What are the extremes of the function $ f\left(x,y\right)=xy\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}} $?

Question

$ f\left(x,y\right)=xy\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}} $ ; where a,b are constants.

I have tried to differentiate the equation, but I get stuck in a very hard system after this.

The answer of the problem is:

$ f_{max}=\frac{ab}{3\sqrt{3}};f_{min}=-\frac{ab}{3\sqrt{3}} $

Answer 1

$\frac{f(x,y)}{ab}=\pm\sqrt{g(x^2/a^2,y^2/b^2)}$ where $$g(u,v)=uv(1-u-v)\quad(u,v\ge0,u+v\le1).$$ $g_\min=0$ is attained all along the boundary of the triangle, so that $g_\max$ must be attained at some point of the interior ($u,v>0,u+v<1$).

$$\frac{\partial g}{\partial u}=v(1-2u-v),$$ $$\frac{\partial g}{\partial v}=u(1-u-2v),$$ and solving $$1-2u-v=1-u-2v=0$$ you find that $$g_\max=g(1/3,1/3)=1/27$$ hence $$f_\max=\frac{ab}{\sqrt{27}}$$ (and $f_\min=-f_\max$).

Answer 2

I'll prove that $(f(x,y) - \frac{ab}{3\sqrt{3}})(f(x,y) + \frac{ab}{3\sqrt{3}}) \leq 0$ which is equivalent.

$(f(x,y) - \frac{ab}{3\sqrt{3}})(f(x,y) + \frac{ab}{3\sqrt{3}}) \leq 0 \iff$

$f(x,y)^2 - \frac{(ab)^2}{27}\leq 0 \iff$

$x^2y^2(1 -\frac{x^2}{a^2} - \frac{y^2}{b^2}) \leq \frac{(ab)^2}{27} \iff$

$x^2y^2 \leq \frac{x^4y^2}{a^2} + \frac{x^2y^4}{b^2} + \frac{(ab)^2}{27}$ which is true by $AM-GM $.

Equality is achieved at $x = \pm\frac{a}{\sqrt{3}}$ and $y = \pm\frac{b}{\sqrt{3}}$.

EDIT:

$AM-GM$ for 3 terms:

$3\sqrt[3]{pqr} \leq p+q+r$.

Take $p = \frac{x^4y^2}{a^2}, q=\frac{x^2y^4}{b^2}, r=\frac{(ab)^2}{27}$.