#### If the sum of the surface areas of cube and a sphere is constant, what is the ratio of an edge of the cube to the diameter of the sphere, when the sums of their volumes are minimum?

Given: The combined surface area of a cube and sphere are constant

To find:  the ratio of an edge of the cube to the diameter of the sphere, when the sums of their volumes are minimum

Explanation: Let ‘a’ be the side of the cube

Then surface area of the cube = $6a^2$….(i)

Take ‘r’ as  the radius of the sphere

Then the surface area of the sphere = $4\pi r^2$…(ii)

According to the question, the surface area of both the figures is added, thus adding the equation (i) and (ii), we get

$6a^{2} + 4\pi r^{2} = k \quad (\text{where } k \text{ is the constant}) \\ \Rightarrow 6a^{2} = k - 4\pi r^{2}$

$\\\Rightarrow \mathrm{a}^{2}=\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\\\Rightarrow \mathrm{a}=\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{1}{2}} \ldots (iii)$
As the formula of volume of cube is
$\mathrm{v}_{c}=\mathrm{a}^{3}$
Plus the volume of a sphere is
$\mathrm{V}_{\mathrm{s}}=\frac{4}{3} \pi \mathrm{r}^{3}$
Hence adding both the volumes will result into,
$\mathrm{V}=\mathrm{a}^{3}+\frac{4}{3} \pi \mathrm{r}^{3}$

Then putting the values from equation (iii) in above equation,
\begin{aligned} \mathrm{V} &=\left(\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{1}{2}}\right)^{3}+\frac{4}{3} \pi \mathrm{r}^{3} \\ \mathrm{~V} &=\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{3}{2}}+\frac{4}{3} \pi \mathrm{r}^{3} \end{aligned}
After finding the first derivative of the volume, we get
$\mathrm{V}^{\prime}=\frac{\mathrm{d}\left(\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{3}{2}}+\frac{4}{3} \pi \mathrm{r}^{3}\right)}{\mathrm{dr}}$
After taking out the constant terms along with using the sum rule of differentiating,
$\mathrm{V}^{\prime}=\frac{\mathrm{d}\left(\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{3}{2}}\right)}{\mathrm{dr}}+\frac{4}{3} \pi \frac{\mathrm{d}\left(\mathrm{r}^{3}\right)}{\mathrm{dr}}$

Using the power rule of differentiation,

\begin{aligned} \mathrm{V}^{\prime} &=\frac{3}{2} \cdot\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{3}{2}-1} \cdot \frac{\mathrm{d}\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)}{\mathrm{dr}}+\frac{4}{3} \pi\left(3 \mathrm{r}^{2}\right) \\ \mathrm{V}^{\prime} &=\frac{3}{2} \cdot\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{1}{2}}\left[\frac{1}{6} \cdot \frac{\mathrm{d}\left(\mathrm{k}-4 \pi \mathrm{r}^{2}\right)}{\mathrm{dr}}\right]+\left(4 \pi \mathrm{r}^{2}\right) \\ \mathrm{V}^{\prime} &=\frac{3}{2} \cdot\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{1}{2}}\left[\frac{1}{6} \cdot(-4 \pi(2 \mathrm{r}))\right]+\left(4 \pi \mathrm{r}^{2}\right) \\ \mathrm{V}^{\prime} &=\frac{3}{2} \cdot\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{1}{2}}\left[-\frac{4}{3} \pi r\right]+\left(4 \pi r^{2}\right) \\ \end{aligned}

$\mathrm{V}^{\prime} =-2 \pi \mathrm{r} \cdot\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{1}{2}}+\left(4 \pi \mathrm{r}^{2}\right)$

\begin{aligned} &\mathrm{V}^{\prime}=-2 \pi \mathrm{r}\left[\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{1}{2}}-2 \mathrm{r}\right] \ldots . \text { (iv) }\\ &\text { Critical point can be calculated by equating the first derivative with } 0 \text { , }\\ &V^{\prime}=0\\ &\Rightarrow-2 \pi r\left[\left(\frac{\mathrm{k}-4 \pi r^{2}}{6}\right)^{\frac{1}{2}}-2 \mathrm{r}\right]=0\\ &\Rightarrow-2 \pi r=0 \text { or }\left[\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{1}{2}}-2 \mathrm{r}\right]=0\\ &\Rightarrow \mathrm{r}=0 \text { or }\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{1}{2}}=2 \mathrm{r}\\ &\Rightarrow \mathrm{r}=0 \text { or } \frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}=4 \mathrm{r}^{2} \end{aligned}

$\\ { \Rightarrow r = 0\ or\ (k-4 \pi r\textsuperscript{2}) = 24r\textsuperscript{2}}\\ { \Rightarrow r = 0\ or\ k = 4 \pi r\textsuperscript{2}+24r\textsuperscript{2}}\\ { \Rightarrow r = 0\ or\ (4 \pi +24)r\textsuperscript{2} = k}\\$

$\\\Rightarrow \mathrm{r}=0 \ or \ \mathrm{r}^{2}=\frac{\mathrm{k}}{(4 \pi+24)} \\\Rightarrow \mathrm{r}=0 \ or \ \mathrm{r}=\pm \sqrt{\frac{\mathrm{k}}{(4 \pi+24)}}$
Now we know, $r \neq 0$
Hence
$\mathrm{r}=\sqrt{\frac{\mathrm{k}}{(4 \pi+24)}}$
To find the second derivative of this volume equation, we cam simply differentiate the equation (ii),
$\mathrm{V}^{\prime \prime}=\frac{\mathrm{d}\left(-2 \pi \mathrm{r}\left[\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{1}{2}}-2 \mathrm{r}\right]\right)}{\mathrm{dx}}$

$\mathrm{V}^{\prime \prime}=\frac{\mathrm{d}\left(-2 \pi \mathrm{r} \cdot\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{1}{2}}+\left(4 \pi \mathrm{r}^{2}\right)\right)}{\mathrm{dx}}$
After removing the constant terms, we apply the sum rule of differentiation,
$\mathrm{V}^{\prime \prime}=-2 \pi \frac{\mathrm{d}\left(\mathrm{r} \cdot\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{1}{2}}\right)}{\mathrm{dx}}+4 \pi \frac{\mathrm{d}\left(\mathrm{r}^{2}\right)}{\mathrm{dx}}$
Using the product rule of differentiation,
$\mathrm{V}^{\prime \prime}=-2 \pi\left[\mathrm{r} \frac{\mathrm{d}\left(\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{1}{2}}\right)}{\mathrm{dx}}+\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{1}{2}} \cdot \frac{\mathrm{dr}}{\mathrm{dr}}\right]+4 \pi \frac{\mathrm{d}\left(\mathrm{r}^{2}\right)}{\mathrm{dx}}$
Again, the power rule of differentiation is used,
$\mathrm{V}^{\prime \prime}=-2 \pi\left[\mathrm{r} \cdot \frac{1}{2} \cdot\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{1}{2}-1} \frac{\mathrm{d}\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)}{\mathrm{dx}}+\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{1}{2}}\right]+4 \pi(2 \mathrm{r})$

Differentiating the equation, we get

$V"=-2\pi\left [ r.\frac{1}{2}\left ( \frac{k-4\pi r^2}{6} \right )^\frac{-1}{2} \left ( \frac{1}{6}\left ( -4\pi(2r) \right ) \right )+\left ( \frac{k-4\pi r^2}{6} \right )^\frac{1}{2} \right ]+8\pi r$

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

$\\ \mathrm{V}^{\prime \prime}=-2 \pi\left[\frac{-2 \pi r^{2}+3\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)}{3\left(\left(\frac{\mathrm{k}-4 \pi r^{2}}{6}\right)^{\frac{1}{2}}\right)}\right]+8 \mathrm{\pi} \\ \left.\mathrm{~V}^{\prime \prime}=-2 \pi \left[ \frac{-2 \pi \mathrm{r}^{2}+\left(\frac{\mathrm{k}-4 \pi r^{2}}{2}\right)}{3\left(\left(\frac{\mathrm{k}-4 \pi r^{2}}{6}\right)^{\frac{1}{2}}\right)}\right]\right]+8 \pi \mathrm{r}$

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

$\mathrm{V}^{\prime \prime}=-2 \pi\left[\frac{\mathrm{k}-8 \pi \mathrm{r}^{2}}{6\left(\left(\frac{\mathrm{k}-4 \mathrm{\pi r}^{2}}{6}\right)^{\frac{1}{2}}\right)}\right]+8 \mathrm{\pi r}$

$\mathrm{V}^{\prime \prime}=-\pi\left[\frac{\mathrm{k}-8 \mathrm{\pi r}^{2}}{6\left(\left(\frac{\mathrm{k}-4 \mathrm{\pi r}^{2}}{6}\right)^{\frac{1}{2}}\right)}\right]+8 \pi \mathrm{r}>0$

$r=\sqrt{\frac{k}{(4 \pi+24)}}$
Hence for
The substituting$r=\sqrt{\frac{k}{(4 \pi+24)}}$, in equation (iii), we get
$\Rightarrow \mathrm{a}=\left(\frac{\mathrm{k}-4 \pi\left(\sqrt{\frac{\mathrm{k}}{(4 \pi+24)}}\right)^{2}}{6}\right)^{\frac{1}{2}}$

$\\ \Rightarrow a=\left(\frac{k-4 \pi\left(\frac{k}{4(\pi+6)}\right)}{6}\right)^{\frac{1}{2}} \\ \Rightarrow a=\left(\frac{k-\left(\frac{k \pi}{(\pi+6)}\right)}{6}\right)^\frac{1}{2} \\ \Rightarrow a=\left(\frac{k(\pi+6)-k \pi}{6(\pi+6)}\right)^{\frac{1}{2}} \\ \Rightarrow a=\left(\frac{k \pi+6 k-k \pi}{6(\pi+6)}\right)^{\frac{1}{2}} \\ \Rightarrow a=\left(\frac{k}{(\pi+6)}\right)^{\frac{1}{2}}$

Now we will find the ratio of an edge of the cube to the diameter of the sphere, when the sums of their volumes are minimum, i.e.,

a:2r

$\\ \Rightarrow\left(\frac{\mathrm{k}}{(\pi+6)}\right)^{\frac{1}{2}}: 2\left(\left(\frac{\mathrm{k}}{(4 \pi+24)}\right)^{\frac{1}{2}}\right) \\ \\ \Rightarrow\left(\frac{\mathrm{k}}{(\pi+6)}\right)^{\frac{1}{2}}: 2\left(\frac{1}{4^{\frac{1}{2}}}\left(\frac{\mathrm{k}}{(\pi+6)}\right)^{\frac{1}{2}}\right) \\ \Rightarrow\left(\frac{\mathrm{k}}{(\pi+6)}\right)^{\frac{1}{2}}: 2\left(\frac{1}{2}\left(\frac{\mathrm{k}}{(\pi+6)}\right)^{\frac{1}{2}}\right)$

$\Rightarrow\left(\frac{\mathrm{k}}{(\pi+6)}\right)^{\frac{1}{2}}:\left(\left(\frac{\mathrm{k}}{(\pi+6)}\right)^{\frac{1}{2}}\right)$

Hence the required ratio is

a:2r = 1:1