constrained optimization using lagrange multipliers

[burt]

I was given the following question:
Find the maximum and minimum of the function \(f(x,y)=x^2y^2\) subject to \(x^2+4y^2=24\).
I tried using lagrange multipliers and got the following:


I was assuming that I would get different answers, and that the biggest one would be the maximum and the smallest would be the minimum. This did not happen. Where did I go wrong, and what steps can I take to solve this problem?

 

[MarkFL]

If I were to work this problem, I would write:

[MATH]2xy^2=\lambda(2x)[/MATH]
[MATH]2x^2y=\lambda(8y)[/MATH]
This implies:

[MATH]\frac{2xy^2}{2x}=\frac{2x^2y}{8y}[/MATH]
[MATH]4xy^3-x^3y=0[/MATH]
[MATH]xy(2y+x)(2y-x)=0[/MATH]
When you divided out \(x\) and \(y\) you eliminated potential critical points.

1.) [MATH]x=0\implies y=\pm\sqrt{6}[/MATH]
[MATH]f(0,\pm\sqrt{6})=0[/MATH]
2.) [MATH]y=0\implies x=\pm2\sqrt{6}[/MATH]
[MATH]f(\pm2\sqrt{6},0)=0[/MATH]
3.) [MATH]2y+x=0\implies (x,y)=(\pm2\sqrt{3},\mp\sqrt{3})[/MATH]
[MATH]f(\pm2\sqrt{3},\mp\sqrt{3})=36[/MATH]
3.) [MATH]2y-x=0\implies (x,y)=(\pm2\sqrt{3},\pm\sqrt{3})[/MATH]
[MATH]f(\pm2\sqrt{3},\pm\sqrt{3})=36[/MATH]
And so, I would conclude that:

[MATH]f_{\min}=0[/MATH]
[MATH]f_{\max}=36[/MATH]

 

[burt]

If I were to work this problem, I would write:

[MATH]2xy^2=\lambda(2x)[/MATH]
[MATH]2x^2y=\lambda(8y)[/MATH]
This implies:

[MATH]\frac{2xy^2}{2x}=\frac{2x^2y}{8y}[/MATH]
[MATH]4xy^3-x^3y=0[/MATH]
[MATH]xy(2y+x)(2y-x)=0[/MATH]
When you divided out \(x\) and \(y\) you eliminated potential critical points.

1.) [MATH]x=0\implies y=\pm\sqrt{6}[/MATH]
[MATH]f(0,\pm\sqrt{6})=0[/MATH]
2.) [MATH]y=0\implies x=\pm2\sqrt{6}[/MATH]
[MATH]f(\pm2\sqrt{6},0)=0[/MATH]
3.) [MATH]2y+x=0\implies (x,y)=(\pm2\sqrt{3},\mp\sqrt{3})[/MATH]
[MATH]f(\pm2\sqrt{3},\mp\sqrt{3})=36[/MATH]
3.) [MATH]2y-x=0\implies (x,y)=(\pm2\sqrt{3},\pm\sqrt{3})[/MATH]
[MATH]f(\pm2\sqrt{3},\pm\sqrt{3})=36[/MATH]
And so, I would conclude that:

[MATH]f_{\min}=0[/MATH]
[MATH]f_{\max}=36[/MATH]

Thank you! Now I see the proper way to eliminate the \(\lambda\).
This makes a lot of sense!