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

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

Answers (1)

\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

 

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