Open-Top Can: Constrained Volume Maximisation
A small metalwork unit is asked to fabricate open-top cylindrical cans from a single sheet of metal with usable area 600 cm². The base and curved side are cut from this sheet; overlap and cutting waste are ignored. The workshop's roller can handle radii only from 2 cm to 6 cm inclusive. For a can of radius r cm and height h cm, the metal area is πr² + 2πrh, and the volume is πr²h. The supervisor writes the area constraint as h = (600 - πr²)/(2πr), then proposes to maximise V(r) = πr²h on the allowed interval. A worker objects that the formula gives a critical radius near 8 cm, outside the roller range, so 'the maximum must be at the critical point' cannot be right. Before cutting, the unit must choose r, report the resulting h and volume, and decide whether the area constraint or the roller limit is controlling the design. A second order for a narrower can with r = 4 cm is also under discussion.
1. Write V as a function of r alone, state its domain, and compute V at the two endpoints r = 2 and r = 6.
Read the worked answer
Substitute h into V:
V(r) = πr² · (600 - πr²)/(2πr) = r(600 - πr²)/2 = 300r - (π/2)r³.
The roller restriction gives the domain 2 ≤ r ≤ 6.
At r = 2: h = (600 - 4π)/(4π) = 150/π - 1 ≈ 46.74 cm, V = 300(2) - (π/2)(8) = 600 - 4π ≈ 587.4 cm³.
At r = 6: h = (600 - 36π)/(12π) = 50/π - 3 ≈ 12.92 cm, V = 300(6) - (π/2)(216) = 1800 - 108π ≈ 1460.7 cm³.
So the endpoint volumes are about 587 cm³ and 1461 cm³. The formula is valid throughout the interval because h remains positive.
2. Differentiate V, locate the critical point, and use the sign of V' to determine the maximum on [2, 6]. Does the worker's objection change the answer?
Read the worked answer
Differentiate:
V'(r) = 300 - (3π/2)r².
Set V'(r) = 0:
(3π/2)r² = 300 ⇒ r² = 200/π ⇒ r = sqrt(200/π) ≈ 7.98 cm.
This critical point is outside [2, 6]. On the allowed interval, r² ≤ 36, so (3π/2)r² ≤ 54π ≈ 169.6 < 300. Hence V'(r) > 0 throughout [2, 6].
A continuous function increasing on a closed interval attains its maximum at the right endpoint. Therefore r = 6 cm, h ≈ 12.92 cm, V ≈ 1460.7 cm³.
The worker is right that the critical point cannot be used directly, but that does not mean there is no maximum. The maximum is controlled by the roller limit, not by the stationary point.
3. For r = 4, compute h and V, compare with r = 6, and explain what this shows about the phrase 'a larger radius always gives a larger volume'.
Read the worked answer
At r = 4:
h = (600 - 16π)/(8π) = 75/π - 2 ≈ 21.87 cm, V = 300(4) - (π/2)(64) = 1200 - 32π ≈ 1099.5 cm³.
This is less than the r = 6 volume of about 1460.7 cm³, so over the permitted range increasing r increases V. That is consistent with V'(r) > 0 on [2, 6].
However, 'larger radius always gives larger volume' is not true globally: the unconstrained critical point at r ≈ 7.98 cm is a maximum, and for r beyond that V decreases. The correct statement is local to the allowed interval and follows from the sign of V', not from intuition about radius alone.