Cheapest Curd Can: When the Ends Cost More Than the Side
A small dairy packs 500 cm³ of curd into a closed right circular cylinder. The metal for the curved side costs ₹0.04 per cm², while the two circular ends use a thicker gauge costing ₹0.06 per cm². The curved surface area is 2πrh and the two ends together have area 2πr². The production manager wants the cheapest can that still holds exactly 500 cm³, and the label printer needs the final radius and height. Let r cm be the base radius and h cm the height; the volume constraint is πr²h = 500. A trainee proposes r = 4 cm because it gives a convenient height of about 9.95 cm, but no optimisation has been done. The manager also asks whether switching to a single sheet of uniform metal at ₹0.05 per cm² would change the best shape; the supplier can cut either design. You are asked to model the cost, locate the minimum, and interpret the result for the production team.
1. Write the total metal cost C(r) as a function of r only, and find the positive value of r that makes dC/dr = 0.
Read the worked answer
Let the side cost be a = ₹0.04/cm² and the end cost be b = ₹0.06/cm². From πr²h = 500, h = 500/(πr²). The total cost is C(r) = 0.04(2πrh) + 0.06(2πr²). Substitute h:
C(r) = 0.08πr·500/(πr²) + 0.12πr² = 40/r + 0.12πr², r > 0.
Differentiate: dC/dr = -40/r² + 0.24πr. Setting dC/dr = 0 gives 0.24πr³ = 40, so r³ = 40/(0.24π) = 500/(3π) ≈ 53.05. Hence r ≈ 3.76 cm.
2. Show that this critical point gives the minimum cost, and report r, h and h/r to two decimal places. How does the trainee's r = 4 cm compare?
Read the worked answer
C''(r) = 80/r³ + 0.24π, which is positive for every r > 0, so the critical point is a minimum. At r = 3.76 cm,
h = 500/(π×3.76²) ≈ 500/44.41 ≈ 11.26 cm,
so h/r ≈ 3.00. The exact relation is h = 3r: for C = 2πa r h + 2πb r² with h = V/(πr²), the first-order condition gives h = 2br/a = 2(0.06)r/0.04 = 3r.
The trainee's r = 4 cm gives h ≈ 9.95 cm and C(4) = 10 + 0.12π(16) ≈ ₹16.03, while C(3.76) ≈ 40/3.76 + 0.12π(3.76)² ≈ 10.64 + 5.33 = ₹15.97. The trainee is close but not optimal; because the ends are the expensive part, the true optimum is taller and narrower.
3. For the uniform ₹0.05/cm² sheet, find the new optimal r and h, and explain why the optimal height-to-radius ratio changes.
Read the worked answer
With uniform cost c = ₹0.05/cm²,
C(r) = 0.05(2πrh + 2πr²) = 0.1πr h + 0.1πr².
Substitute h = 500/(πr²): C(r) = 50/r + 0.1πr². Then dC/dr = -50/r² + 0.2πr = 0, so r³ = 50/(0.2π) = 250/π ≈ 79.58, giving r ≈ 4.30 cm. Then h = 500/(π×4.30²) ≈ 8.61 cm, and h/r ≈ 2.00.
The ratio changes from 3 to 2 because now every square centimetre costs the same, so the problem is simply to minimise total surface area; the standard closed-cylinder condition h = 2r applies. When ends are more expensive, the optimum shrinks the end area by making the can taller and narrower.