# Math Help - Easy Lagrange optimization problem help

1. ## Easy Lagrange optimization problem help

Find the max/min for $3xy$ subject to $x^2 + y^2 = 8$

$L = 3xy - x^2\lambda - y^2\lambda + 8\lambda$

$L'(x) = 3y - 2x\lambda = 0$
$L'(y) = 3x -2y\lambda = 0$
$L'(\lambda) = -x^2 - y^2 + 8 = 0$

$3y=2x\lambda$
$y=(2x\lambda)/3$

$3x=3y\lambda$
$3x=3((2x\lambda)/3)\lambda$
$x=4\lambda$

Am I doing anything wrong so far? I can't seem to get the right answer. All help appreciated!

2. ## Re: Easy Lagrange optimization problem help

I think that we can solve this problem without Lagrange optimization tools:

We have a circle center (0,0) with radius of sqrt(8), our mission is to find point, (x,y), on that circle, with max{x*y}, try to show that such point have the following property -- x=y.

Therefor, if x=y, we'll get,

2x^2=8

x=2
y=2

and 3xy=12.

3. ## Re: Easy Lagrange optimization problem help

$\lambda=\frac{3y}{2x}=\frac{3x}{2y}$

$x^2=y^2$

Substituting this in $L'(\lambda)=-x^2-y^2+8=0$, we get $x=\pm 2$ and $y=\pm 2$. Clearly, the maximum is 12 and minimum is -12.

4. ## Re: Easy Lagrange optimization problem help

Originally Posted by alexmahone
$\lambda=\frac{3y}{2x}=\frac{3x}{2y}$

$x^2=y^2$

Substituting this in $L'(\lambda)=-x^2-y^2+8=0$, we get $x=\pm 2$ and $y=\pm 2$. Clearly, the maximum is 12 and minimum is -12.
Ah thanks a lot!