<s>[INST] Finding maxima values of a multivariate function 
Find the maximal value of the function for $a=24.3$, $b=41.5$:
  $$f(x,y)=xy\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}$$

Using the second derivative test for partial derivatives, I find the critical point in terms of $a$ and $b$ by taking partial derivatives of $x$ and $y$ and equating them to $0$. 
$$f_y=0$$
$$f_x=0$$
Getting
$$y \left ( 1-\frac{x^2}{a^2}-\frac{y^2}{b^2} \right ) -\frac{yx^2}{a^2}=0$$
and
$$x \left ( 1-\frac{x^2}{a^2}-\frac{y^2}{b^2} \right ) - \frac{xy^2}{b^2}=0$$
Then i combine both equations together to get
$$\left ( 1-\frac{y^2}{b^2} \right ) =\frac{2x^2}{a^2}$$
and
$$\left ( 1-\frac{x^2}{a^2} \right ) =\frac{2y^2}{b^2}$$
Solving both equations to get the critical points in terms of a and b. I got
$$(0,0)$$
$$\left(\frac{a}{\sqrt{3}},\frac{b}{\sqrt{3}}\right)$$
Hence to get the maximum value I substitute the second critical point back into the original function. And I let $a=24.3$, $b=41.5$ to get the maximal point. However, I don't seem to get the right answer.
Is my method correct?
 [/INST] Another way is to consider equivalently the maximum of 
$$\frac{f^2}{a^2b^2}=\frac{x^2}{a^2} \frac{y^2}{b^2} \left(1-\frac{x^2}{a^2} -\frac{y^2}{b^2} \right)$$
which is the product of three positive terms with constant sum, so each term must be equal to (in this case) $\frac13$ at maximum.
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