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Find the critical point of $$ f(x,y) = 3x^3 + 3y^3 + x^3y^3 $$

To do this, I know that I need to set $$f_y = 0, f_x = 0 $$

So $$f_x= 9x^2 + 3x^2y^3$$ $$f_y = 9y^2 + 3y^2x^3$$

Then you solve for x, but substituting these two equations into each other.

But somehow I ended up with $$x = y$$ and thats not very helpful.

Is there something I did wrong or misunderstood?

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You could solve $f_{x} = 0$ so that you get $3x^2(3 + y^3) = 0$ then substitute that in your $f_{y}$ equation. That will give you the common roots of $f_{x}$ and $f_{y}$. – acyrl Sep 30 '12 at 19:26
If you did indeed end up with $x=y$ as a necessary condition for being a critical point, use that observation in one of the two derivatives you found to solve for, say, $x$, then go back and find $y$. – bwsullivan Sep 30 '12 at 19:41

1 Answer

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Well, $f_x=0$ is $3x^2(3-y^3)=0$, so either $x=0$ or $y^3=3$ (among reals it's $\sqrt[3]3$). And, at the same time $f_y=0$ must also hold, that is ($y=0$ or $x=\sqrt[3]3$).

It gives you the $(0,0)$ and $(\sqrt[3]3,\sqrt[3]3)$ solutions.

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It's $3+y^3$, not $3-y^3$. – Hans Lundmark Sep 30 '12 at 20:16

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