#### Need solution for RD Sharma maths class 12 chapter Maxima and Minima exercise 17.5 question 17.

$h=\frac{2R}{\sqrt{3}}$

Hint: For maxima and minima value of A, we must have $\frac{dA}{dx}=0$

Given: $h = 2\sqrt{R^2-r^2}$

Solution:  Let the height and radius of the base of the cylinder be h and r

$\frac{h^2}{4}+r^2=R^2$

$h = 2\sqrt{R^2-r^2}$               .....(1)

Volume of the cylinder V $=\pi r^2 h$

Square on both the sides

$V^2=\pi ^2 r^4h^2$

$V^2=4\pi ^2 r^4 (R^2-r^2)$                       ..........from (1)

Now,

\begin{aligned} &Z=4 \pi^{2}\left(r^{4} R^{2}-r^{6}\right) \\ &\frac{d Z}{d x}=4 \pi^{2}\left(4 r^{3} R^{2}-6 r^{5}\right) \\ &\frac{d Z}{d x}=0 \\ &4 \pi^{2}\left(4 r^{3} R^{2}-6 r^{5}\right)=0 \end{aligned}

\begin{aligned} &4 r^{3} R^{2}=6 r^{5} \\ &6 r^{2}=4 R^{2} \\ &r^{2}=\frac{4 R^{2}}{6} \\ &r=\frac{2 R}{\sqrt{6}} \end{aligned}

Subsitute r value in eqn (1)

\begin{aligned} &h=2 \sqrt{R^{2}-\left(\frac{2 R}{\sqrt{6}}\right)^{2}} \\ &=2 \sqrt{\frac{6 R^{2}-4 R^{2}}{6}}=2 \sqrt{\frac{R^{2}}{3}}=\frac{2 R}{\sqrt{3}} \\ &\frac{d^{2} Z}{d x^{2}}=4 \pi^{2}\left(12 r^{2} R^{2}-30 r^{4}\right) \end{aligned}

\begin{aligned} &=4 \pi^{2}\left(12\left(\frac{2 R}{\sqrt{6}}\right)^{2} R^{2}-30\left(\frac{2 R}{\sqrt{6}}\right)^{4}\right) \\ &=4 \pi^{2}\left(8 R^{4}-\frac{80 R^{4}}{6}\right) \end{aligned}

\begin{aligned} &=4 \pi^{2}\left(\frac{48 R^{4}-80 R^{4}}{6}\right) \\ &=4 \pi^{2}\left(\frac{-16 R^{4}}{3}\right)<0 \end{aligned}

Volume of the cylinder is max when, $h=\frac{2R}{\sqrt{3}}$