#### Provide solution for RD Sharma maths class12 Chapter Maxima and Minima exercise 17.5 question 40.

x = 2r

Hint: For maximum or minimum value of V, we must have $\frac{dV}{dr}=0$

Given:

Let r be the radius of the sphere, x is side of the cube and S be the sum of the surface area of both of them

$S= 4 \pi r^2 + 6x^2$

$x= \left ( \frac{S-4\pi r^2}{6} \right )$                    ......(1)

Solution:

Sum of Volumes, $V=\frac{4}{3}\pi r^3+x^3$

$V=\frac{4}{3} \pi r^{3}+\left[\left(\frac{S-4 \pi r^{2}}{6}\right)\right]^{\frac{3}{2}}$

From eqn (1)

$\frac{dV}{dr}=4 \pi r^{2}-2 \pi r\left[\left(\frac{S-4 \pi r^{2}}{6}\right)\right]^{\frac{1}{2}}$

$\frac{dV}{dr}=0$                    .........(2)

\begin{aligned} &4 \pi r^{2}-2 \pi r\left[\frac{\left(S-4 \pi r^{2}\right)}{6}\right]^{\frac{1}{2}}=0 \\ &4 \pi r^{2}=2 \pi r\left[\frac{\left(S-4 \pi r^{2}\right)}{6}\right]^{\frac{1}{2}} \end{aligned}

$4 \pi r^2 -2\pi rx$ from eqn (1),

\begin{aligned} &\frac{d^{2} V}{d r^{2}}=8 \pi r-2 \pi\left[\frac{\left(S-4 \pi r^{2}\right)}{6}\right]^{\frac{1}{2}}-\frac{2 \pi r}{2}\left[\frac{\left(S-4 \pi r^{2}\right)}{6}\right]^{-\frac{1}{2}}\left(\frac{-8 \pi r}{b}\right) \\ &=8 \pi r-2 \pi\left[\frac{\left(S-4 \pi r^{2}\right)}{6}\right]^{\frac{1}{2}}+\frac{4 \pi^{2} r^{2}}{3}\left[\frac{6}{\left(S-4 \pi r^{2}\right)}\right]^{\frac{1}{2}} \\ &=8 \pi r-2 \pi x+\frac{4 \pi^{2} r^{2}}{3} \frac{1}{x}=8 \pi r-4 \pi r+\frac{2}{3} \pi^{2} r \\ &\frac{d^{2} V}{d r^{2}}=4 \pi r+\frac{2}{3} \pi^{2} r>0 \end{aligned}

Volume is minimum when x = 2r