Find Maximal Edge of rectangle trapped in a triangle with a given surface using derivative

Question

Inside $\triangle EFG$ with surface of $9 \ cm^2$ there is a rectangle $ABCD$ shuch as $AB=2CD$.
I need to find the maximal length of BC.
I've tried putting $BC=y$ and $x$ to several other stuff in order to find $y(x)$ and then solving $y'(x)=0$ in order to maximize $x$, but I'm getting equations that makes is really hard to solve for $y$.
Any ideas maybe? Thanks

Answer 1

First we have $\Delta EAB$ and $\Delta EFG$ are similars. Let's call $GF=a$ and the height of $\Delta EFG$ equal to $h$. Then:

$$\frac{h-x}{h}=\frac{2x}{a}$$

Once the area of $\Delta EFG$ is $9$ then $a\cdot h=18 \to h=18/a$ we get

$$x=\frac{18a}{36+a^2}=f(a)$$

For the maximum $x$ we have $f'(a)=0 \to a=6$ and then

$$x=\frac{3}{2}$$

Answer 2

The dual problem is to find the smallest triangle circumscribed to a $2\times 1$ rectangle, but if triangle is circumscribed to a $2\times 1$ rectangle, by "paper folding" we get that the area of the triangle is at least twice the area of the rectangle. It follows that the largest inscribed rectangle with such proportions inscribed in a triangle with area $9$ has area $\frac{9}{2}$, hence shortest side with length $\color{red}{\frac{3}{2}}$.