#### Need solution for RD Sharma Maths Class 12 Chapter 17 Maxima and Minima Excercise Case Study Based Questions question 6 subquestion (ii)

Answer:  $\frac{1}{\pi}\left[(\pi+4) x^{2}-56 x+196\right]$

Hint: use formulas.

Solution:

\begin{aligned} &\text { Area }=\pi(\text { raduis })^{2}+(\text { side })^{2} \\ & \end{aligned}

$A=\pi y^{2}+x^{2} \rightarrow(1)$

We know that

\begin{aligned} 2 x+\pi y &=14 \\ \end{aligned}

$\pi y =14-2 x \\$

$y =\frac{14-2 x}{\pi}$

Substituting  value of   $\mathrm{y} \text { in (1) } \pi\left(\frac{14-2 \mathrm{x}}{\pi}\right)^{2}+\frac{\mathrm{x}^{2}}{4^{2}}$

\begin{aligned} A &=\pi\left(\frac{14-2 x}{\pi}\right)^{2}+\frac{x^{2}}{16} \\ \end{aligned}

$=\frac{1}{\pi}(14-2 x)^{2}+\frac{x^{2}}{16} \\$

$=\frac{1}{\pi}\left(14^{2}-2(14)(2 x)+(2 x)^{2}\right)+\frac{x^{2}}{16} .$

\begin{aligned} &=\frac{1}{\pi}\left[196-56 \mathrm{x}+4 \mathrm{x}^{2}\right]+\frac{\mathrm{x}^{2}}{16} \\ \end{aligned}

$=\frac{1}{\pi}\left[(\pi+4) \mathrm{x}^{2}-56 \mathrm{x}+196\right]$