Can Geometry at the Tender-Coconut Water Unit
A small unit in Alappuzha packs tender coconut water in cylindrical aluminium cans of volume exactly 500 cm³. The purchase committee has two rates: the curved surface costs ₹0.02 per cm², while each of the two circular end discs costs ₹0.03 per cm² because they are stamped from thicker sheet. The seaming machine adds a rule - the height must be at least 1.5 times the radius, or the seam opens on the filling line.
A trainee proposes r = 5 cm with h = 6.37 cm, arguing that a wider can 'uses less metal for the same volume'. The production manager wants the cost of that proposal worked out, the true cost-minimising dimensions, and a check that the seaming rule holds at the optimum. She also wants to know how the optimum shifts if end discs become costlier than the curved surface.
1. Write the cost per can as a function of the radius r alone, locate the stationary point, justify that it is a minimum, and compare its cost with the trainee's proposal. Does the trainee's design satisfy the seaming rule?
Read the worked answer
Volume gives $h=\dfrac{500}{\pi r^{2}}$. Cost is curved surface at ₹0.02/cm² plus two ends at ₹0.03/cm²:
$$C(r)=0.02(2\pi rh)+0.03(2\pi r^{2})=\frac{20}{r}+0.06\pi r^{2}$$
$$C'(r)=-\frac{20}{r^{2}}+0.12\pi r=0 \Rightarrow r^{3}=\frac{20}{0.12\pi}\approx 53.05 \Rightarrow r\approx 3.76\text{ cm}$$
$C''(r)=\dfrac{40}{r^{3}}+0.12\pi>0$ for all $r>0$, so this stationary point is the global minimum.
Then $h=\dfrac{500}{\pi(3.76)^{2}}\approx 11.26$ cm, so $h/r\approx 3.0\ge 1.5$: the seaming rule holds.
Cost at the optimum: $C=\dfrac{20}{3.76}+0.06\pi(14.13)\approx 5.32+2.66=₹7.98$.
Trainee: $h=\dfrac{500}{\pi(25)}\approx 6.37$ cm, $h/r=1.27<1.5$, so the design fails the seaming rule outright. Even ignoring that, $C=\dfrac{20}{5}+0.06\pi(25)\approx 4.00+4.71=₹8.71$, about 73 paise costlier. A wider can does not use less metal for a fixed volume; the trainee confused volume share with surface area.
2. The end discs are now quoted at ₹0.05 per cm² while the curved surface stays at ₹0.02 per cm². Find the new optimal radius, height and cost, and explain the direction of the change.
Read the worked answer
Now $C(r)=\dfrac{20}{r}+0.10\pi r^{2}$.
$$C'(r)=-\frac{20}{r^{2}}+0.20\pi r=0 \Rightarrow r^{3}=\frac{20}{0.2\pi}\approx 31.83 \Rightarrow r\approx 3.17\text{ cm}$$
$C''>0$, so this is the minimum. Then $h=\dfrac{500}{\pi(3.17)^{2}}\approx 15.84$ cm and $h/r\approx 5.0\ge 1.5$, so the seaming rule is still met.
$$C\approx \frac{20}{3.17}+0.10\pi(10.05)\approx 6.31+3.16=₹9.47$$
The radius falls from 3.76 cm to 3.17 cm while the can grows taller. The economics is straightforward: disc area grows like $r^{2}$ and the discs are now the expensive component, so the optimum trades disc area for curved area, which grows only like $rh$. With equal rates the balance would sit near $h=2r$; unequal rates push the height-to-radius ratio away from 2, here to about 5.
3. Suppose the seaming rule were tightened to 'height at least 4 times the radius'. Find the best feasible design and its cost, and state whether the rule is now binding.
Read the worked answer
The constraint $h\ge 4r$ becomes $\dfrac{500}{\pi r^{2}}\ge 4r$, i.e. $r^{3}\le \dfrac{500}{4\pi}\approx 39.79$, so $r\le 3.415$ cm.
Since $C$ decreases on $(0,,3.76)$ and rises afterwards, the unconstrained optimum $r=3.76$ is now infeasible and the best feasible point is the largest allowed radius:
$$r=3.415\text{ cm},\quad h=\frac{500}{\pi(3.415)^{2}}\approx 13.65\text{ cm},\quad h/r=4.0$$
$$C=\frac{20}{3.415}+0.06\pi(11.66)\approx 5.86+2.20=₹8.06$$
The rule is binding: cost rises from ₹7.98 to ₹8.06, and the optimum sits exactly on the constraint boundary rather than at $C'=0$. If the rule were loosened again, the design would return to $r\approx 3.76$ cm.