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This is the function: $$f(x,y)=xye ^{-x-y}$$ I have trouble finding critical points and fxx, fyy and fxy. I think fx is $$fx=(x-1)y(-e ^{-x-y)}$$ $$fy=x(y-1)(-e ^{-x-y)}$$

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up vote 1 down vote accepted

The partials are $${f_x} = y{e^{ - x - y}}(1 - x)$$ $${f_y} = x{e^{ - x - y}}(1 - y)$$ so your answers are correct. With these you can find ${f_{xx}}$, ${f_{yy}}$ and ${f_{xy}}$ using the product rule and the chain rule.

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so i got fxx=e^(-x-y), fyy=e^(-x-y)+y-1 and fxy=e^(-x-y)+y-1 – Jack F Oct 22 '12 at 22:23
D(x,y)=(e^(-x-y)+y)(e^(-x-y)+y-1) -(e^(-x-y)+y-1)^2 – Jack F Oct 22 '12 at 22:24
I think ${f_{xx}} = y{e^{ - x - y}}(x - 2) = {f_{yy}}$. – glebovg Oct 22 '12 at 22:29
are you sure about it? – Jack F Oct 22 '12 at 22:34
Yes. Take $y{e^{ - x - y}}$ and $(1 - x)$, then use the product rule and the chain rule. Then take $x{e^{ - x - y}}$ and $(1 - y)$ to compute ${f_{yy}}$. Do not forget to factor. – glebovg Oct 22 '12 at 22:40

Notice the exponential function is always positive(or negative). So, for $f_x = 0$, you have either $x = 0$ or $y =0$ and in the $f_y = 0$, you have $x = 0$ or $y=1$. Now, what are the critical points???? You know should be able to find them.

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(0,0), (1,1)? but i have a hard time finding fxx and fyy and fxy now. – Jack F Oct 22 '12 at 22:13

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