Open-Top Can with Two Material Costs
A metal fabricator has an order for 500 open-top cylindrical cans, each with volume 500 cm³. The side wall is cut from one sheet and costs c per cm². The circular base must be cut from a thicker sheet that costs 2c per cm². A junior draughtsman minimised the total surface area of an open cylinder of this volume and obtained the familiar result radius = height, then used that shape for the quote. The production manager suspects the quote is too high because the base material is twice as expensive as the side material, so the cheapest can may not be the one with least surface area. The shop can form any radius r > 0 and the corresponding height h, but the volume must remain exactly 500 cm³ and the can must have a base and no lid. The manager wants the dimensions that minimise material cost, a check that the result is a true minimum rather than a stationary point, and a comparison with the junior’s radius = height design.
1. Express the material cost C in terms of r alone, using the volume constraint, and state the domain of r.
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Volume constraint: πr²h = 500, so h = 500/(πr²). Base area = πr², side area = 2πrh. With base cost 2c per cm² and side cost c per cm², C = 2c(πr²) + c(2πrh) = 2cπr² + 2cπr·500/(πr²) = 2cπr² + 1000c/r. The only geometric restriction is r > 0; h is then determined and positive.
2. Find the radius and height that minimise C. Justify that your stationary point gives the global minimum.
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Differentiate: dC/dr = 4cπr − 1000c/r². Set dC/dr = 0: 4πr³ = 1000 ⇒ r³ = 250/π ⇒ r = (250/π)^(1/3) ≈ 4.30 cm. Then h = 500/(πr²). Using πr³ = 250, h/r = 500/(πr³) = 500/250 = 2, so h ≈ 8.60 cm. Second derivative: C'' = 4cπ + 2000c/r³ > 0 for r > 0, so the stationary point is a local minimum. Also C → ∞ as r → 0+ and as r → ∞, so this local minimum is the global minimum on r > 0.
3. The junior’s design has h = r. Compare its cost with the minimum cost, and explain in cost terms why the two answers differ.
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If h = r, then πr³ = 500 and r = (500/π)^(1/3) ≈ 5.42 cm. The cost is C_jun/c = 2πr² + 1000/r ≈ 369.1. At the true optimum, C_min/c ≈ 348.7. The saving is about 20.4c per can, or about 5.5%. The cost-optimal can is taller and narrower (h = 2r) because the base costs twice as much per cm²; the design shifts area from the expensive base to the cheaper side.