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I'm a bit confused on the procedures for finding the zeros of the partial derivatives of these kind of functions. If you could correct what I am doing wrong it would be much appreciated. $$ f(x,y)=8xy-\frac {1}{4}(x+y)^{4}$$ I took the partial derivatives as follows. $$ \frac{\partial}{\partial x}=8y-(x+y)^{3}$$ $$ \frac{\partial}{\partial y}=8x-(x+y)^{3}$$ When I set both partials to zero I got $0=8x-(x+y)^{3}=8y-(x+y)^{3}$ In which i concluded that $x=y$. When I plug that into one of the partial derivatives I get that $y=y^{3}$. I don't see how that helps me in any way. I know that the critical points occur at $(-1,-1)$, $(0,0)$, and $(1,1)$ but I don't see how to get there.

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You need to solve the equation $y=y^3$, or $y-y^3=0$. Factor out the $y$ to get $$ y(y^2-1)=0 $$ Can you see what values $y$ must have now?

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  • $\begingroup$ Yes, thank you so much that is very clear thank you :) $\endgroup$ – Jeremy Correa Mar 2 '15 at 5:49

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