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)
 A: 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*}
A: 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....
A: Without calculus:
Let $r:=\left(\dfrac xy\right)^2$. We maximize $ry^3$ under $(r+1)y^2=100$, i.e. we maximize
$$r\left(\frac{100}{r+1}\right)^{3/2}$$ or simply
$$\frac{r^2}{(r+1)^3}.$$
The options give
$$\frac{16}{125},\frac9{64},\frac{324}{2197},\frac4{27}.$$
The largest value is the fourth, corresponding to $\sqrt2$.
A: 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:
$$r^2 = \dfrac{w^4}{(1 + w^2)^{3}} = \dfrac{1}{w^2 + 3 + 3w^{-2} + w^{-4}} > 0$$
$$d(r^2) = -r^{2}(2w - 6w^{-3} - 4w^{-5}) = 0$$
$$w^6 - 3w^{2} - 2 = 0$$
$$(w^2 - 2)(w^2+1)^2 = 0$$
$$w = \sqrt 2$$
