Open-Top Cylindrical Bin: Cost Minimization with a Height Cap
A fabricator is quoting a batch of 500 open-top cylindrical steel bins. Each bin must hold 2.0 L (2000 cm^3). The side wall uses 1.2 mm sheet priced at 0.85 currency units per 100 cm^2; the base uses 2.0 mm sheet priced at 1.40 per 100 cm^2. A single weld joins the base to the side and costs 0.30 per cm of circumference. Cutting and rolling can produce any radius, but the shop's press brake can set heights only in 0.5 cm increments; radius is continuously adjustable. The customer requires that the finished bin stand no taller than 30 cm and no shorter than 12 cm. The shop currently proposes r = 10.0 cm and h = 20.0 cm because it gives a round number. The estimator suspects this is not cost-optimal but must also account for the discrete height step and the height limits. The batch is large enough that per-bin savings matter; a 1% reduction in unit cost is worth about 120 currency units over the order.
1. Formulate the cost per bin as C(r,h), eliminate h using the volume constraint, and find the continuous unconstrained minimizer over r > 0. Then impose the 12 cm <= h <= 30 cm limits and locate the constrained optimum. State which constraint binds.
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The volume constraint is pi r^2 h = 2000, so h = 2000/(pi r^2). The side area is 2 pi r h = 4000/r and the base area is pi r^2. Converting prices: side costs 0.0085 per cm^2, base costs 0.014 per cm^2, and the weld costs 0.30 per cm of circumference. Thus C(r) = 0.0085(4000/r) + 0.014 pi r^2 + 0.30(2 pi r) = 34/r + 0.014 pi r^2 + 0.6 pi r. Differentiate: C'(r) = -34/r^2 + 0.028 pi r + 0.6 pi. Setting C' = 0 gives 0.028 pi r^3 + 0.6 pi r^2 - 34 = 0, or approximately 0.08796 r^3 + 1.88496 r^2 - 34 = 0. Numerically r ≈ 3.91 cm, giving h ≈ 41.6 cm. This violates h <= 30 cm. For fixed volume, h <= 30 means r >= sqrt(2000/(30 pi)) ≈ 4.61 cm. Since C'(r) > 0 for r > 3.91, the cost is increasing over the feasible interval to the right, so the constrained minimum is at the boundary h = 30 cm, with r ≈ 4.61 cm. The lower limit h >= 12 cm is not active.
2. Compute the unit cost at the current proposal r = 10.0 cm, h = 20.0 cm and at the constrained optimum. What is the saving over the 500-bin batch? Does the 0.5 cm height increment change the constrained choice?
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At r = 10.0, h = 20.0: side area = 2 pi(10)(20) = 1256.64 cm^2, side cost = 0.0085(1256.64) = 10.68; base area = pi(10)^2 = 314.16, base cost = 0.014(314.16) = 4.40; circumference = 2 pi(10) = 62.83, weld cost = 0.30(62.83) = 18.85. Total ≈ 33.93 per bin. At h = 30, r = sqrt(2000/(30 pi)) ≈ 4.607 cm: side area = 2 pi(4.607)(30) ≈ 868.3, side cost ≈ 7.38; base area = pi(4.607)^2 ≈ 66.7, base cost ≈ 0.93; circumference = 2 pi(4.607) ≈ 28.95, weld cost ≈ 8.69. Total ≈ 17.00 per bin. Saving ≈ 16.93 per bin, or about 8,465 over 500 bins. The constrained height h = 30 cm is already on the 0.5 cm grid and is the best allowed height: h = 29.5 cm would be worse because cost decreases as h increases up to the unconstrained 41.6 cm. So the discrete step does not change the constrained choice here.
3. If the customer relaxes the height cap to 40 cm, recompute the best allowed height on the 0.5 cm grid. Compare h = 39.5 cm and h = 40.0 cm and state whether the continuous unconstrained height is now feasible.
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The continuous unconstrained height was about 41.6 cm, which is still above the relaxed cap of 40 cm, so the cap still binds. The grid candidates near the boundary are h = 40.0 cm and h = 39.5 cm. For h = 40.0: r = sqrt(2000/(40 pi)) ≈ 3.989 cm. Cost = 34/r + 0.014(pi r^2) + 0.6 pi r. Since pi r^2 = 2000/40 = 50, the base cost is 0.014(50) = 0.70; side cost = 34/3.989 ≈ 8.52; weld cost = 0.6 pi(3.989) ≈ 7.52; total ≈ 16.74. For h = 39.5: r = sqrt(2000/(39.5 pi)) ≈ 4.015 cm; base area = 2000/39.5 ≈ 50.63, base cost ≈ 0.709; side cost = 34/4.015 ≈ 8.47; weld cost = 0.6 pi(4.015) ≈ 7.57; total ≈ 16.75. Thus h = 40.0 cm is slightly better. The continuous unconstrained height of 41.6 cm is not feasible under the 40 cm cap.