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

Answer:  $10 \mathrm{x}-(\pi+2) \mathrm{x}^{2}$

Hint: use the formulas of area and perimeter.

Solution:

Perimeter of the window is given, by

$\mathrm{P}=2 \mathrm{x}+2 \mathrm{y}+\pi \mathrm{x} \\$

\begin{aligned} & &\mathrm{p}=2 \mathrm{x}+4 \mathrm{y}+\pi \mathrm{x} \end{aligned}                                                                              $(\therefore \mathrm{p}=\mathrm{1 0} \mathrm{~ m})$

\begin{aligned} 2 x+4 y+\pi x &=10 \\ \end{aligned}

$y =\frac{10-\pi x-2 x}{4}-(1) \\$

$\text { Area }=(2 x)(2 y) +\frac{\pi x^{2}}{2}$

[∴Area of the window = Area of the Rectangle + Area of the Semicircle]

\begin{aligned} &A=4 x y+\frac{\pi x^2}{2} \ \ \ ---(2)\\ & \end{aligned}

$\mathrm{y} \text { value in (2) }$

\begin{aligned} &A=4 x\left(\frac{10-\pi x-2 x}{4}\right)+\frac{\pi x^{2}}{2} \\ & \end{aligned}

$=10 x-\pi x^{2}-2 x^{2}+\frac{\pi x^{2}}{2}$

\begin{aligned} &=10 \mathrm{x}-2 \mathrm{x}^{2}-\frac{\pi \mathrm{x}^{2}}{2} \\ & \end{aligned}

$=10 \mathrm{x}-\frac{4 \mathrm{x}^{2}+\pi \mathrm{x}^{2}}{2} \\$

$\quad=10 \mathrm{x}-(2 \mathrm{x}^{2}+\pi \mathrm{x}^{2}) \\$

$\mathrm{~A}=10 \mathrm{x}-\mathrm{x}^{2}(\pi+2)$