Cylindrical can of minimum surface area

The third standard problem: at fixed volume, the cylindrical can of minimum surface area has height equal to the diameter.

Example — Cylindrical can of minimum surface area

A can of cylindrical shape must contain a given volume $V$. Find the dimensions $(r,h)$ that minimise the total surface area (wall + two bases).

Solution

Constraint: $\pi r^2 h = V \Rightarrow h = V/(\pi r^2)$.

Total surface area: $S(r) = 2\pi r^2 + 2\pi r h = 2\pi r^2 + \dfrac{2V}{r}$, with $r>0$. $S'(r) = 4\pi r - \frac{2V}{r^2} = 0 \iff 4\pi r^3 = 2V \iff r^3 = \frac{V}{2\pi}.$ Substituting: $h=V/(\pi r^2) = 2r$. The optimal can has height equal to the diameter ($\boxed{h=2r}$). This is the geometry of the “ideal” can; real commercial cans are generally a little taller because there is also a cost of sealing at the edges.

Links

Topics: Derivatives
Concepts: Objective function · Maximum and minimum
Methods: Maximum minimum via derivative
Skills: Differentiating · Modelling