Explain solution RD Sharma class 12 Chapter 17 Maxmima and Minima Exercise Case Study Questions question 4 subquestion (iv)

Answer:  $\frac{20}{\pi+4}, \frac{10}{\pi+4}$

Hint: use concept of maxima and minima.

Solution:
Let x, y be the length and breadth of the rectangle ABCD, Let  $\frac{x}{2}$  be the radius of semicircle with center at O

Perimeter = 10 m

$x+2 y+\frac{\pi x}{2}=10 \\$

$2 x+4 y+\pi x=20 \\$

$4 y=20-(\pi+2) x \$

$\ y=\frac{20-(\pi+2) x}{4} \\$

\begin{aligned} & &A=x y+\frac{1}{2} \pi\left(\frac{x}{2}\right)^{2} \end{aligned}

$=x\left[\frac{20-(\pi+2) x}{4}\right]+\left[\frac{x^{2} \pi}{8}\right] \\$

$A=1 / 4\left[20 x-(\pi+2) x^{2}\right]+\left[\frac{\pi x^{2}}{8}\right] \\$

\begin{aligned} & &\frac{d A}{d x}=\frac{1}{4}[20-2(\pi+2) x]+\frac{\pi x}{4} \end{aligned}

\begin{aligned} \frac{\mathrm{d} \mathrm{A}}{\mathrm{dx}} &=0 \end{aligned}

$\frac{1}{4}[20-2(\pi+2) x]+\frac{\pi x}{4}=0 . \\$

$20-2 \pi x-4 x+\pi x=0 \\$

$20-(\pi+4) x=0 \\$

\begin{aligned} & &x=\frac{20}{\pi+4} \end{aligned}

And,

$y=\frac{20-(\pi+2)\left(\frac{20}{\pi+4}\right)}{4} \$

\begin{aligned} &\ &=\frac{-20 \pi+80-20 \pi-40}{4(\pi+4)} \end{aligned}

$\mathrm{y}=\frac{10}{\pi+4}$