Minimising Metal for an Open Cylindrical Tank
A metal workshop in Douala has been asked to make an open-top cylindrical tank with a capacity of 500 litres. The tank will be cut and welded from a single rectangular sheet of metal with area 3.5 m². The workshop manager wants to use the least possible amount of metal, but the sheet is already bought, so the practical question is whether the sheet is large enough and what radius and height should be chosen. The tank has no lid. Its volume is 500 dm³, and its surface area consists of a circular base and a curved side. The manager suspects that a tall narrow tank uses less metal than a short wide one, while an apprentice argues the opposite. The workshop works in decimetres because 1 litre = 1 dm³ and the sheet area is 350 dm². The manager asks for a calculus-based recommendation, including a check that the stationary value is a minimum, not a maximum, and a clear statement about leftover metal.
1. Express the surface area S of the open tank as a function of base radius r, and state the domain.
Read the worked answer
Let the base radius be r dm and the height be h dm. The volume is V = πr²h = 500, so h = 500/(πr²).
The open tank has a circular base of area πr² and a curved side of area 2πrh. Therefore S = πr² + 2πrh = πr² + 2πr(500/(πr²)) = πr² + 1000/r.
The radius must be positive, so the domain is r > 0. The surface area is in dm² because r and h are in dm.
2. Find the radius and height that minimise S, and justify that this is a minimum.
Read the worked answer
Differentiate: S(r) = πr² + 1000/r S′(r) = 2πr − 1000/r².
Set S′(r) = 0: 2πr = 1000/r² 2πr³ = 1000 r³ = 500/π r = (500/π)^(1/3) ≈ 5.42 dm.
Then h = 500/(πr²) = 500/(π × 5.42²) ≈ 5.42 dm.
So the minimum occurs when h = r. To justify a minimum, S″(r) = 2π + 2000/r³ > 0 for r > 0, so the stationary point is a minimum. The minimum surface area is S_min = πr² + 1000/r = πr² + 2πr² = 3πr² ≈ 3π(5.42²) ≈ 276.8 dm².
3. Decide whether the available 350 dm² sheet is sufficient, and comment on the apprentice's claim that a short wide tank always uses less metal.
Read the worked answer
The minimum required area is about 276.8 dm², which is less than the available 350 dm², so the sheet is sufficient. The leftover is about 350 − 276.8 = 73.2 dm², allowing for seams and waste.
The apprentice's claim is too simple. The minimum occurs at r = h ≈ 5.42 dm, not at an extreme shape. For example, r = 8 dm gives h = 500/(π × 64) ≈ 2.49 dm and S ≈ π(64) + 1000/8 ≈ 201.1 + 125 = 326.1 dm². This is still under 350 dm² but larger than the minimum. A very wide tank, such as r = 10 dm, gives h ≈ 1.59 dm and S ≈ π(100) + 100 = 414.2 dm², which exceeds the sheet area. So some short wide tanks use more metal, not less; the calculus identifies the true optimum.