Cylindrical Can Cost Minimization with Thicker Ends
A food charity wants a closed cylindrical can with volume 1.2 liters (1200 cm^3). Metal for the curved side costs 0.04 currency units per cm^2, while metal for each circular end costs 0.06 per cm^2 because it is thicker. The can must have radius between 4 cm and 8 cm inclusive and height between 8 cm and 20 cm inclusive. A supplier charges a fixed setup fee of 12 units per order plus material cost. The charity expects to order 500 cans. Determine dimensions minimizing material cost per can, check constraints, and compare with a proposed can of radius 5 cm and height 15.3 cm. The proposed can is allowed to deviate from 1200 cm^3 by at most 1%.
1. Express the material cost per can as a function of radius, find the radius that minimizes it, and check the radius and height constraints.
Read the worked answer
Let r be radius and h height. Volume gives πr²h=1200, so h=1200/(πr²). The side area is 2πrh, and the two ends have total area 2πr². Thus C(r)=0.04(2πrh)+0.06(2πr²)=0.08πr·1200/(πr²)+0.12πr²=96/r+0.12πr².
Differentiate: C'(r)=−96/r²+0.24πr. Set C'(r)=0: 0.24πr³=96, so r³=400/π≈127.32 and r≈5.03 cm. Then h=1200/(π·5.03²)≈15.1 cm. These satisfy 4≤r≤8 and 8≤h≤20. Also C''(r)=192/r³+0.24π>0, so the critical point is a minimum. The minimum material cost is about 0.12π·25.30+96/5.03≈9.54+19.09=28.63.
2. Check the proposed can with radius 5 cm and height 15.3 cm. Is it within the 1% volume tolerance, and how does its material cost compare with the optimum?
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For r=5 and h=15.3, volume=π·25·15.3≈1201.7 cm³. The deviation from 1200 is 1.7 cm³, or about 1.7/1200≈0.14%, which is within the allowed 1%.
Material cost=0.04(2π·5·15.3)+0.06(2π·25)=0.04·153π+0.06·50π=6.12π+3π=9.12π≈28.65. This is only about 0.02–0.03 above the optimum 28.63, so the proposed can is feasible and nearly optimal.
3. Suppose the supplier offers thinner end metal at 0.05 per cm² instead of 0.06. Recompute the optimal radius and height, check constraints, and estimate the saving per can.
Read the worked answer
With end cost 0.05, the cost becomes C(r)=96/r+0.10πr². Differentiate: C'(r)=−96/r²+0.20πr. Setting C'(r)=0 gives 0.20πr³=96, so r³=480/π≈152.79 and r≈5.35 cm. Then h=1200/(π·5.35²)≈13.35 cm. These values are still within 4≤r≤8 and 8≤h≤20.
The new cost is about 0.10π·28.6+96/5.35≈8.98+17.95=26.93. Compared with 28.63, the saving is about 1.70 per can. Since the minimizing radius is inside the allowed interval, the constraints do not force a boundary solution.