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?
 A: I think you are missing some solutions: for instance, if $y=0$ also the points $(a,0)$ and $(-a,0)$ are solutions. Maybe during the calculations you divided by $y$ and/or $x$ without checking what happens when they vanish.
A thing that often helps me "seeing" the problem before actually calculating the solution is searching for structures. Your function is defined on the elliptic region $D=\{(x,y)\in\mathbb{R}^2|\frac{x^2}{a^2}+\frac{y^2}{b^2}\leq1\}$, it is zero on the border of $D$ (i.e. on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$) and on the axes. This might help visualize the graph of $f$ and could give an idea of where the maxima/minima might be.
As for the calculations, you correctly get the system
\begin{equation}
y\left(1-\frac{x^2}{(a/2)^2}-\frac{y^2}{b^2}\right)=0\quad\text{and}\quad x\left(1-\frac{x^2}{a^2}-\frac{y^2}{(b/2)^2}\right)=0
\end{equation}
Here again you can spot two ellipses on which your derivatives are constantly zero. Let's call $E_1$ the ellipse $\frac{x^2}{(a/2)^2}+\frac{y^2}{b^2}=1$ and $E_2$ the ellipse $\frac{x^2}{a^2}+\frac{y^2}{(b/2)^2}=1$. When intersecting ellipses you can have up to 4 solutions; in your case you have to intersect the set $S_1=\{y=0\}\cup E_1$ with the set $S_2=\{x=0\}\cup E_2$ (you can see them as "barred" ellipses) so the solutions can be more (up to 9). If you get only two solutions maybe you'd like to check your calculations again (I'm not saying that you $must$ have more than two solutions, only saying that you $could$).
For instance, if $y=0$, then for $x$ you get the values $0,a,-a$ so all the points $(0,0),(a,0)$ and $(-a,0)$ are solutions. Similarly, if $x=0$ you have the solutions $(0,0),(0,b)$ and $(0,-b)$. For other solutions you have to intersect the two ellipses $E_1$ and $E_2$. Hope that helps
A: 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.
