The sum of surface areas of a rectangular parallelopiped with sides $x$, $2x$ and $\frac{x}{3}$ and a sphere is given to be constant. Prove that the sum of their volumes is minimum, if $x$ is equal to three times the radius of the sphere. Also, find the minimum value of the sum of their volumes.

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

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

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

$k = 2\left[ {2{x^2} + \frac{{2{x^2}}}{3} + \frac{{{x^2}}}{3}} \right] + 4\pi {r^2}$

$= 2\left[ {3{x^2}} \right] + 4\pi {r^2} = 6{x^2} + 4\pi {r^2}$
$\Rightarrow$ $4\pi {r^2} = k - 6{x^2}$

$\Rightarrow$ ${r^2} = \frac{{k - 6{x^2}}}{{4\pi }}$

$\Rightarrow$ $r = \sqrt {\frac{{k - 6{x^2}}}{{4\pi }}}$ ……..(i)

Let $V$ denotes the volume of both the parallelopiped and the sphere.
Then $V = 2x \cdot x \cdot \frac{x}{3} + \frac{4}{3}\pi {r^3} = \frac{2}{3}{x^3} + \frac{4}{3}\pi {r^3}$

$= \frac{2}{3}{x^3} + \frac{4}{3}\pi {\left( {\frac{{k - 6{x^2}}}{{4\pi }}} \right)^{3/2}}$

$= \frac{2}{3}{x^3} + \frac{4}{3}\pi \cdot \frac{1}{{8{\pi ^{3/2}}}}{\left( {k - 6{x^2}} \right)^{3/2}}$

$= \frac{2}{3}{x^3} + \frac{1}{{6\sqrt \pi }}{\left( {k - 6{x^2}} \right)^{3/2}}$ ………(ii)

On differentiating both sides w.r.t. $x$, we get
$\frac{{dV}}{{dx}} = \frac{2}{3} \cdot 3{x^2} + \frac{1}{{6\sqrt \pi }} \cdot \frac{3}{2}{\left( {k - 6{x^2}} \right)^{1/2}} \cdot ( - 12x)$
$= 2{x^2} - \frac{{12x}}{{4\sqrt \pi }}\sqrt {k - 6{x^2}}$

$= 2{x^2} - \frac{{3x}}{{\sqrt \pi }}{\left( {k - 6{x^2}} \right)^{1/2}}$ ……(iii)

$\Rightarrow$ $2{x^2} = \frac{{3x}}{{\sqrt \pi }}{\left( {k - 6{x^2}} \right)^{1/2}}$

$\Rightarrow$ $4{x^4} = \frac{{9{x^2}}}{\pi }\left( {k - 6{x^2}} \right)$

$\Rightarrow$ $4\pi {x^4} + 54{x^4} = 9k{x^2}$

$\Rightarrow$ ${x^4}[4\pi + 54] = 9 \cdot k \cdot {x^2}$

$\Rightarrow$ ${x^2} = \frac{{9k}}{{4\pi + 54}}$

$\Rightarrow$ $\quad x = 3 \cdot \sqrt {\frac{k}{{4\pi + 54}}}$ ……..(iv)

Again, differentiating Eq. (iii) w.r.t. $x$, we get

$\frac{{{d^2}V}}{{d{x^2}}} = 4x - \frac{3}{{\sqrt \pi }}\left[ {x \cdot \frac{1}{2}{{\left( {k - 6{x^2}} \right)}^{ - 1/2}} \cdot ( - 12x) + {{\left( {k - 6{x^2}} \right)}^{1/2}} \cdot 1} \right]$

$= 4x - \frac{3}{{\sqrt \pi }}\left[ { - 6{x^2} \cdot {{\left( {k - 6{x^2}} \right)}^{ - 1/2}} + {{\left( {k - 6{x^2}} \right)}^{1/2}}} \right]$

$= 4x - \frac{3}{{\sqrt \pi }}\left[ {\frac{{ - 6{x^2} + k - 6{x^2}}}{{\sqrt {k - 6{x^2}} }}} \right]$

$= 4x - \frac{3}{{\sqrt \pi }}\left[ {\frac{{k - 12{x^2}}}{{\sqrt {k - 6{x^2}} }}} \right]$

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

$= 12\sqrt {\frac{k}{{4\pi + 54}}} - \frac{3}{{\sqrt \pi }}\left[ {\frac{{k - \frac{{108k}}{{4\pi + 54}}}}{{\sqrt {k - \frac{{54k}}{{4\pi + 54}}} }}} \right]$

$= 12\sqrt {\frac{k}{{4\pi + 54}}} - \frac{3}{{\sqrt \pi }}\left[ {\frac{{4k\pi + 54k - 108k/4\pi + 54}}{{\sqrt {4k\pi + 54k - 54k/4\pi + 54} }}} \right]$

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

$\left[ {{\rm{since,}}\,(2\pi - 27) < 0 \Rightarrow \frac{{{d^2}V}}{{d{x^2}}} > 0;k > 0} \right]$

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 $r = \sqrt {\frac{{k - 6{x^2}}}{{4\pi }}}$ [using Eq.(i)]

$= \frac{1}{{2\sqrt \pi }}\sqrt {k - 6 \cdot \frac{{9k}}{{4\pi + 54}}}$

$= \frac{1}{{2\sqrt \pi }} \cdot \sqrt {\frac{{4k\pi + 54k - 54k}}{{4\pi + 54}}}$

$= \frac{1}{{2\sqrt \pi }}\sqrt {\frac{{4k\pi }}{{4\pi + 54}}} = \frac{{\sqrt k }}{{\sqrt {4\pi + 54} }} = \frac{1}{3}x$
$\Rightarrow$ $x = 3r$ Hence proved.

Therefore the minimum sum of volume,

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

$= \frac{2}{3}{x^3} + \frac{4}{3}\pi \cdot \frac{{{x^3}}}{{27}} = \frac{2}{3}{x^3}\left( {1 + \frac{{2\pi }}{{27}}} \right)$