Minimising the cost of a cylindrical vat under a height limit
A dairy cooperative needs a closed cylindrical vat with capacity 8.00 m³. The curved side is made from stainless sheet costing $45 per m², while each flat circular end costs $70 per m² because of extra stiffening. The vat must fit under a shed, so its height h must satisfy 1.20 m ≤ h ≤ 2.50 m. A supplier proposes a radius of 1.20 m and height calculated from the volume, arguing that 'a wider vat always uses less expensive metal per litre'. The cooperative asks for the dimensions that minimise total material cost within the shed constraint, and for a check of whether the unconstrained minimum is usable. Weld and labour costs are ignored, and the sheet thickness is constant. Use π as required and give dimensions to three significant figures.
1. Express the total material cost as a function of the radius r, find the stationary point, and determine the corresponding height.
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Let r be the radius and h the height. The volume condition gives πr²h = 8, so h = 8/(πr²). The cost is C(r) = 45(2πrh) + 70(2πr²) = 90πrh + 140πr². Substituting h: 90πr·8/(πr²) = 720/r, so C(r) = 720/r + 140πr². Differentiate: C′(r) = −720/r² + 280πr. Setting C′ = 0 gives 280πr³ = 720, so r³ = 720/(280π) = 18/(7π), and r ≈ 0.935 m. Then h = 8/(πr²) ≈ 2.91 m. Also C″ = 1440/r³ + 280π > 0, so this is a minimum of the unconstrained model. It is not feasible because h ≈ 2.91 m exceeds the shed limit 2.50 m.
2. Find the cost-minimising dimensions within the height constraint and check the supplier's claim.
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The height limits 1.20 ≤ h ≤ 2.50 correspond, using h = 8/(πr²), to r between √(8/(2.5π)) ≈ 1.01 m and √(8/(1.2π)) ≈ 1.46 m. The stationary radius 0.935 m lies below this interval. For r > 0.935, C′(r) > 0, so C increases throughout the feasible interval. Hence the constrained minimum is at the smallest feasible radius, r ≈ 1.01 m, with h = 2.50 m. The cost is about C ≈ 720/1.009 + 140π(1.009)² ≈ 714 + 448 = 1162 dollars. The supplier's r = 1.20 m gives h = 8/(π(1.20)²) ≈ 1.77 m and cost ≈ 600 + 633 = 1233 dollars, so it is more expensive. The claim 'wider always cheaper' is false: beyond the unconstrained optimum, extra radius raises cost because the expensive circular ends grow.