The sum of the surface areas of a sphere and a cube is given.

Solution:

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

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

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

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

$\Rightarrow V=\frac{4 \pi r^{3}}{3}+\left[\frac{\left(S-4 \pi r^{2}\right)}{6}\right]^{\frac{3}{2}}$        [From eq. (1)]

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

For the minimum or maximum values of $V$, we must have

$\frac{d V}{d r}=0$        ....(2)

$\Rightarrow 4 \pi r^{2}-2 \pi r\left[\frac{\left(S-4 \pi r^{2}\right)}{6}\right]^{\frac{1}{2}}=0$          [From eq. (2)]

 

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

$\Rightarrow 4 \pi r^{2}=2 \pi r x$           [From eq. (1)]

 

$\Rightarrow x=2 r$

Now,

$\frac{d^{2} V}{d r^{2}}=8 \pi r-2 \pi\left[\frac{\left(S-4 \pi r^{2}\right)}{6}\right]^{\frac{1}{2}}-\frac{2 \pi r}{2}\left[\frac{\left(S-4 \pi r^{2}\right)}{6}\right]^{-\frac{1}{2}} \frac{(-8 \pi r)}{6}$

$\Rightarrow \frac{d^{2} V}{d r^{2}}=8 \pi r-2 \pi\left[\frac{\left(S-4 \pi r^{2}\right)}{6}\right]^{\frac{1}{2}}+\frac{4}{3} \pi^{2} r^{2}\left[\frac{6}{\left(S-4 \pi r^{2}\right)}\right]^{\frac{1}{2}}$

$\Rightarrow \frac{d^{2} V}{d r^{2}}=8 \pi r-2 \pi x+\frac{4}{3} \pi^{2} r^{2} \frac{1}{x}=8 \pi r-4 \pi r+\frac{2}{3} \pi^{2} r$

 

$\Rightarrow \frac{d^{2} V}{d r^{2}}=4 \pi r+\frac{2}{3} \pi^{2} r>0$

So, volume is minimum when x = 2r.