<s>[INST] Extrema of $f(x,y) = xy\sqrt{1 - \frac{x^2}{a^2} - \frac{y^2}{b^2}}$ From Demidovich

Find the extrema of the function of $z = xy \sqrt{1 - \frac{x^2}{a^2} - \frac{y^2}{b^2}}$

What I've done so far:
I've make $z_{x}'= 0$ and $z_{y}' = 0$, wich gives me:
$$\frac{\partial z}{\partial x} = y \sqrt{1 - \frac{x^2}{a^2} - \frac{y^2}{b^2}} - \frac{yx}{a^2 \sqrt{1 - \frac{x^2}{a^2} - \frac{y^2}{b^2}}} = 0$$
$$\frac{\partial z}{\partial y} = x \sqrt{1 - \frac{x^2}{a^2} - \frac{y^2}{b^2}} - \frac{yx}{b^2 \sqrt{1 - \frac{x^2}{a^2} - \frac{y^2}{b^2}}} = 0$$
And I've tried to simplify, and I got:
$$\frac{y a^2 (1 - \frac{x^2}{a^2} - \frac{y^2}{b^2} ) - xy}{a^2 \sqrt{1 - \frac{x^2}{a^2} - \frac{y^2}{b^2}}} = 0$$
$$\frac{x b^2 (1 - \frac{x^2}{a^2} - \frac{y^2}{b^2} ) - xy}{b^2 \sqrt{1 - \frac{x^2}{a^2} - \frac{y^2}{b^2}}} = 0$$
Now the problem is, some cubic terms appear and I don't know how to solve this and find the critical points to proceed with the exercise. Of course the denominator cannot be zero, so my system is composed only by the numerators.
I'm pretty sure I'm doing something wrong, 'cause this seems too much trouble for an introductory exercise (it's one of the firsts in the chapter)
 [/INST] With the polar coordinates $\frac xa = r\cos t$ and $\frac yb  =r \sin t$,
$$z=\frac12ab r^2\sin 2t \sqrt{1-r^2}$$
For any given $r\le 1$, 
$$|z(r)|\le \frac12abr^2 \sqrt{1-r^2}$$
Then, set $(r^2 \sqrt{1-r^2})'=0$ to locate the extreme at $r^2=\frac23$. Thus, 
$$-\frac{ab}{3\sqrt3}\le z \le \frac{ab}{3\sqrt3}$$
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