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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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