Mathematics - Application of Derivative Question with Solution | TestHub

$2\sin 2{\theta }_1=2\sin 2{\theta }_2$$\Rightarrow 2{\theta }_1=\pi -2{\theta }_2$$\Rightarrow {\theta }_2=\frac{\pi }{2}-{\theta }_1\dots .\left(1\right)$
Now perimeter$p\left({\theta }_1, {\theta }_2\right)=2\left\{\left({\theta }_2-{\theta }_1\right)+2\sin 2{\theta }_1\right\}$$\Rightarrow p\left({\theta }_1\right)=2\left[\frac{\pi }{2}-2{\theta }_1+2\sin 2{\theta }_1\right]$$\Rightarrow p^{\text{'}}\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^{\text{'}}\left({\theta }_1\right)=0\& p^{"}\left({\theta }_1\right)<0$$\Rightarrow \cos 2{\theta }_1=\frac{1}{2}\Rightarrow 2{\theta }_1=\frac{\pi }{3}\Rightarrow {\theta }_1=\frac{\pi }{6}$Now area at${\theta }_1=\frac{\pi }{6}$will be
$\left({\theta }_2-{\theta }_1\right)\times 2\sin 2{\theta }_1$$=\left(\frac{\pi }{2}-2{\theta }_1\right)\cdot 2\sin 2{\theta }_1$$=\left(\frac{\pi }{2}-\frac{\pi }{3}\right)\times 2\sin \frac{\pi }{3}$$=\frac{\pi }{6}\cdot \sqrt{3}=\frac{\pi }{2\sqrt{3}}$