Sizing an Open-Top Tank Under a Height Code
A fabricator in Kanpur has won a contract to supply an open-top cylindrical storage tank of capacity 10 m³ for a dairy. Sheet metal for the circular base costs ₹500 per m², while the curved side wall costs ₹320 per m² because it needs rolling and welding; the open top costs nothing. The workshop can roll any radius, but the dairy's safety code requires the tank's height to be at least 2.5 m so that a standard stirrer can be lowered in without fouling. The supervisor's thumb rule is to 'make the base and the side cost roughly equal', which gives a tank about 1.27 m in radius and 1.98 m tall. The safety inspector has refused to certify that shape, and the contractor now wants to know the cheapest shape that is actually legal, and whether the extra cost is worth arguing about with the dairy.
1. Write the total metal cost C as a function of the base radius r alone. Find the radius and height that minimise C when the height rule is ignored, and show that the stationary point is a minimum.
Read the worked answer
Let r = base radius and h = height in metres. The volume is fixed:
πr²h = 10 ⇒ h = 10/(πr²).
Base area = πr²; curved wall area = 2πrh. Hence
C(r) = 500πr² + 320(2πrh) = 500πr² + 640πr·10/(πr²) = 500πr² + 6400/r, r > 0.
Differentiate: C'(r) = 1000πr − 6400/r². Setting C' = 0 gives
1000πr³ = 6400 ⇒ r³ = 6.4/π = 2.0372 ⇒ r = 1.267 m, then h = 10/(π·1.6058) = 1.983 m.
C''(r) = 1000π + 12800/r³ > 0 for every r > 0, so C is convex on (0, ∞); since C → ∞ as r → 0⁺ and as r → ∞, this stationary point is the global minimum. Cost ≈ 500(5.045) + 6400/1.267 ≈ 2522 + 5051 = ₹7573.
2. The inspector insists on h ≥ 2.5 m. Find the cheapest legal design and the extra cost per tank compared with the unconstrained answer.
Read the worked answer
Fix the volume: πr²h = 10, so h ≥ 2.5 ⇔ πr² ≤ 10/2.5 = 4 ⇔ r ≤ 2/√π = 1.1284 m.
The unconstrained optimum r = 1.267 m lies outside this interval. On (0, 1.1284] we have r < 1.267, so C'(r) = 1000πr − 6400/r² < 0, i.e. C is strictly decreasing throughout the allowed interval. A decreasing function on an interval is minimised at the right end, so the constrained optimum sits on the boundary:
r = 2/√π = 1.1284 m, h = 2.5 m.
Base area = πr² = 4 m², base cost = ₹2000. Curved area = 2πrh = 2π(1.1284)(2.5) = 17.72 m², wall cost = 320 × 17.72 = ₹5671. Total ≈ ₹7672.
Extra cost = 7672 − 7573 ≈ ₹99 per tank, under 1.4%. The code is expensive only in the sense that it forces the design off the true optimum; the penalty itself is small.
3. The dairy offers to relax the code to h ≥ 2.0 m. Should the contractor press for 2.0 m, or for something lower still? Quantify before answering.
Read the worked answer
Relaxing to h ≥ 2.0 gives πr² ≤ 10/2 = 5 ⇔ r ≤ √(5/π) = 1.2616 m. This is still just short of the unconstrained 1.267 m, so the constraint still binds, but only barely:
base cost = 500 × 5 = ₹2500; curved area = 2π(1.2616)(2.0) = 15.85 m², cost = ₹5073; total ≈ ₹7573, i.e. the unconstrained optimum to the nearest rupee.
So the whole ₹99 penalty is recovered by moving the rule from 2.5 m to 2.0 m, and pressing for anything below 2.0 m buys nothing further - the design is already at its free optimum. The judgement is to argue for exactly 2.0 m and to check first that a stirrer short enough for a 2.0 m tank can be procured; asking for 1.8 m would win no money and would cost negotiating goodwill.