Mathematics - Application of Derivative Question with Solution | TestHub

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

$$\begin{array}{rl}& \mathrm{DA}=\mathrm{CB}=\mathrm{R}\sin \theta ,\mathrm{OD}=\mathrm{R}\cos \theta \\ & \Rightarrow \mathrm{CO}=\mathrm{CB}\cot \alpha =\mathrm{R}\sin \theta \cot \alpha \end{array}$$

Now, $\mathrm{CD}=\mathrm{OD}-\mathrm{OC}=\mathrm{R}\cos \theta -\mathrm{R}\sin \theta \cot \alpha =\mathrm{R}(\cos \theta -\sin \theta \cot \alpha )$.

Area of rectangle ABCD, $\mathrm{S}=\mathrm{CD}\cdot \mathrm{CB}$

$$\begin{array}{rl}& \mathrm{S}=\mathrm{R}(\cos \theta -\sin \theta \cot \alpha )\mathrm{R}\sin \theta ={\mathrm{R}}^2\left(\sin \theta \cos \theta -{\sin }^2\theta \cot \alpha \right)\\ & =\frac{{\mathrm{R}}^2}{2}(\sin 2\theta -(1-\cos 2\theta )\cot \alpha )\\ & =\frac{{\mathrm{R}}^2}{2\sin \alpha }[\cos (2\theta -\alpha )]\\ & {\mathrm{S}}_{\max }=\frac{{\mathrm{R}}^2}{\sin \alpha }(1-\cos \alpha )\left(\text{ for }\theta =\alpha /2\right)\end{array}$$

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