Consider all rectangles lying in the region \ (x, y) ∈ R × R: 0 ≤ x...

$2\sin 2{\theta}_{1} = 2\sin 2{\theta}_{2}$ $⇒ 2{\theta}_{1} = \pi - 2{\theta}_{2}$ $⇒ {\theta}_{2} = \dfrac{\pi}{2} - {\theta}_{1} \ldots . ( 1 )$
Now perimeter $p\left({\theta}_{1}, {\theta}_{2}\right) = 2\left\{\left({\theta}_{2} - {\theta}_{1}\right) + 2\sin 2{\theta}_{1}\right\}$ $⇒ p\left({\theta}_{1}\right) = 2\left[\dfrac{\pi}{2} - 2{\theta}_{1} + 2\sin 2{\theta}_{1}\right]$ $⇒ {p}^{'}\left({\theta}_{1}\right) = 2\left(- 2 + 4\cos 2{\theta}_{1}\right)$ Also, ${p}^{"}\left({\theta}_{1}\right) = 2\left(- 8\sin 2{\theta}_{1}\right) < 0$ For maximum perimeter, {p}^{'}\left({\theta}_{1}\right) = 0 & {p}^{"}\left({\theta}_{1}\right) < 0 $⇒ \cos 2{\theta}_{1} = \dfrac{1}{2} ⇒ 2{\theta}_{1} = \dfrac{\pi}{3} ⇒ {\theta}_{1} = \dfrac{\pi}{6}$ Now area at ${\theta}_{1} = \dfrac{\pi}{6}$ will be 
\left({\theta}_{2} - {\theta}_{1}\right) \times 2\sin 2{\theta}_{1} = \left(\dfrac{\pi}{2} - 2{\theta}_{1}\right) ⋅ 2\sin 2{\theta}_{1} = \left(\dfrac{\pi}{2} - \dfrac{\pi}{3}\right) \times 2sin\dfrac{\pi}{3} = \dfrac{\pi}{6} ⋅ \sqrt{3} = \dfrac{\pi}{2\sqrt{3}}