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)

One approach could be to parameterize the curve $x^2+y^2=100$. Define the functions $x$ and $y$ by $x(t)=10\cos(t)$ and $y(t)=10\sin(t)$ on the interval $(0,\pi/2)$ (this constraint ensures that $x$ and $y$ are bijective and strictly positive, since $\cos(t)>0$ and $\sin(t)>0$ for all $t\in(0,\pi/2)$). The quantity $x^2y$ is then simply $\cos^2(t)\sin(t)$, and we can find its maximum by finding the maximum of $\cos^2(t)\sin(t)$ on $(0,\pi/2)$.
Differentiating and equating the resulting expression to $0$ gives
$$\frac{d}{dt}\left[\cos^2(t)\sin(t)\right]=-2\cos(t)\sin^2(t)+\cos^3(t)=0$$
$$\iff \cos^3(t)=2\cos(t)\sin^2(t)$$
$$\iff \frac{1}{2}=\tan^2(t)$$
$$\iff t=\tan^{-1}\left(\frac{1}{\sqrt{2}}\right)\text{, since tan(t) is bijective on }(0,\pi/2)$$
Once you've proven that the critical point $t=\tan^{-1}\left(\frac{1}{\sqrt{2}}\right)$ does indeed correspond to a maximum for $\cos^2(t)\sin(t)$, it follows that the ratio $\frac{x}{y}$ that maximizes $x^2 y$ is
\begin{align*}
\frac{\cos\left(\tan^{-1}\frac{1}{\sqrt{2}}\right)}{\sin\left(\tan^{-1}\frac{1}{\sqrt{2}}\right)} &= \frac{1}{\tan\left(\tan^{-1}\frac{1}{\sqrt{2}}\right)}\\
&= \frac{1}{1/\sqrt{2}}\\
&= \sqrt{2}
\end{align*}
