#### Need solution for RD Sharma maths class 12 chapter Maxima and Minima exercise 17.5 question 20.

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

Hint: For maxima or minima f'(x) must be zero f'(x) = 0

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

Solution:  Let h, r and R be in the height and radius and base of the cone

\begin{aligned} &h=R+\sqrt{R^{2}-r^{2}} \\ &(h-R)^{2}=R^{2}-r^{2} \\ &h^{2}+R^{2}-2 h r=R^{2}-r^{2} \end{aligned}

$r^2=2hR-h^2$               .......(1)

Volume of cone $= \frac{1}{3}\pi r^2h$

$V= \frac{1}{3}\pi h \left (2hR -h^2 \right )$                 ..... from (1)

$V= \frac{1}{3}\pi \left (2h^2R -h^3 \right )$

Now,

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

$\frac{d V}{d h}=0$

$\frac{\pi }{3}(4hR-3h^2)=0$

$4hR=3h^2$

$h=\frac{4R}{3}$

Substituting y value in eqn (1)

\begin{aligned} &r^{2}=4\left(r^{2}-\left(\frac{r}{\sqrt{2}}\right)^{2}\right) \\ &=4\left(r^{2}-\frac{r^{2}}{2}\right)=4\left(\frac{r^{2}}{2}\right)=2 r^{2} \end{aligned}

$x=r\sqrt{2}$

\begin{aligned} &\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{4 \pi R}{3}<0 \end{aligned}

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

$h =\frac{2}{3}$  (Diameter of sphere)

Hence proved