Mathematics - Application of Derivative Question with Solution | TestHub

Let $\mathrm{AD}=x$. Let $\mathrm{DM}=y_1$. $\mathrm{BD}=(2a-x)$. $\mathrm{DE}=2x_1$. In $△\mathrm{DBM}$, $\angle \mathrm{B}=\frac{\pi }{3}$.

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

${\mathrm{y}}_1=(2a-x)\times \frac{\sqrt{3}}{2}$. In $△\mathrm{ADP}$, $\angle \mathrm{D}=\frac{\pi }{3}$. $\cos 60^{\circ }=\frac{x_1}{x}$.

$x_1=x\times \frac{1}{2}$

$2x_1=x$

$\Delta (x)=Areaofrectangle=2x_{1}y$

$\Delta (x)=x\times (2a-x)\frac{\sqrt{3}}{2}$

${\Delta }^{\prime }(x)=\frac{\sqrt{3}}{2}(2a-2x)=0\Rightarrow x=a$

${\Delta }^{\prime \prime }(a)=-ve$

$x=a$ is the point of maxima.

The maximum area is $a\times \frac{a\sqrt{3}}{2}=\frac{\sqrt{3}{a}^2}{2}$.