Minimizing Metal Cost for a Cylindrical Container
A pharmaceutical company needs a closed cylindrical container of volume 1000 cm³. The metal for the curved side costs ₹0.02 per cm², while the metal for each flat circular end costs ₹0.04 per cm² because it must be thicker. The supplier quotes a standard container with radius 5.00 cm and height 12.73 cm, claiming it is 'nearly optimal.' The production engineer must verify the claim and also respect a machine constraint: the finished height must lie between 10.0 cm and 15.0 cm inclusive. The container is a right circular cylinder, the metal thickness is neglected in the volume calculation, and no overlap or waste is considered. The company wants the minimum material cost per container that satisfies the height constraint.
1. Write the material cost as a function of radius and find the unconstrained radius and height that minimize cost.
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Let r be radius (cm) and h height (cm). Volume: πr²h = 1000, so h = 1000/(πr²). Cost: side area 2πrh at ₹0.02/cm² gives 0.04πrh; two ends area 2πr² at ₹0.04/cm² gives 0.08πr². Thus C(r) = 0.04πr[1000/(πr²)] + 0.08πr² = 40/r + 0.08πr². Differentiate: C'(r) = −40/r² + 0.16πr. Set C' = 0: 0.16πr³ = 40, r³ = 250/π = 79.58, r = 4.30 cm. Then h = 1000/(π×4.30²) = 17.2 cm. C = 40/4.30 + 0.08π(4.30²) = 9.30 + 4.65 = ₹13.95. Since C'' = 80/r³ + 0.16π > 0, this is a minimum. So the unconstrained optimum is r ≈ 4.30 cm, h ≈ 17.2 cm.
2. Evaluate the supplier's claim for r = 5.00 cm and h = 12.73 cm. Is the container feasible, and how far is its cost from the true unconstrained minimum?
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Supplier: r = 5.00 cm, h = 12.73 cm. Check volume: π(5.00²)(12.73) = π×25×12.73 = 999.8 cm³, essentially 1000 cm³. Cost = 40/5.00 + 0.08π(25) = 8.00 + 6.28 = ₹14.28. Compared with ₹13.95, it is ₹0.33 more expensive, about 2.4% above the true minimum. So 'nearly optimal' is fair in cost terms, but the claim of optimality is false. The supplier's height satisfies the machine limit (10–15 cm), so it is feasible, but it is not the minimum-cost design.
3. Now apply the machine constraint that height must be between 10.0 cm and 15.0 cm. Find the constrained optimum and compare it with the supplier's container.
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The unconstrained height 17.2 cm exceeds the 15.0 cm limit, so the constraint binds. For fixed volume, h decreases as r increases. The admissible range is where 10 ≤ 1000/(πr²) ≤ 15, i.e. r between sqrt(1000/(15π)) = 4.61 cm and sqrt(1000/(10π)) = 5.64 cm. Because C(r) is increasing for r > 4.30 cm, the lowest cost in this interval is at the smallest allowed r, r = 4.61 cm, which corresponds to h = 15.0 cm. Cost = 40/4.61 + 0.08π(4.61²) = 8.68 + 5.34 = ₹14.02. At the other endpoint r = 5.64 cm, h = 10.0 cm and cost = 40/5.64 + 0.08π(31.8) = 7.09 + 7.99 = ₹15.08. Therefore the constrained optimum is r ≈ 4.61 cm, h = 15.0 cm, at about ₹14.02; the supplier's can is feasible but ₹0.26 costlier.