Open-Top Box: Constrained Optimization and a Costly Intuition
A packaging engineer is designing an open-top rectangular box with a square base to hold 32 ft³ of material. The base material costs $3.00 per square foot and the four side panels cost $1.50 per square foot. The box must fit on a shelf with 3 ft of clearance, so its height cannot exceed 3 ft. A junior engineer sets the derivative of the cost function equal to zero, obtains a height of about 3.17 ft, and recommends that design. A senior engineer objects that the recommendation violates the shelf constraint and that the true minimum may lie on the boundary. The team also wants to know whether reducing the height to 2.5 ft, which 'uses less material,' would save money.
1. Write the cost as a function of the base edge x, find the unconstrained critical point, and classify it.
Read the worked answer
Constraint. Volume x²h = 32, so h = 32/x², with x > 0.
Cost. Base area is x² at $3.00; side area is 4xh at $1.50. Thus C(x) = 3x² + 6xh = 3x² + 6x(32/x²) = 3x² + 192/x.
Critical point. C'(x) = 6x − 192/x² = 0 → 6x³ = 192 → x³ = 32 → x = 32^(1/3) ≈ 3.175 ft. Then h = 32/3.175² ≈ 3.175 ft, so the unconstrained optimum is a box with h = x.
Classification. C''(x) = 6 + 384/x³ > 0 for all x > 0, so the critical point is a minimum. C(3.175) = 3(10.079) + 192/3.175 ≈ 30.24 + 60.47 = $90.71.
This design has h ≈ 3.17 ft, which violates the 3-ft shelf limit.
2. Impose the height constraint and find the true minimum cost. Compare it with the unconstrained value and explain where the optimum occurs.
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Translate the constraint. h ≤ 3 means 32/x² ≤ 3, so x² ≥ 32/3 and x ≥ 3.266 ft.
Locate the optimum. C is decreasing on (0, 3.175) and increasing on (3.175, ∞) because C'(x) has the sign of x³ − 32. The feasible interval [3.266, ∞) lies entirely in the increasing region, so the minimum is at the left endpoint x = 3.266, h = 3.
Cost. C(3.266) = 3(10.667) + 6(3.266)(3) = 32.00 + 58.79 = $90.79.
Comparison. The constrained optimum is only about $0.08 above the unconstrained value, so the cost penalty is negligible - but the junior's design is physically infeasible, and the correct answer is a boundary optimum, not the interior critical point. Reporting the interior solution would be an error even though its cost looks marginally better.
3. Evaluate the proposal to set the height to 2.5 ft. Compute its cost and area, and explain why the 'less material' reasoning fails.
Read the worked answer
At h = 2.5. x² = 32/2.5 = 12.8, so x = 3.578 ft.
- Cost = 3(12.8) + 6(3.578)(2.5) = 38.40 + 53.67 = $92.07.
- Surface area = x² + 4xh = 12.8 + 35.78 = 48.58 ft².
Compare with h = 3. Cost $90.79, area = 10.667 + 39.19 = 49.86 ft².
Why the intuition fails. The shorter box does use less total material (48.58 vs 49.86 ft²), but material composition matters as much as amount: base area rises from 10.67 to 12.80 ft² at $3.00 per ft², while side area falls only from 39.19 to 35.78 ft² at $1.50 per ft². The expensive base grows faster than the cheap sides shrink, so cost rises by about $1.28. A correct recommendation minimizes the weighted cost, not the raw area.