Sizing a Cylindrical Can When Body and End-Cap Sheet Priced Differently
A fabrication unit must supply closed cylindrical cans of exactly 500 cm³ internal volume. The body is rolled from 0.4 mm aluminium sheet costing ₹2 per cm² of surface, while the two end discs are stamped from 0.6 mm sheet costing ₹3 per cm². The production engineer must choose radius r (cm) and height h (cm) to minimise material cost subject to πr²h = 500. The shop's roller handles radii only between 2.5 cm and 6.0 cm, and the seaming machine requires a height of at least 6.0 cm. A junior engineer proposes r = 5 cm because a 10 cm die is already in stock, and argues the cost difference from the true optimum will be 'negligible'. Before signing off, the plant manager wants the true optimum, the actual cost penalty of the junior engineer's proposal, and a general rule connecting the optimal height-to-radius ratio to the ratio of end-cap cost to body cost, so future jobs with different sheet prices can be quoted quickly without redoing the calculus.
1. Determine the radius and height that minimise material cost, and confirm that the result satisfies the shop constraints and is a true minimum.
Read the worked answer
Cost = body area × 2 + end area × 3:
C(r) = 2(2πrh) + 3(2πr²) = 4πrh + 6πr²
Eliminate h using h = 500/(πr²): 4πr·500/(πr²) = 2000/r, so
C(r) = 2000/r + 6πr²
dC/dr = −2000/r² + 12πr = 0 ⇒ r³ = 500/(3π) = 53.05 ⇒ r ≈ 3.76 cm
h = 500/(π × 14.12) ≈ 11.3 cm; C ≈ 532 + 266 ≈ ₹798.
Second derivative 4000/r³ + 12π > 0 for all r > 0, so this stationary point is the global minimum on the positive axis. Constraints: 2.5 ≤ 3.76 ≤ 6.0 ✓ and h = 11.3 ≥ 6.0 ✓, so the optimum is feasible, not a boundary case. The large h/r ratio is the expected consequence of the ends being 1.5 times dearer per cm² than the body: the optimum buys less end area by shrinking the radius and paying for it in body area.
2. Compute the percentage cost penalty of the junior engineer's r = 5 cm proposal and assess whether 'negligible' is a fair description.
Read the worked answer
At r = 5 cm, h = 500/(π × 25) = 6.37 cm, which still clears the 6.0 cm seaming limit, so the proposal is manufacturable.
C(5) = 2000/5 + 6π(25) = 400 + 471.2 = ₹871.2
C(3.76) ≈ ₹798.4
Excess = 871.2 − 798.4 = ₹72.8 per can, i.e. 72.8/798.4 = 9.1%.
The phrase is wrong on two counts. First, 9.1% of material cost is far above normal can-making margins, so at volume (say 200,000 cans) the penalty is roughly ₹1.46 crore. Second, the cost curve is flat only very near the optimum; at r = 5 the radius is 33% above optimum, which is well outside that flat region. A defensible decision to use the stock die would have to rest on tooling cost or lead time, explicitly traded against ₹72.8 per can - not on an unquantified claim of negligibility.
3. Derive a general rule for the optimal h/r when ends cost k times as much per cm² as the body, and state what the rule implies for quoting.
Read the worked answer
Let body cost be c per cm² and end cost kc per cm²:
C = c(2πrh) + kc(2πr²), with V = πr²h.
Divide by c and substitute h = V/(πr²):
C/c = 2πr·V/(πr²) + 2kπr² = 2V/r + 2kπr²
d(C/c)/dr = −2V/r² + 4kπr = 0 ⇒ r³ = V/(2kπ)
Then h/r = V/(πr³) = V/(π · V/(2kπ)) = 2k.
So the optimal shape obeys h = 2kr, and this ratio is independent of the volume - volume sets the scale, the price ratio sets the shape. For k = 1.5 the rule gives h = 3r, matching the numerical answer above (11.3/3.76 ≈ 3.0). For a job with equal sheet prices (k = 1) the rule gives h = 2r, the classic closed-cylinder result.
Quoting implication: the estimator never needs calculus again. Measure the price ratio, set h = 2kr, then r = (V/(2kπ))^{1/3}. The rule also flags when the die on the shelf is worth using: the penalty grows as the shop's fixed radius departs from (V/(2kπ))^{1/3}, and is worst for small, high-k jobs.