Can design with a filling-line height constraint
A drinks company is designing a closed cylindrical can to hold 330 ml, which it will fill on an existing line. Metal for the circular top and base costs 0.060 cents per cm², while metal for the curved side costs 0.035 cents per cm² because it is a different alloy. The company's designer first ignores the filling line and simply minimises the total material cost per can, on the assumption that any radius is possible. The production manager then points out a hard constraint: the filling nozzles and the shrink-wrap sleeve require the height to be between 1.6 and 2.4 times the radius, otherwise the cans jam. The designer must check whether the mathematically cheapest can is actually buildable, and if not, work out the cheapest can that satisfies the constraint, together with the extra cost per can and over a run of two million cans.
1. Express the material cost C, in cents, as a function of the radius r alone, and state a sensible domain.
Read the worked answer
Let h be the height. Volume: πr²h = 330, so h = 330/(πr²).
Top and base area = 2πr², at 0.060 cents/cm² → 0.12πr² cents. Side area = 2πrh = 2πr × 330/(πr²) = 660/r cm², at 0.035 cents/cm² → 23.1/r cents.
C(r) = 0.12πr² + 23.1/r (cents), with 0.12π = 0.3770.
Domain: r > 0 mathematically. In practice a hand-held 330 ml can has a radius roughly between 2.5 cm and 4.5 cm, so answers far outside that should be treated with suspicion. The volume condition has already been substituted in, so h is no longer free - that substitution is the step that makes this a one-variable problem.
2. Find the radius and height that minimise the material cost with no constraint, and verify it is a minimum.
Read the worked answer
Differentiate: C'(r) = 0.24πr − 23.1/r².
Set C'(r) = 0: 0.24πr = 23.1/r² r³ = 23.1/(0.24π) = 23.1/0.75398 = 30.64 r = 3.13 cm
Then h = 330/(π × 3.13²) = 330/30.78 = 10.7 cm.
Cost: C = 0.3770 × 9.80 + 23.1/3.13 = 3.69 + 7.38 = 11.07 cents per can.
Verification: C'(r) is negative for small r (the 1/r term dominates) and positive for large r, so the stationary point is a minimum. The optimum is a tall, thin can with h/r = 10.7/3.13 = 3.42, which already hints that it will clash with the filling line.
3. Apply the height constraint, find the cheapest buildable can, and quantify the cost penalty over two million cans.
Read the worked answer
The optimum has h/r = 3.42, which exceeds the 2.4 limit, so the cheapest can cannot be built. The cost function is increasing for r beyond 3.13 cm, so the best allowed design sits on the boundary h = 2.4r.
Substitute into the volume: πr²(2.4r) = 330 r³ = 330/(2.4π) = 43.77 r = 3.52 cm, h = 2.4 × 3.52 = 8.46 cm
Check volume: π × 3.52² × 8.46 = 329.9 ml. ✓
Cost: C = 0.3770 × 3.52² + 23.1/3.52 = 4.68 + 6.56 = 11.24 cents per can.
Penalty = 11.24 − 11.07 = 0.17 cents per can, about 1.5%. Over 2,000,000 cans: 0.17 × 2,000,000 = 340,000 cents = €3,400.
The judgement point is that the constraint forces a squatter can; the unconstrained answer is not merely slightly off, it is infeasible.