Mathematics - Application of Derivative Question with Solution | TestHub

Let $\mathrm{AD}=\mathrm{x}$. Let $\mathrm{DM}={\mathrm{y}}_1$. $\mathrm{BD}=(2a-x)$. $\mathrm{DE}=2{\mathrm{x}}_1$. In $△\mathrm{DBM}$, $\angle \mathrm{B}=\frac{\pi }{3}$.

$\sin 60^{\circ }=\frac{y_1}{2a-x}$

$y_1=(2a-x)\times \frac{\sqrt{3}}{2}$. In $△\mathrm{ADP}$, $\angle \mathrm{D}=\frac{\pi }{3}$.

$$\begin{array}{rl}\cos 60^{\circ }& =\frac{x_1}{x}\\ x_1& =x\times \frac{1}{2}\\ 2x_1& =x\end{array}$$

$$\begin{array}{rl}\Delta (\mathrm{x})& =Areaofrectangle=2{\mathrm{x}}_1\mathrm{y}\\ \Delta (\mathrm{x})& =\mathrm{x}\times (2\mathrm{a}-\mathrm{x})\frac{\sqrt{3}}{2}\\ {\Delta }^{\prime }(\mathrm{x})& =\frac{\sqrt{3}}{2}(2\mathrm{a}-2\mathrm{x})=0\Rightarrow \mathrm{x}=\mathrm{a}\\ {\Delta }^{\prime \prime }(\mathrm{a})& =-\mathrm{ve}\\ \mathrm{x}& =\mathrm{a}pointofmaxima\end{array}$$

Maximum area $=\mathrm{a}\times \frac{\mathrm{a}\sqrt{3}}{2}=\frac{\sqrt{3}{\mathrm{a}}^2}{2}$.