#### Provide solution for RD Sharma maths class12 Chapter Maxima and Minima exercise 17.5 question 41.

$\frac{l}{D}=\frac{\pi}{\pi +2}$

Hint: For maximum or minimum value of S, we must have $\frac{dS}{dD}=0$

Given: Volume, $V =\frac{1}{2}\pi l \left ( \frac{D}{2} \right )^2$

$V =\frac{\pi lD^2}{8}$

$I=\frac{8V}{\pi D^2}$                        .........(1)

Solution:

Total surface area = $\frac{\pi D^2}{4}+lD+\frac{\pi Dl}{2}$

$S=\frac{\pi D^2}{4}+\frac{8V}{\pi D}+\frac{8V}{2D}$                    .....from (1)

\begin{aligned} &\frac{d S}{d D}=\frac{\pi D}{2}-\frac{8 V}{\pi D^{2}}-\frac{8 V}{2 D^{2}} \\ &\frac{d S}{d D}=0 \\ &\frac{\pi D}{2}-\frac{8 V}{\pi D^{2}}-\frac{8 V}{2 D^{2}}=0 \end{aligned}

\begin{aligned} &\frac{\pi D}{2}=\frac{8 V}{D^{2}}\left(\frac{1}{\pi}+\frac{1}{2}\right) \\ &D^{3}=\frac{16 V}{\pi}\left(\frac{1}{\pi}+\frac{1}{2}\right) \\ &\frac{d^{2} S}{d D^{2}}=\frac{\pi}{2}+\frac{16 V}{D^{3}}\left(\frac{1}{\pi}+\frac{1}{2}\right) \\ &=\frac{\pi}{2}+\pi>0 \end{aligned}

\begin{aligned} &l=\frac{8 V}{\pi D^{2}} \\ &l=\frac{8}{\pi D^{2}}\left[\frac{\pi D^{3}}{16}\left[\frac{2 \pi}{\pi+2}\right]\right] \\ &l=D\left(\frac{\pi}{\pi+2}\right), \frac{l}{D}=\frac{\pi}{\pi+2} \end{aligned}

Hence proved