- #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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