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Minimum Distance Using Lagrange Multipliers


Date: 11/17/1999 at 20:27:48
From: Gloria Compise
Subject: Lagrange Multipliers

I'm trying to solve the following problem. Find the minimum distance 
from (0,0,c) to the cone z^2 = (x^2/a^2)+(y^2/b^2).

So far what I've done is,

     f(x,y,z) = x^2 + y^2 + (z-c)^2

              = x^2 + y^2 + (x^2/a^2) + (y^2/b^2) + c^2
                                      - 2c{(x^2/a^2) + (y^2/b^2)}^1/2

I know that I'm supposed to use Lagrange Multipliers, but I'm having a 
difficult time. Can you help me?

Thank you.


Date: 11/18/1999 at 12:08:01
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 our problem is to find stationary points on

     f(x,y,z) = x^2 + y^2 + (z-c)^2  subject to constraint

     g(x,y,z) = x^2/a^2 + y^2/b^2 - z^2 = 0

At stationary points of f(x,y,z) 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)= x^2 + y^2 +(z-c)^2

and 

     g(x,y,z) = x^2/a^2 + y^2/b^2 -z^2

The auxiliary function is

     phi(x,y,z) = f(x,y,z) - kg(x,y,z)
                = x^2+y^2+(z-c)^2 - k(x^2/a^2 + y^2/b^2 - z^2)

Then:

     part(d(phi)/dx) = 2x -k(2x/a^2) = 0   ......................(1)

     part(d(phi)/dy) = 2y -k(2y/b^2) = 0   ......................(2)

     part(d(phi)/dz) = 2(z-c) -k(-2z) = 0  ......................(3)

            g(x,y,z) = x^2/a^2 + y^2/b^2 - z^2 = 0   ............(4)

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

From (1) and (2) x and y can take any values and we could have:

From (1) 1 - k/a^2 = 0, so  k = a^2

     then from (2) y = 0

From (3) z-c + kz = 0, so  z(1+k) = c, z = c/(1+k)

          z = c/(1+a^2)

From (4) x^2/a^2 + y^2/b^2 - c^2/(1+k)^2 = 0

             x^2/a^2 + 0 - c^2/(1+a^2)^2 = 0

                                 x^2/a^2 = c^2/(1+a^2)^2
and taking square roots
                                     x/a = c/(1+a^2)

                                       x = ac/(1+a^2)

So a point on the cone nearest to (0,0,c) is

     x = ac/(1+a^2)
     y = 0
     z = c/(1+a^2).

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

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