Mathematics - Application of Derivative Question with Solution | TestHub

Let A be any point on the arc such that $\angle \mathrm{YOA}=\theta$ Where $0\le \theta \le \alpha$

$$\begin{array}{rl}& \mathrm{DA}=\mathrm{CB}=\mathrm{R}\sin \theta ,\mathrm{OD}=\mathrm{R}\cos \theta \\ & \Rightarrow \mathrm{CO}=\mathrm{CB}\cot \alpha =\mathrm{R}\sin \theta \cot \alpha \\ & \text{ Now, }\mathrm{CD}=\mathrm{OD}-\mathrm{OC}=\mathrm{R}\cos \theta -\mathrm{R}\sin \theta \cot \alpha \\ & =\mathrm{R}(\cos \theta -\sin \theta \cot \alpha )\\ & \text{ Area of rectangle }\mathrm{ABCD},\mathrm{S}=2\cdot \mathrm{CD}\cdot \mathrm{CB}\\ & =2\mathrm{R}(\cos \theta -\sin \theta \cot \alpha )\mathrm{R}\sin \theta =2{\mathrm{R}}^2\left(\sin \theta \cos \theta -{\sin }^2\theta \cot \alpha \right)\\ & {\mathrm{R}}^2(\sin 2\theta -(1-\cos 2\theta )\cot \alpha )=\frac{{\mathrm{R}}^2}{\sin \alpha }[\cos (2\theta -\alpha )-\cos \alpha ]\\ & {\mathrm{S}}_{\max }=\frac{{\mathrm{R}}^2}{\sin \alpha }(1-\cos \alpha )(\text{ for }\theta =\alpha /2)\end{array}$$

Hence, greatest area of the rectangle $={\mathrm{R}}^2\tan \frac{\alpha }{2}$