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

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

Hint: use concept of maxima and minima.

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}


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}


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