) to see what happens to f(x,y,z)

Consider

     (x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy +yz + zx)
             0 =   1  + 2(xy + yz + zx)
            -1 =  2.f(x,y,z)

and another stationary value is f(x,y,z) = -1/2.

Going back to the first solution we have:

     f(x,y,z) = 1/3 + 1/3 + 1/3 = 1

So the two stationary values are 1 and -1/2

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