Tin Can Design: Calculus Optimum Versus the Production Line
A student team is designing a cylindrical tin for a local food producer. The tin must hold 500 cm³. The supplier quotes Rs 0.02 per cm² for the side wall and Rs 0.04 per cm² for the lacquered top and bottom discs, so the ends cost twice as much per unit area as the side. The team's first design is r = 4.0 cm, h = 9.95 cm, chosen because it looks like a standard can. The production engineer can form any radius, but the filling line's shelf bracket accepts a maximum diameter of 8.0 cm, and the available tooling gives radii only in 0.5 cm steps starting at 2.5 cm. A teammate argues the mathematical optimum is r = 3.4 cm, h = 13.7 cm, but marketing wants a container that is stable on the shelf. The team must recommend a radius from the available set r = 3.0, 3.5 or 4.0 cm and justify the cost penalty rather than simply quoting the calculus answer.
1. Derive the cost function C(r) for fixed volume 500 cm³, find the unconstrained optimum, and show that h = 4r there.
Read the worked answer
Volume gives h = 500/(πr²). Side area = 2πrh, end area = 2πr². Cost C = 0.02(2πrh) + 0.04(2πr²) = 0.04πrh + 0.08πr². Substituting h: C = 0.04πr × 500/(πr²) + 0.08πr² = 20/r + 0.08πr². Differentiate: dC/dr = −20/r² + 0.16πr. Setting zero gives 0.16πr³ = 20, so r³ = 20/(0.16π) = 39.79 and r = 3.414 cm. Then h = 500/(π × 11.66) = 13.66 cm, so h/r = 4.00. The second derivative is positive, so this is a minimum. The result makes sense: because the ends are the expensive surface, the optimum is taller and narrower than the equal-cost case h = 2r.
2. Compute the cost at r = 3.0, 3.5 and 4.0 cm, choose one, and quantify the penalty relative to the true optimum.
Read the worked answer
Use C = 20/r + 0.08πr². At r = 3.0: 6.667 + 2.262 = Rs 8.929. At r = 3.5: 5.714 + 3.079 = Rs 8.793. At r = 4.0: 5.000 + 4.021 = Rs 9.021. The unconstrained optimum at r = 3.414 costs about Rs 8.787. So r = 3.5 cm is only about Rs 0.006 per tin (0.07%) above the optimum, while r = 4.0 cm is about Rs 0.234 (2.7%) above. Choose r = 3.5 cm: it is effectively optimal, fits the 8.0 cm bracket with a 7.0 cm diameter, and is more stable than the tall 3.4 cm design. The 4.0 cm can is acceptable only if the shelf bracket and marketing stability dominate, because its cost penalty is real but modest.
3. Suppose the lacquer surcharge doubles, so the ends cost Rs 0.08 per cm² while the side remains Rs 0.02. How does the optimum change, and what does this show about the design rule?
Read the worked answer
Now C = 0.02(2πrh) + 0.08(2πr²) = 0.04πrh + 0.16πr². Substituting h gives C = 20/r + 0.16πr². dC/dr = −20/r² + 0.32πr = 0, so r³ = 20/(0.32π) = 19.89 and r = 2.71 cm. Then h = 500/(π × 7.34) = 21.7 cm, and h/r = 8.0. The can becomes much taller and narrower because the expensive ends are reduced. This shows there is no universal 'best can shape'; the optimum depends on the relative cost of side and end surfaces. The team should re-optimize whenever the supplier's price ratio changes, then snap to the nearest tooling size and check stability and handling.