Cylindrical can: cutting labour with a kink at r = 5 cm
A metal-fabrication unit at a polytechnic makes open-top cylindrical oil cans of fixed volume 64π cm³ (about 201 cm³). Sheet metal is the only material cost and is priced at ₹0.50 per cm². The unit's contract with its cutter fixes a labour charge of ₹20 per can for any base radius up to and including 5 cm; beyond 5 cm the cutter must change blades mid-run, and labour becomes 20 + 6(r − 5) rupees, where r is the base radius in cm. The foreman currently sets r = 5 cm, arguing that “a wider can holds the same oil with less metal”. The purchase officer wants the radius that minimises total cost per can, and also wants to know whether the total-cost function is smooth at r = 5, because the accounts software estimates marginal cost with a numerical derivative at whichever radius the foreman enters.
1. Ignoring labour for the moment, express the metal cost per can as a function of r and find the radius that minimises it. State the corresponding height.
Read the worked answer
Step 1 - height from the volume constraint.
V = πr²h = 64π ⇒ h = 64/r².
Step 2 - surface area (open top: one circular base plus the curved side).
S(r) = πr² + 2πrh = πr² + 2πr(64/r²) = πr² + 128π/r.
Step 3 - cost.
M(r) = 0.50·S(r) = 0.5π(r² + 128/r). M′(r) = 0.5π(2r − 128/r²) = π(r − 64/r²).
Set M′(r) = 0: r³ = 64 ⇒ r = 4 cm.
Step 4 - confirm a minimum.
M″(r) = π(1 + 128/r³) > 0 for all r > 0, so r = 4 gives the minimum on (0, ∞).
Step 5 - values. h = 64/16 = 4 cm, so the cheapest can is the one with height equal to radius. M(4) = 0.5π(16 + 32) = 24π ≈ ₹75.40, against M(5) = 0.5π(25 + 25.6) = 25.3π ≈ ₹79.48. The foreman's instinct is wrong in this range: going from r = 4 to r = 5 adds about ₹4.08 of metal, because the growing base outweighs the shorter side.
2. Test the labour charge L(r) for continuity and differentiability at r = 5. What does your conclusion mean for the software that computes a numerical derivative at r = 5?
Read the worked answer
Step 1 - the piecewise definition.
L(r) = 20 for 0 < r ≤ 5; L(r) = 20 + 6(r − 5) for r > 5.
Step 2 - continuity at 5.
L(5) = 20. Left limit: lim(r→5⁻) 20 = 20. Right limit: 20 + 6(5 − 5) = 20. All three agree, so L is continuous at r = 5 - the contract has no jump in price.
Step 3 - differentiability at 5.
Left derivative: d/dr(20) = 0. Right derivative: d/dr[20 + 6(r − 5)] = 6.
0 ≠ 6, so L′(5) does not exist. Geometrically the graph has a corner at r = 5: it leaves the point flat and then rises with slope 6.
Step 4 - consequence for the software. A numerical derivative at r = 5 uses values on both sides of 5 (or a one-sided choice), so it will return something between 0 and 6 depending on the step size - a meaningless number for a marginal cost. The officer should either evaluate marginal cost at a radius away from 5 or replace the corner with a smooth (differentiable) labour formula.
3. Now minimise the total cost T(r) = M(r) + L(r) over all r > 0. Does r = 4 survive the labour step, and how much per can is the foreman's r = 5 costing?
Read the worked answer
Step 1 - split at the kink, because the formula changes there.
Piece (0, 5]: T(r) = 0.5π(r² + 128/r) + 20. The labour part is constant, so the minimum of M on this piece applies. M′(r) = π(r − 64/r²) is negative for r < 4 and positive for r > 4, so T decreases to r = 4 and then increases up to r = 5. T(4) = 24π + 20 ≈ 75.40 + 20 = ₹95.40.
Piece (5, ∞): T′(r) = π(r − 64/r²) + 6. At r = 5 this is π(5 − 2.56) + 6 ≈ 7.67 + 6 = 13.67 > 0, and both terms increase with r, so T is increasing throughout (5, ∞). Its smallest value there is approached as r → 5⁺: T(5) = 25.3π + 20 ≈ ₹99.48.
Step 2 - compare. ₹95.40 at r = 4 beats every radius above 5, so the global minimum is r = 4 cm, h = 4 cm, total cost ≈ ₹95.40 per can.
Step 3 - answer the foreman. His r = 5 costs about 99.48 − 95.40 = ₹4.08 more per can, and the corner at r = 5 does not rescue it. Note the check that mattered: because the labour rule changes at 5, the boundary had to be evaluated separately - a stationary point alone would not have settled the question.