Cylindrical Container Cost under a Press Limit
A metalworks company makes open-top cylindrical containers of volume 1,000 cm³. The base is cut from a stronger alloy costing 3 paisa per cm², while the curved side uses a cheaper sheet costing 2 paisa per cm². The production engineer models the total material cost C(r) in paisa as a function of base radius r cm. An existing press can form radii only between 4 cm and 6 cm, and the company wants the cheapest container within that range. A customer proposes doubling the volume to 2,000 cm³ while keeping the same height-to-radius ratio as the current optimum. The engineer must find the radius and height that minimise cost for 1,000 cm³, check whether the press constraint is active, and determine the new dimensions and cost scaling for the double-volume request.
1. Derive C(r) for volume 1,000 cm³, find the radius and height that minimise cost, and check whether the press limit is active.
Read the worked answer
The volume is πr²h = 1000, so h = 1000/(πr²). The base area is πr² and the curved side area is 2πrh. Thus C(r) = 3(πr²) + 2(2πrh) = 3πr² + 4πr·1000/(πr²) = 3πr² + 4000/r.
Differentiate: C'(r) = 6πr − 4000/r². Setting C'(r) = 0 gives 6πr³ = 4000, so r³ = 4000/(6π) ≈ 212.21 and r ≈ 5.96 cm. Also C''(r) = 6π + 8000/r³ > 0, so this is a minimum.
The corresponding height is h = 1000/(π×5.96²) ≈ 8.96 cm. Since 5.96 cm lies between 4 cm and 6 cm, the press constraint is not active. The minimum cost is about 3π(5.96²) + 4000/5.96 ≈ 334.6 + 671.1 ≈ 1005.7 paisa, or about Rs 10.06.
2. For the double-volume request, find the new radius and height if the height-to-radius ratio is kept the same, and state whether the existing press can make it.
Read the worked answer
For a fixed height-to-radius ratio, volume is proportional to r³. Doubling the volume multiplies both r and h by 2^(1/3) ≈ 1.260.
New radius ≈ 5.96 × 1.260 ≈ 7.51 cm, and new height ≈ 8.96 × 1.260 ≈ 11.29 cm. Surface area, and hence material cost at fixed prices, scales by 2^(2/3) ≈ 1.587, so the ideal cost would be about 1005.7 × 1.587 ≈ 1596 paisa.
However, the press can form radii only up to 6 cm. Since 7.51 cm > 6 cm, the existing press cannot make the requested container while keeping the same ratio. The press constraint becomes active, and the company must either use a larger press or change the shape, which will cost more than the ideal scaling.
3. If the press is limited to a 6 cm radius, find the height needed for 2,000 cm³ and compare the resulting cost with the ideal scaled cost.
Read the worked answer
With r = 6 cm and V = 2000 cm³, h = 2000/(π×6²) = 2000/(36π) ≈ 17.68 cm. The cost is C = 3π(6²) + 4π(6)(17.68) = 108π + 424.32π ≈ 339.3 + 1333.1 = 1672.4 paisa.
The ideal same-ratio cost was about 1596 paisa, so the press limit adds roughly 76 paisa, about 4.8%. The container is much taller and narrower than the ideal double-volume shape. This shows that the press constraint forces a cost penalty, even though the container is still manufacturable.