Let the inner dimensions be:

• inner radius $= r$
• inner height $= h$

Given fixed inner volume, $\pi r^2 h = V$

Thickness information:

• wall thickness $= 2$ mm
• bottom thickness $= 2$ mm
• open at top

Hence:

• outer radius $= r+2$
• outer height $= h+2$

But since the top is open, the material consists of:

• cylindrical side wall
• bottom solid disc

Volume of material used

(i) Side wall volume This is the volume difference of two cylinders of height $h$: $V_{\text{side}} = \pi\big((r+2)^2-r^2\big)h$ $= \pi(4r+4)h = 4\pi(r+1)h$

(ii) Bottom disc volume Bottom is a solid disc of thickness $2$ mm and radius equal to outer radius $(r+2)$: $V_{\text{bottom}} = \pi(r+2)^2\cdot 2$

Therefore total material volume: $M = 4\pi(r+1)h + 2\pi(r+2)^2$

Use the fixed inner volume constraint $h = \frac{V}{\pi r^2}$

Substitute into $M$: $M(r)=4\pi(r+1)\cdot \frac{V}{\pi r^2}+2\pi(r+2)^2$ $M(r)=\frac{4V(r+1)}{r^2}+2\pi(r+2)^2$

Minimize $M(r)$

Differentiate with respect to $r$: $M(r)=4V\left(\frac{r+1}{r^2}\right)+2\pi(r+2)^2$

Write $\frac{r+1}{r^2}=\frac{1}{r}+\frac{1}{r^2}$

So, $\frac{d}{dr}\left(\frac{r+1}{r^2}\right)=-\frac{1}{r^2}-\frac{2}{r^3}=-\frac{r+2}{r^3}$

Hence, $M'(r)=4V\left(-\frac{r+2}{r^3}\right)+4\pi(r+2)$

$M'(r)=4(r+2)\left(\pi-\frac{V}{r^3}\right)$

For minimum, given it occurs at $r=10$ mm, $M'(10)=0$

Since $10+2\neq 0$, $\pi-\frac{V}{10^3}=0$ $V=1000\pi$

Compute the asked value $\frac{V}{250\pi}=\frac{1000\pi}{250\pi}=4$

Comparison with stored answer Derived answer $=4$, which matches the stored correct answer.