We have given that, the sum of the surface areas of a rectangular parallelopiped with sides $x, 2 x$ and $\frac{x}{3}$ and a sphere is constant.

Let $S$ be the sum of both the surface area.

$\therefore \quad S=2\left(x \cdot 2 x+2 x \cdot \frac{x}{3}+\frac{x}{3} \cdot x\right)+4 \pi r^2=k$

$$\begin{aligned} k & =2\left[2 x^2+\frac{2 x^2}{3}+\frac{x^2}{3}\right]+4 \pi r^2 \\ & =2\left[3 x^2\right]+4 \pi r^2=6 x^2+4 \pi r^2 \end{aligned}$$

$$\begin{array}{lr} \Rightarrow & 4 \pi r^2=k-6 x^2 \\ \Rightarrow & r^2=\frac{k-6 x^2}{4 \pi} \\ \Rightarrow & r=\sqrt{\frac{k-6 x^2}{4 \pi}}\quad\text{.... (i)} \end{array}$$

Let V denotes the volume of both the parallelopiped and the sphere.

Then,

$$\begin{aligned} V & =2 x \cdot x \cdot \frac{x}{3}+\frac{4}{3} \pi r^3=\frac{2}{3} x^3+\frac{4}{3} \pi r^3 \\ & =\frac{2}{3} x^3+\frac{4}{3} \pi\left(\frac{k-6 x^2}{4 \pi}\right)^{3 / 2} \\ & =\frac{2}{3} x^3+\frac{4}{3} \pi \cdot \frac{1}{8 \pi^{3 / 2}}\left(k-6 x^2\right)^{3 / 2} \\ & =\frac{2}{3} x^3+\frac{1}{6 \sqrt{\pi}}\left(k-6 x^2\right)^{3 / 2}\quad\text{.... (ii)} \end{aligned}$$

On differentiating both sides w.r.t. $x$, we get

$$\begin{aligned} \frac{d V}{d x} & =\frac{2}{3} \cdot 3 x^2+\frac{1}{6 \sqrt{\pi}} \cdot \frac{3}{2}\left(k-6 x^2\right)^{1 / 2} \cdot(-12 x) \\ & =2 x^2-\frac{12 x}{4 \sqrt{\pi}} \sqrt{k-6 x^2} \\ & =2 x^2-\frac{3 x}{\sqrt{\pi}}\left(k-6 x^2\right)^{1 / 2}\quad\text{.... (iii)} \end{aligned}$$

$$\begin{array}{ll} \because & \frac{d V}{d x}=0 \\ \Rightarrow & 2 x^2=\frac{3 x}{\sqrt{\pi}}\left(k-6 x^2\right)^{1 / 2} \end{array}$$

$$\begin{aligned} \Rightarrow \quad & 4 x^4 =\frac{9 x^2}{\pi}\left(k-6 x^2\right) \\ \Rightarrow \quad & 4 \pi x^4 =9 k x^2-54 x^4 \\ \Rightarrow \quad & 4 \pi x^4+54 x^4 =9 k x^2 \\ \Rightarrow \quad & x^4[4 \pi+54] =9 \cdot k \cdot x^2 \\ \Rightarrow \quad& x^2 =\frac{9 k}{4 \pi+54} \\ \Rightarrow \quad & x =3 \cdot \sqrt{\frac{k}{4 \pi+54}}\quad\text{.... (iv)} \end{aligned}$$

$$\begin{aligned} &\text { Again, differentiating Eq. (iii) w.r.t. } x \text {, we get }\\ &\begin{aligned} \frac{d^2 V}{d x^2} & =4 x-\frac{3}{\sqrt{\pi}}\left[x \cdot \frac{1}{2}\left(k-6 x^2\right)^{-1 / 2} \cdot(-12 x)+\left(k-6 x^2\right)^{1 / 2} \cdot 1\right] \\ & =4 x-\frac{3}{\sqrt{\pi}}\left[-6 x^2 \cdot\left(k-6 x^2\right)^{-1 / 2}+\left(k-6 x^2\right)^{1 / 2}\right] \\ & =4 x-\frac{3}{\sqrt{\pi}}\left[\frac{-6 x^2+k-6 x^2}{\sqrt{k-6 x^2}}\right] \\ & =4 x-\frac{3}{\sqrt{\pi}}\left[\frac{k-12 x^2}{\sqrt{k-6 x^2}}\right] \end{aligned} \end{aligned}$$

Now, $\left(\frac{d^2 V}{d x^2}\right)_{x=3 \cdot \sqrt{\frac{k}{4 \pi+54}}}=4 \cdot 3 \sqrt{\frac{k}{4 \pi+54}}-\frac{3}{\sqrt{\pi}}\left[\frac{k-12 \cdot 9 \cdot \frac{k}{4 \pi+54}}{\sqrt{k-\frac{6 \cdot 9 \cdot k}{4 \pi+54}}}\right]$

$$\begin{aligned} & =12 \sqrt{\frac{k}{4 \pi+54}}-\frac{3}{\sqrt{\pi}}\left[\frac{k-\frac{108 k}{4 \pi+54}}{\sqrt{k-\frac{54 k}{4 \pi+54}}}\right] \\ & =12 \sqrt{\frac{k}{4 \pi+54}}-\frac{3}{\sqrt{\pi}}\left[\frac{4 k \pi+54 k-108 k / 4 \pi+54}{\sqrt{4 k \pi+54 k-54 k / 4 \pi+54}}\right] \\ & =12 \sqrt{\frac{k}{4 \pi+54}}-\frac{3}{\sqrt{\pi}}\left[\frac{4 k \pi-54 k}{\sqrt{4 k \pi} \sqrt{4 \pi+54}}\right] \\ & =12 \sqrt{\frac{k}{4 \pi+54}}-\frac{6}{\sqrt{\pi}}\left[\frac{k(2 \pi-27)}{\sqrt{k} \sqrt{16 \pi^2+216 \pi}}\right] \end{aligned}$$

For $x=3 \sqrt{\frac{k}{4 \pi+54}}$, the sum of volumes is minimum.

For $x=3 \sqrt{\frac{k}{4 \pi+54}}$, then $\quad r=\sqrt{\frac{k-6 x^2}{4 \pi}}\quad\text{[using Eq. (i)]}$

$$\begin{aligned} & =\frac{1}{2 \sqrt{\pi}} \sqrt{k-6 \cdot \frac{9 k}{4 \pi+54}} \\ & =\frac{1}{2 \sqrt{\pi}} \cdot \sqrt{\frac{4 k \pi+54 k-54 k}{4 \pi+54}} \\ & =\frac{1}{2 \sqrt{\pi}} \cdot \sqrt{\frac{4 k \pi}{4 \pi+54}}=\frac{\sqrt{k}}{\sqrt{4 \pi+54}}=\frac{1}{3} x \end{aligned}$$

$$ \begin{aligned} & \Rightarrow \quad x=3 r \quad\text{Hence proved.}\\ & \therefore \text { Minimum sum of volume, } \end{aligned}$$

$\therefore$ Minimum sum of volume,

$$\begin{aligned} \left.V_{\left(x=3 \cdot \sqrt{\frac{k}{4 \pi+54}}\right.}\right) & =\frac{2}{3} x^3+\frac{4}{3} \pi r^3=\frac{2}{3} x^3+\frac{4}{3} \pi \cdot\left(\frac{1}{3} x\right)^3 \\ & =\frac{2}{3} x^3+\frac{4}{3} \pi \cdot \frac{x^3}{27}=\frac{2}{3} x^3\left(1+\frac{2 \pi}{27}\right) \end{aligned}$$