- #1

goatsebear

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## Homework Statement

Find the stationary points and local extreme values of f(x,y) = xy^2e^-([tex]\frac{x^2 + y^2)}{2}[/tex]

## Homework Equations

You need to find the gradient for the function and then set it equal to 0. So then df/dy equals df/dx and you can solve for a solution set.

## The Attempt at a Solution

df/dx = y^2(-x^2 + 1)*e^-([tex]\frac{x^2 + y^2)}{2}[/tex]

df/dy = y*x(-y^2 + 2)*e^-([tex]\frac{x^2 + y^2)}{2}[/tex]

So since e^-([tex]\frac{x^2 + y^2)}{2}[/tex] cannot equal zero, y^2(-x^2 + 1) = 0

and y*x(-y^2 + 2) = 0

So then, y^2(-x^2 + 1) goes to y^2 - x^2y^2 = 0 and then [tex]\pm[/tex]1 = x

Plugging this into the dy/dx equation gives me the following x, y critical points

(0,0) (1,-[tex]\sqrt{2}[/tex]) (1, -[tex]\sqrt{2}[/tex]) (-1, [tex]\sqrt{2}[/tex]) (-1, [tex]\sqrt{-2}[/tex])

So then to find out if they are max or mins or saddle points, I solve for the second partials of fx, fy, fxy

fxx=A = x(-1 + 2y^2 - x^2y^2) * e^-([tex]\frac{x^2 + y^2)}{2}[/tex]

fyy=C = y(x - 2y + 3yx) * e^-([tex]\frac{x^2 + y^2)}{2}[/tex]

fxy=B = y(2- 2x^2 + y^2 = Y2x^2) * e^-([tex]\frac{x^2 + y^2)}{2}[/tex]Now I use the equation D = AC - B^2

Can someone check my math out? It seems that something is wrong somewhere. I know this is a huge problem so thanks for any help.

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