Step 1: Set up the geometry.
A closed cylinder has radius $r$ and height $h$, with fixed volume $V=\pi r^2 h$. We want the dimensions minimising the total surface area $S=2\pi r^2+2\pi r h$.
Step 2: Reduce to one variable.
From the volume constraint, $h=\frac{V}{\pi r^2}$. Substituting, $S(r)=2\pi r^2+\frac{2V}{r}$.
Step 3: Differentiate and set to zero.
Then $\frac{dS}{dr}=4\pi r-\frac{2V}{r^2}$. Setting $\frac{dS}{dr}=0$ gives $4\pi r=\frac{2V}{r^2}$, so $V=2\pi r^3$.
Step 4: Confirm it is a minimum.
The second derivative $\frac{d^2S}{dr^2}=4\pi+\frac{4V}{r^3}>0$, so this critical point is indeed a minimum.
Step 5: Translate $V=2\pi r^3$ back to $h$.
Using $V=\pi r^2 h$, we get $\pi r^2 h=2\pi r^3$, hence $h=2r$.
Step 6: Write the required ratio.
Therefore $\frac{h}{r}=\frac{2}{1}$, that is, height to base radius is $2:1$. \[ \boxed{2:1} \]