Finding critical points of f(x,y)

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 $y$. – brendansullivan07 Sep 30 '12 at 19:41

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

If you solve the system $f_x=0=f_y$ , you will the following four points:

         [0, -1.44, -9], [-1.44, 0, -9], [0, 0, 0], [-1.44, -1.44, -9]


These are all critical points. Now, we should examine the signs of $f_{xx}$ and $$f_{xx}f_{yy}-f^2_{xy}$$ at these points. If the Hessian gets zero at any of them, then we don't know whether that point is max, min or a saddle point and so we need additional considerations.

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Needs a thumb's up! +1 – amWhy Aug 23 '13 at 11:03

Getting $x=y$ is very useful! This is because say you have two equations: $$1. f_x=9x^2+3x^2y^3$$ $$2. f_y=9y^2+3y^2x^3$$ Substituting $x=y$, or $y=x$ into both equations and making the left side equal to zero will yield the same result: $$0=9x^2+3x^2x^3$$ $$0=9x^2+3x^5$$ $$0=3x^2(3+x^3)$$ $$(0=3x^2) or (3+x^3=0)$$ $$(0=x^2) or (x^3=-3)$$ $$(x=0) or (x=-(3)^{1/3})$$

Again, since we know that $x=y$, we know that the critical points are $(0,0,0)$ and $(-3^{1/3},-3^{1/3},-9)$

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The two equations $$f_x=3x^2(3+y^3)=0,\qquad f_y=3y^2(3+x^3)=0$$ lead to the two critical points $(0,0)$, $\>(-\sqrt[3]3,-\sqrt[3]3)$.

In order to qualify the critical point $(0,0)$ we consider the function $\phi(x):=f(x,0)=3x^3$. Since $\phi$ assumes as well positive as negative values in the immediate neighborhood of $0$ we can conclude that $f$ does not assume a local extremum at $(0,0)$.

To analyze the critical point $(-\sqrt[3]3,-\sqrt[3]3)$ we compute the Hessian $$\left[\matrix{18x+6xy^3 &9x^2y^2\cr 9x^2y^2 &18y+6yx^3\cr}\right]\ .$$ Its determinant is $$9xy\bigl(36+12(x^3+y^3)-5x^3y^3\bigr)\ ,$$ which is negative at $(-\sqrt[3]3,-\sqrt[3]3)$. Therefore we don't have a local extremum at $(-\sqrt[3]3,-\sqrt[3]3)$ either.

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