Hot-water cylinder: constrained surface-area minimisation
A manufacturer produces stainless-steel hot-water cylinders. Each is a right circular cylinder of radius $r$ metres and height $h$ metres with a flat base and a hemispherical dome on top, so the overall height is $h + r$. The volume must be exactly $2.0\ \text{m}^3$. Base, curved side and dome are cut from the same sheet and cost the same per square metre, so the manufacturer wants to minimise total external surface area. Finished units must pass through the delivery hatch, which limits the external diameter to $1.4\ \text{m}$; height is unrestricted. A junior engineer differentiates the surface-area formula, obtains $r = 0.726\ \text{m}$ and $h = 0.725\ \text{m}$, and reports a minimum area of $8.27\ \text{m}^2$. The production manager is uneasy: that shape is wider than the hatch allows, and the predicted saving over the factory's current $r = 0.70\ \text{m}$ design looks too good to be true.
1. Show that the total surface area can be written as $S(r) = \tfrac{5}{3}\pi r^2 + \dfrac{4}{r}$, and find the radius that minimises $S$ when the $1.4\ \text{m}$ diameter limit is ignored. Justify that your value gives a minimum.
Read the worked answer
Volume constraint. $V = \pi r^2 h + \tfrac{2}{3}\pi r^3 = 2$, so
$$h = \frac{2}{\pi r^2} - \frac{2}{3}r .$$
Surface area. Base $\pi r^2$, curved side $2\pi r h$, dome $2\pi r^2$, so $S = 3\pi r^2 + 2\pi r h$. Substituting $h$:
$$S = 3\pi r^2 + 2\pi r\left(\frac{2}{\pi r^2} - \frac{2}{3}r\right) = 3\pi r^2 + \frac{4}{r} - \frac{4}{3}\pi r^2 = \frac{5}{3}\pi r^2 + \frac{4}{r}.$$
Stationary point. $\dfrac{dS}{dr} = \dfrac{10}{3}\pi r - \dfrac{4}{r^2} = 0 \Rightarrow \dfrac{10}{3}\pi r^3 = 4 \Rightarrow r^3 = \dfrac{1.2}{\pi}$, giving $r = 0.7258\ \text{m}$ (and $h = 0.7246\ \text{m}$, matching the junior engineer).
Justification. $\dfrac{d^2S}{dr^2} = \dfrac{10}{3}\pi + \dfrac{8}{r^3} > 0$ for all $r > 0$, so the curve is concave up and this stationary point is the global minimum on $r > 0$. Note the domain $r > 0$ matters: $S \to \infty$ as $r \to 0^+$ and as $r \to \infty$, so a single interior minimum is expected.
2. Now apply the $1.4\ \text{m}$ diameter limit. Determine the radius that actually minimises surface area and the resulting area (3 s.f.). Explain carefully why the stationary point is no longer used.
Read the worked answer
Restricted domain. Diameter $\le 1.4$ means $r \le 0.70$, so the feasible domain is $0 < r \le 0.70$.
Where is the stationary point? At $r = 0.7258$, which lies outside the feasible interval. Since $\dfrac{dS}{dr} = \dfrac{10}{3}\pi r - \dfrac{4}{r^2}$ has exactly one positive zero (at $r = 0.7258$) and $\dfrac{dS}{dr} < 0$ for $0 < r < 0.7258$, the area is strictly decreasing on the whole interval $(0, 0.70]$.
Conclusion. A strictly decreasing function on $(0, 0.70]$ is smallest at the right endpoint, so the constrained minimum is at $r = 0.70\ \text{m}$:
$$S(0.70) = \tfrac{5}{3}\pi(0.49) + \frac{4}{0.70} = 2.565 + 5.714 = 8.279 \approx 8.28\ \text{m}^2,$$
with $h = \dfrac{2}{\pi(0.49)} - \tfrac{2}{3}(0.70) = 1.299 - 0.467 = 0.83\ \text{m}$. The hatch limit costs only $8.28 - 8.27 = 0.01\ \text{m}^2$, so the manager's suspicion is about feasibility, not about a lost saving.
3. Write the recommendation the engineer should give, and state the main assumptions built into the surface-area model that affect how the numbers should be read.
Read the worked answer
Recommendation. Build to $r = 0.70\ \text{m}$, $h = 0.83\ \text{m}$ (overall height about $1.53\ \text{m}$), giving $S \approx 8.28\ \text{m}^2$. Do not relax the hatch: the unconstrained optimum is only $0.01\ \text{m}^2$ cheaper, well inside any reasonable uncertainty in the model, so the restriction is effectively free.
Assumptions to flag.
- The 'surface area' model treats the steel as a zero-thickness mathematical surface and takes cost as strictly proportional to area; in practice sheet is bought in rectangles, so cutting waste and seam overlap add area that the formula ignores. The true optimum for nesting patterns could shift slightly.
- The dome is assumed to be an exact hemisphere welded to the cylinder with no additional material, and the base is flat rather than dished.
- The volume is treated as exactly $2.0\ \text{m}^3$; real capacity tolerances and the space taken by insulation, fittings and the element are not in the model.
Judgement point: a stationary point that falls outside the physical constraint must be discarded, and the correct conclusion comes from the sign of $S'(r)$ on the feasible interval, not from the second derivative.