Tin Can with a Lid-Stock Limit
A small tin factory has an order for closed cylindrical cans of volume 800 cm³. The only lid-making die in the workshop accepts diameters up to 8.0 cm, and a new die is too expensive. The labelling machine can handle can heights from 8 cm to 17 cm; outside that range labels wrinkle or the conveyor jams. Body sheet costs ₹0.02 per cm², while the circular top and bottom sheet costs ₹0.035 per cm² because it must be coated. The manager first ignores the die and machine limits and uses calculus to minimise material cost, obtaining a radius of about 4.17 cm. A supervisor objects that this radius cannot be used. The production team must decide the actual radius and height, and explain whether the cost difference from the unconstrained answer is worth arguing about. They also want a check that the chosen can fits both the die and the labelling machine.
1. Write the cost C as a function of r alone for volume 800 cm³, and find the unconstrained stationary radius. Classify the stationary point.
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With h = 800/(πr²), the body area is 2πrh and the two ends have total area 2πr². Thus
C(r) = 0.02(2πrh) + 0.035(2πr²) = 0.04πr(800/(πr²)) + 0.07πr² = 32/r + 0.07πr².
C′(r) = −32/r² + 0.14πr. Setting C′ = 0 gives r³ = 32/(0.14π) ≈ 72.76, so r ≈ 4.174 cm and h ≈ 800/(π(4.174)²) ≈ 14.61 cm.
C″(r) = 64/r³ + 0.14π > 0, so this is a minimum of the unconstrained model.
2. Apply the die and labelling-machine constraints. What radius and height should be used? Compute the cost at this choice and compare it with the unconstrained cost.
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The lid die requires 2r ≤ 8, so r ≤ 4. The labelling machine requires 8 ≤ h ≤ 17. With h = 800/(πr²), h ≤ 17 gives r ≥ sqrt(800/(17π)) ≈ 3.87, and h ≥ 8 gives r ≤ sqrt(800/(8π)) ≈ 5.64. The feasible radii are therefore about 3.87 ≤ r ≤ 4.00.
On (0, 4.174) the derivative is negative, so C decreases all the way to the unconstrained point. Over the feasible interval the minimum is at the largest allowed radius, r = 4.00 cm. Then h = 800/(16π) ≈ 15.92 cm, which is within 8–17 cm.
C(4) = 32/4 + 0.07π(16) = 8 + 3.52 = 11.52 rupees per can. The unconstrained C(4.174) ≈ 7.67 + 3.83 = 11.50, so the constraint costs only about ₹0.02 per can.
3. A supplier offers a 10 cm die. Would that make the unconstrained radius feasible? Would it reduce cost materially? What limit would then matter?
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A 10 cm die allows r ≤ 5 cm, so the unconstrained r ≈ 4.17 cm and h ≈ 14.61 cm become feasible; both are inside the labelling limits. The cost is essentially the unconstrained minimum, about ₹11.50 versus ₹11.52 per can.
The saving is ₹0.02 per can, so on 100,000 cans it is about ₹2,000, almost certainly less than the cost of a new die and its installation. Buying the die is not justified by material saving alone. The binding limit would no longer be the die; the height range would remain, but at r ≈ 4.17 the height 14.61 is comfortably inside 8–17, so the current 8 cm limit is the one that actually changes the design.