<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] Alternatively, by AM-GM inequality, $x^2y = 2\sqrt{\dfrac{x^2}{2}\cdot \dfrac{x^2}{2}\cdot y^2}\le 2\sqrt{\dfrac{\left(\dfrac{x^2}{2}+\dfrac{x^2}{2}+ y^2\right)^3}{27}}= 2\cdot \sqrt{\dfrac{100^3}{27}}= \dfrac{2,000}{3\sqrt{3}}=\dfrac{2,000\sqrt{3}}{9}\implies (x^2y)_{\text{max}} = \dfrac{2,000\sqrt{3}}{9}$ at $\dfrac{x^2}{2} = y^2\implies x^2=2y^2\implies 3y^2=100\implies y = \dfrac{10\sqrt{3}}{3}, x = y\sqrt{2}=\dfrac{10\sqrt{6}}{3}\implies \boxed{\dfrac{x}{y} = {\sqrt{2}}}$.
Note: I just saw the comment by @Albus above after posting this answer. Any way, just leave it here....
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