Closed Can Design With Two Material Prices
A beverage company wants a closed cylindrical can of volume 500 cm³. The side-wall material costs ₹0.02 per cm², while the top and bottom discs cost ₹0.03 per cm² because they are printed. The production engineer must choose radius r and height h. A shipping constraint requires r ≤ 5 cm, and the filling nozzle requires h ≥ 8 cm. The company currently uses r = 4 cm and h ≈ 9.95 cm because it looks proportionate, but the cost accountant suspects this is not optimal. A supplier offers a cheaper side-wall alloy that would reduce the side material price to ₹0.015 per cm², but only if the can's height stays below 10 cm. The engineer needs to know the true cost-minimising dimensions, whether the current can is close, and whether the supplier's offer should be accepted.
1. Set up the total material cost as a function of r alone, find the stationary point, and verify it is a minimum under the stated constraints.
Read the worked answer
From πr²h = 500, h = 500/(πr²). The side area is 2πrh = 1000/r, and the top-plus-bottom area is 2πr². Thus C(r) = 0.02(1000/r) + 0.03(2πr²) = 20/r + 0.06πr².
Differentiate: C'(r) = −20/r² + 0.12πr. Setting C' = 0 gives 0.12πr³ = 20, so r³ = 20/(0.12π) ≈ 53.05 and r ≈ 3.757 cm. Then h = 500/(π(3.757)²) ≈ 11.28 cm. The constraints r ≤ 5 and h ≥ 8 are satisfied.
C''(r) = 40/r³ + 0.12π > 0, so the stationary point is a minimum. The minimum cost is C ≈ 20/3.757 + 0.06π(3.757)² ≈ 5.323 + 2.660 = ₹7.98 per can.
At r = 4, h = 500/(16π) ≈ 9.95 cm and C = 20/4 + 0.06π(16) ≈ 5.000 + 3.016 = ₹8.016. The current can is close to optimal but costs about ₹0.03 more per can.
2. The supplier offers side material at ₹0.015/cm² provided h < 10 cm. Re-optimise and decide whether to accept. Include the constraint check.
Read the worked answer
With the cheaper side material, C(r) = 0.015(1000/r) + 0.03(2πr²) = 15/r + 0.06πr². Then C'(r) = −15/r² + 0.12πr = 0 gives r³ = 15/(0.12π) ≈ 39.79 and r ≈ 3.415 cm, with h = 500/(π(3.415)²) ≈ 13.65 cm. This violates h < 10 cm, so the unconstrained optimum is not available.
The height constraint binds at the boundary h = 10 (or just below it). From πr²(10) = 500, r² = 50/π ≈ 15.915, so r ≈ 3.989 cm. This satisfies r ≤ 5. The cost is approximately 15/3.989 + 0.06π(15.915) ≈ 3.760 + 3.000 = ₹6.76 per can.
The constrained cost ₹6.76 is well below the original ₹7.98, a saving of about ₹1.22 per can. The supplier's offer should be accepted if a height of 10 cm is acceptable; because the condition is strict, use h slightly below 10 cm and r slightly above 3.989 cm. The unconstrained cheaper-material minimum would have been about ₹6.59, so the height cap costs about ₹0.17 per can.
3. If only the original material is available, how sensitive is the optimal radius to a 50% rise in the side-material price? Does the optimal height move closer to the 8 cm nozzle limit?
Read the worked answer
A 50% rise changes the side price from ₹0.02 to ₹0.03 per cm². Then C(r) = 0.03(1000/r) + 0.06πr² = 30/r + 0.06πr².
C'(r) = −30/r² + 0.12πr = 0, so r³ = 30/(0.12π) ≈ 79.58 and r ≈ 4.30 cm. The height is h = 500/(π(4.30)²) ≈ 8.61 cm.
The optimal radius rises from about 3.76 cm to 4.30 cm, and the height falls from about 11.28 cm to 8.61 cm. The height is still above the 8 cm nozzle limit, so the constraint does not bind. The higher side price makes the company substitute away from side area toward the relatively cheaper top and bottom area, which means a wider, shorter can. The nozzle constraint is closer but not active.