Show that the maximum area of a rectangle which can be inscribed in a triangle of area $A$ is $\dfrac{A}{2}$.
I was trying to solve this as an application of maxima/minima, but it becomes a little clumsy.
From the figure, $AH=\frac{x}{\tan A}$ and $BK=\frac{x}{\tan B}$ , where $FH=GK=x$ (say)
$\therefore$ Area $(\Delta)$ of rectangle $FGKH=x\times HK=x(AB-(AH+BK))$
$=x\left(c-\left(\frac{x}{\tan A}+\frac{x}{\tan B}\right)\right)$ $\qquad$$(AB=c)$
Then simplifying $\dfrac{d(\Delta)}{dx}=0$ to get the answer becomes somewhat tiresome.
This answer gives a nice approach for the solution. Is there any simple yet rigorous alternative proof of this proposition?