#### Need solution for RD Sharma maths class 12 chapter Maxima and Minima exercise Multiple Choice question, question 19.

Answer: option(d)  $\frac{2}{3}$

Hint: For local maxima or minima, we must have $f'(x)=0$.

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

Solution:

We have,

Solution:

$AB=2r$

$OC=r$

$CD =R+h$

We have,

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

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

Squaring on both sides,

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

$h^2+R^2-2hR=R^2-r^2$

$r^2=2hR-h^2$                                                                                                 …(i)

Now,

$V=\frac{1}{3}\pi r ^2h$

$V=\frac{\pi}{3}(2hR-h^2)h$                        …using equation(i)

\begin{aligned} &V=\frac{\pi}{3}\left(2 h^{2} R-h^{3}\right) \\ &\frac{d V}{d h}=\frac{\pi}{3}\left(4 h R-3 h^{2}\right) \end{aligned}

For maxima and minima  $\frac{dV}{dh}=0$

\begin{aligned} &\Rightarrow \frac{\pi}{3}\left(4 h R-3 h^{2}\right)=0 \\ &\Rightarrow 4 h R-3 h^{2}=0 \\ &\Rightarrow 3 h^{2}=4 h R \\ &\Rightarrow h=\frac{4 R}{3} \end{aligned}

Now,

$\frac{d^{2} V}{d h^{2}}=\frac{\pi}{3}(4 R-6 h)=\frac{\pi}{3}\left(4 R-6 \times \frac{4 R}{3}\right)=\frac{\pi}{3} 4 R\left(1-\frac{6}{3}\right)=\frac{\pi}{3} 4 R\left(\frac{-3}{3}\right)=-\frac{4 \pi R}{3}<0$

Volume is maximum, when  $h=\frac{4R}{3}$

$h=\frac{2(2R)}{3}$

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