Mathematics - Application of Derivative Question with Solution | TestHub

$Let\begin{array}{rlr}& AD=x& BD=(2a-x)\end{array}$

Let $\mathrm{DM}={\mathrm{y}}_1$ DE $=2{\mathrm{x}}_1$ In

$\Delta DBM$

$\angle B=\frac{\pi }{3}$

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

$y_1=(2a-x)\times \frac{\sqrt{3}}{2}$

In $△\mathrm{ADP}$

$\angle D=\frac{\pi }{3}$

$\begin{array}{rlr}& x_1=x\times \frac{1}{2}& 2x_1=x\end{array}$

$\Delta (x)=$ Area of rectangle $=2x_{1}y$

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

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

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

$\mathrm{x}=\mathrm{a}$ point of maxima

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