Find the extreme values of the function subject to the given constraint

**downthesun01** · October 20th 2010, 09:51 PM

$f(x,y)=x^{2}y$ , $x^{2}+2y^{2}=6$

Ok, so I'm trying to do this using Lagrange multipliers

So, first I need $\nabla f=\lambda \nabla g$ , right?

Which is:

$2xyi+x^{2}j=\lambda (2xi+4yj)$

So:

$2xy=\lambda 2x$ and $x^{2}=\lambda 4y$

From here, I have no idea what to do. Any help?

Edit:

Ok, I THINK I figured it out. Please someone double check my work

So,

$y=\lambda$ and $x^{2}=4\lambda^{2}$

We plug these into the constraint and get:

$4\lambda^{2}+2\lambda^{2}=6$

$=>\lambda=\pm 1$

Then we plug $\lambda$ back into $y=\lambda$ and $x=2\lambda$

So we get, $y=\pm 1$ and $x=\pm 2$

Then, plugging these numbers into $f(x,y)=x^{2}y$

$(-2)^{2}(-1)=-4$ and $(2)^{2}(1)=4$

Is this all correct?

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