Can Design with Unequal End Costs
A food company wants to design a cylindrical can that holds 16 fluid ounces (28.875 cubic inches). The side wall material costs $0.002 per square inch, while the top and bottom material costs $0.004 per square inch because it must be thicker. The can must have a diameter between 2.0 and 4.0 inches to fit existing filling equipment, and the height must not exceed 6.0 inches. A designer proposes making the can with height equal to twice the radius, citing the 'standard' result that minimizes surface area for a fixed volume. The production manager wants the lowest material cost per can and asks whether the designer's proportions are optimal under these costs and constraints. The company also wants to know how much the material cost changes if the top/bottom material cost were equal to the side cost.
1. Set up the material cost function C(r) using the volume constraint. Find the unconstrained critical radius and height, and compare with the designer's h=2r proposal.
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V=πr^2h=28.875, so h=V/(πr^2). Side area=2πrh=2V/r. End area (two disks)=2πr^2. Cost C(r)=0.002(2V/r)+0.004(2πr^2)=0.004V/r+0.008πr^2. Differentiate: C'(r)=−0.004V/r^2+0.016πr. Set to zero: 0.016πr^3=0.004V, so r^3=V/(4π)=28.875/(12.566)=2.298, r≈1.320 in. Then h=28.875/(π·1.320^2)=28.875/5.474≈5.28 in. The ratio h/r≈4.00. The designer's h=2r gives r^3=V/(2π)=4.596, r≈1.662 in, h≈3.32 in. At the optimum, C≈0.004(28.875)/1.320+0.008π(1.320)^2≈0.0875+0.0438=$0.1313; at h=2r, C≈0.0695+0.0694=$0.1389, about 5.5% more.
2. Check the feasible interval endpoints and confirm whether the critical point is the global minimum. Identify any binding constraints.
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Constraints: diameter 2–4 in means r∈[1,2]; height≤6 in means V/(πr^2)≤6, so r^2≥28.875/(6π)=1.532, r≥1.238. Feasible r∈[1.238,2]. The critical r=1.320 is inside. C''(r)=0.008V/r^3+0.016π>0 for r>0, so C is convex and the critical point is the unique global minimum on the interval. Check endpoints: at r=1.238, C≈0.0933+0.0385=$0.1318; at r=2, C≈0.0578+0.1005=$0.1583. Both exceed $0.1313. At the optimum h≈5.28 in<6 in, so the height limit is not binding; the diameter bounds are not binding. Only the volume constraint is active.
3. If top/bottom material cost equals side cost, derive the new optimal h/r ratio. Explain in one sentence why the ratio changes.
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With equal costs c, C=c(2V/r+2πr^2). C'=c(−2V/r^2+4πr)=0 gives 4πr^3=2V, so r^3=V/(2π). Since V=πr^2h, r^3=πr^2h/(2π)=r^2h/2, so r=h/2, or h/r=2. The ratio changes because cheaper ends make the optimal design shorter and wider: when ends cost more, the optimum is taller with smaller end area (h/r≈4); when ends and sides cost the same, the classic h=2r balances side and end material.