Problem source

A cylindrical biscuit tin has volume $V$ and surface area $S$ (including the ends). Show that the minimum possible surface area for a given value of $V$ is $S=3(2\pi V^{2})^{1/3}.$ For this value of $S$ show that the volume of the largest sphere which can fit inside the tin is $\frac{2}{3}V$, and find the volume of the smallest sphere into which the tin fits.

Solution source

Suppose we have height $h$ and radius $r$, then: $V = \pi r^2 h$ and $S = 2\pi r^2 + 2\pi r h$.

$h = \frac{V}{\pi r^2}$, so 
\begin{align*}
S &= 2 \pi r^2 + 2 \pi r\frac{V}{\pi r^2} \\
&= 2\pi r^2 +V \frac1{r}+V \frac1{r} \\
&\underbrace{ \geq }_{\text{AM-GM}} 3 \sqrt[3]{2\pi r^2 \frac{V^2}{r^2} } = 3 (2 \pi V^2)^{1/3}
\end{align*}

Equality holds when $r = \sqrt[3]{\frac{V}{2 \pi}}, h = \frac{V}{\pi (V/2\pi)^{2/3}} = \sqrt[3]{\frac{4V}{\pi}}$

Since $h > r$ the sphere has a maximum radius of $r$ and so it's largest volume is $\frac43 \pi r^3 = \frac43 \pi \frac{V}{2 \pi} = \frac23 V$.

\begin{center}
    \begin{tikzpicture}[scale=3]
        \draw (-0.542,-0.27)--(-0.542,0.27)--(0.542,0.27)--(0.542,-0.27) -- cycle;
        \draw (0,0) circle ({sqrt(0.542^2+0.27^2)});

        \draw[<->] (-0.542,{-sqrt(0.542^2+0.27^2)-0.05}) -- (0.542,{-sqrt(0.542^2+0.27^2)-0.05});

        \draw[<->] ({sqrt(0.542^2+0.27^2)+0.05}, .27) -- ({sqrt(0.542^2+0.27^2)+0.05}, -.27);

        \node[right] at ({sqrt(0.542^2+0.27^2)+0.05}, 0) {$r$};
        \node[below] at (0 ,{-sqrt(0.542^2+0.27^2)-0.05}) {$h$};
    \end{tikzpicture}
\end{center}

The radius of the sphere is $\sqrt{\left (\frac{r}{2} \right)^2 + \left (\frac{h}{2} \right)^2 } = \frac12 \sqrt{r^2+h^2}$

\begin{align*}
V_{sphere} &= \frac43 \pi (r^2+h^2)^{3/2} \\
&= \frac43 \pi \left (\left( \frac{V}{2 \pi} \right)^{2/3}+\left( \frac{4V}{ \pi} \right)^{2/3} \right)^{3/2} \\
&= \frac43 \pi \frac{V}{ \pi} \left ( 2^{-2/3}+4^{2/3}\right)^{3/2} \\
&= \frac 43 V \left ( \frac{1+4}{2^{2/3}} \right)^{3/2} \\
&= \frac43 \frac{5^{3/2}}{2} V \\
&= \frac{2 \cdot \sqrt{125}}{3} V
\end{align*}