Question: > 
> 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)

Answer: 
**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 $\sqrt2$.
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\*}