Cheapest feasible curd cup: when an active height limit overrides the calculus optimum
A dairy cooperative packs curd in open-top cylindrical cups of capacity 500 cm³, filled to the brim so that the container volume is exactly πr²h = 500 cm³. The side wall is thin food-grade polymer sheet costing ₹0.02 per cm², while the circular base must be a thicker laminate costing ₹0.05 per cm²; there is no lid. The existing filling line imposes two hard constraints: the cup height cannot exceed 9.0 cm because of the tunnel shelf, and the diameter must be at least 6.0 cm so that the nozzle fits. A trial run at r = 5.0 cm and h = 6.37 cm cost about ₹0.39 per cup more than the best feasible design. The plant makes 1.2 million cups a year and treats every ₹0.01 saved per cup as ₹12,000 a year. Production wants the cheapest feasible geometry and a verdict on a one-time ₹40,000 tunnel modification that would raise the height limit to 10.0 cm.
1. Write the cost per cup as a function of r alone and locate the unconstrained minimum. Is that design buildable?
Read the worked answer
Cost function. Side area = 2πrh, base area = πr², so
C(r,h) = 0.02(2πrh) + 0.05(πr²).
Volume constraint πr²h = 500 gives h = 500/(πr²), hence
C(r) = 0.04πr·500/(πr²) + 0.05πr² = 20/r + 0.05πr².
Stationary point. C′(r) = −20/r² + 0.1πr = 0 → r³ = 200/π = 63.66 → r = 3.99 cm, h = 500/(π × 15.94) = 10.0 cm.
Cost: 20/3.99 = ₹5.01 for the wall, 0.05π(15.94) = ₹2.50 for the base, total ₹7.51.
Minimum check. C″(r) = 40/r³ + 0.1π > 0 for every r > 0, so C is strictly convex and this stationary point is the global minimum over r > 0.
Feasibility. h = 10.0 cm exceeds the 9.0 cm tunnel shelf, so the calculus answer cannot be moulded. The optimisation is not finished; it must be re-solved on the feasible interval.
2. Find the true feasible optimum and quantify how badly the trial run performed against it.
Read the worked answer
The height cap h ≤ 9 with πr²h = 500 becomes r ≥ √(500/(9π)) = √17.68 = 4.21 cm. The diameter rule r ≥ 3.0 cm is weaker, so the binding constraint is the height one.
C is convex with its only turning point at r = 3.99, so for r > 3.99 the cost is increasing. The feasible set is r ≥ 4.21, therefore the best feasible point is the boundary itself: r = 4.21 cm, h = 9.0 cm.
Cost: 20/4.21 = ₹4.76; 0.05π(4.21)² = ₹2.78; total ₹7.53 per cup.
Trial run check. r = 5.0, h = 6.37: wall 20/5.0 = ₹4.00, base 0.05π(25) = ₹3.93, total ₹7.93. That is ₹0.40 per cup above the feasible optimum, about ₹4.8 lakh a year at 1.2 million cups.
Diagnosis. The loss did not come from the height cap. It came from a squat, wide cup whose oversized base carries the 2.5× premium; at r = 5 the base alone costs more than the entire optimal cup.
3. Should the plant pay ₹40,000 to raise the height limit to 10.0 cm? Give the number and the decision rule.
Read the worked answer
With a 10.0 cm cap the feasible set becomes r ≥ √(500/(10π)) = 3.99 cm, which just admits the unconstrained optimum at r = 3.99, h = 10.0, C = ₹7.51.
Saving = 7.534 − 7.513 = ₹0.021 per cup → 0.021 × 1.2×10⁶ = ₹25,200 a year.
Payback = 40,000 / 25,200 ≈ 1.6 years.
Judgement. Buy only if the modified tunnel stays in service for roughly two years or more; on a one-year horizon the modification destroys value. The smallness of the gain is the real lesson: at a stationary point the cost surface is flat, so relaxing an active constraint by 1 cm on a 10 cm cup buys under 0.3%. Any claim that the height limit is the main cost driver should be rejected on these numbers.
Sanity check on the other constraint. A 10.0 cm cup at r = 3.99 has diameter 8.0 cm, so the 6.0 cm nozzle requirement still holds; no constraint swaps over.