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)
Best 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 $\sqrt2$.