In a recent question here, the area of an inscribed yellow rectangle is shown maximum when OB bisects the central sector angle POQ using differentiation.
Is it possible to arrive at this result without finding derivative by logic of increasing and decreasing functions ( base/height of the rectangle) ?
According to the figure, $BC=\sin\alpha$ and $DC=\cos\alpha-\frac{\sin\alpha}{\tan\theta}$.
Let $\tan\theta=a$. Now the area is $$A=\sin\alpha\cos\alpha-\frac{\sin^2 \alpha}{a}$$ $$=\frac{a\sin2\alpha +\cos 2\alpha -1}{2a}$$ To maximize area, it is sufficient to maximize $a\sin2\alpha +\cos 2\alpha$. This can be done by writing it as $$\sqrt{a^2+1}\left(\dfrac{a}{\sqrt{a^2+1}}\sin 2\alpha+\dfrac{1}{\sqrt{a^2+1}}\cos 2\alpha\right)$$
Now substitute $a=\tan\theta$ and you obtain $$A=\frac{sec\theta\cos(\theta-2\alpha)-1}{2\tan\theta}$$
For maximum, $\theta=2\alpha$ and we are done!
