1. ## LaGrange Multipliers

Consider the problem of finding the points on the surface $xy+yz+zx=3$ that are closest to the origin.

1) Use the identity $(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$ to prove that $x+y+z$ is not equal to 0 for any point on the given surface.

2) Use the method of Lagrange multipliers to find a system of four equations in $x,y,z$ and $\lambda$ whose solutions will give the closest points.

3) Find the points on $xy+yz+zx=3$ that are closest to the origin.

I'm clueless on what to do for the 1st part (although I imagine it's actually something simple), but I think I have the second part down. Problem is, I think I probably need to use the 1st part for the 3rd somehow.

For the second part, I found the system of four equations to be:

$2x=\lambda(y+z)$
$2y=\lambda(x+z)$
$2z=\lambda(x+y)$
$xy+yz+zx=3$

2. ## Part1

since xz+yx+yz = 3

x^2 +y^2 +z^2 +2(xz+yx+yz) >6 on the surface

3. ## solution

eliminating lambda

you should be able to show

x^2+ xz = y^2 + yz

(x-y)(x+y) = -z(x-y)

if x doesn't equal y then x+y = -z but this yields x+y+z = 0
but this can't happen so x = y

similarly you can show y^2 +xy = z^2 + xz

and use the same argument so x = y = z

use this in your constraint xy + xz +yz =3 to solve

4. Originally Posted by Calculus26
eliminating lambda

you should be able to show

x^2+ xz = y^2 + yz

(x-y)(x+y) = -z(x-y)

if x doesn't equal y then x+y = -z but this yields x+y+z = 0
but this can't happen so x = y

similarly you can show y^2 +xy = z^2 + xz

and use the same argument so x = y = z

use this in your constraint xy + xz +yz =3 to solve
Thank you so much, I understand the identity now, but if you don't mind me asking:

How would I go about eliminating lambda?

5. For example

from 2x = L(y+z)

Solve L =2x/(y+z)

From 2y = L(x+z)

Solve L = 2y/(x+z)

Therefore 2x/(y+z) = 2y/(x+z)

Similarly you can eliminate lambda using the 2d and 3d equations