<s>[INST] Maximize $x^2y$ where $x^2+y^2=100$. (Looking for other approaches) 
For positive numbers $x,y$ we have $x^2+y^2=100$. For which ratio of $x$ to $y$, the value of $x^2y$ will be maximum?
$1)\ 2\qquad\qquad2)\ \sqrt3\qquad\qquad3)\ \frac32\qquad\qquad4)\ \sqrt2$

It is a problem from a timed exam so I'm looking for alternative approaches to solve this problem quickly. Here is my approach:
We have $x^2=100-y^2$, so $x^2y=y(100-y^2)$. By differentiating with respect to $y$ and putting it equal to zero, we have:
$$(100-y^2)-2y^2=0\qquad\qquad y^2=\frac{100}3$$
Hence $x^2=\frac{200}3$ and $\frac{x^2}{y^2}=2$ therefore $\frac{x}{y}=\sqrt2$.
Is there another quick approach to solve it? (like AM-GM inequality or others)
 [/INST] I think your solution is fine, the only change I would make it to point out that the constraint $x^2 + y^2 = 100$ can be changed to $x^2 + y^2 = z$ for any positive $z$ without changing the solution, since it is just ratios.  But here is an alternative:
$$x^2 + y^2 = 1$$
$$r = x^2y$$
$$x = wy$$
Write r in terms of w:
$$r = (wy)^2y = w^2y^3$$
$$(wy)^2 + y^2 = 1 \implies y^2 = (1+w^2)^{-1}$$
Using the old "the critical points of $r$ are included in the critical points of $r^2$" trick:
$$r^2 = \dfrac{w^4}{(1 + w^2)^{3}} = \dfrac{1}{w^2 + 3 + 3w^{-2} + w^{-4}} > 0$$
$$d(r^2) = -r^{2}(2w - 6w^{-3} - 4w^{-5}) = 0$$
$$w^6 - 3w^{2} - 2 = 0$$
$$(w^2 - 2)(w^2+1)^2 = 0$$
$$w = \sqrt 2$$
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