Can design: an optimum that is not allowed
A drinks company fills 500 cm³ cans on a line whose labeller cannot handle a can with height less than three times its radius, so the design is constrained by h ≥ 3r. The engineer's brief is to minimise the sheet metal used, taken as the total surface area of a closed cylinder, S = 2πr² + 2πrh, at fixed volume V = πr²h = 500 cm³. A junior engineer computes the unconstrained optimum, then argues that because the labeller rule is 'only a guideline' it can be ignored and that design built anyway. The production manager objects that the machine is a hard limit. The engineer also notes that the dies for the two ends cut circular blanks of radius r plus 4 mm of trim, and that the body seam is welded along its full height.
1. Find the unconstrained radius and height that minimise surface area, and show whether this design satisfies the labeller constraint.
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Eliminate h using h = 500/(πr²):
S(r) = 2πr² + 2πr · 500/(πr²) = 2πr² + 1000/r.
dS/dr = 4πr − 1000/r² = 0 → r³ = 1000/(4π) = 79.58 → r = 4.30 cm, and h = 500/(π × 18.50) = 8.60 cm.
Check the second derivative: d²S/dr² = 4π + 2000/r³ > 0 for all r > 0, so this stationary point is the unique global minimum. Also h = 2r, the classic result for a closed cylinder.
Now apply the constraint: h/r = 8.60/4.30 = 2.0, but the labeller requires h ≥ 3r. The unconstrained optimum lies outside the feasible set (r must satisfy h ≥ 3r, i.e. r ≤ (500/3π)^(1/3) = 3.76 cm).
So the junior engineer's argument is invalid in a way that matters: the mathematical optimum is real but infeasible, and manufacturing it would stop the line. The correct problem is a constrained minimisation, and the constrained optimum lies on the boundary h = 3r.
2. Find the constrained optimum and quantify the cost of the constraint in surface area.
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Inside the feasible set the unconstrained minimiser is excluded, so the minimum is on the boundary h = 3r. Substitute:
V = πr²(3r) = 3πr³ = 500 → r³ = 500/(3π) = 53.05 → r = 3.76 cm, h = 11.27 cm.
S = 2πr² + 2πrh = 2π(3.757)² + 2π(3.757)(11.27) = 88.7 + 266.1 = 354.8 cm².
Unconstrained minimum for comparison: S = 2π(4.301)² + 1000/4.301 = 116.2 + 232.5 = 348.7 cm².
Penalty = 6.1 cm² per can, i.e. about 1.7% more metal.
Judgement: 1.7% is small. It is almost certainly cheaper than a new labeller or a redesign of the filling line, so the right decision is to accept the constraint and specify r = 3.76 cm, h = 11.27 cm. It is also worth stating the rounding: rounding r to 3.8 cm with V held at 500 cm³ changes S by well under 0.1%, so the design should be rounded to drawable dimensions and the volume checked afterwards rather than the surface area re-optimised.
3. The engineer wants to challenge the objective itself. Using the numbers, argue how much weight the 'minimise surface area' objective deserves.
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The surface-area function is extremely flat near its minimum, so the objective has weak discriminating power. Holding V = 500 cm³ and moving the radius ±10% from 4.30 cm gives S = 352.5 cm² at r = 3.87 cm and 352.0 cm² at r = 4.73 cm - both within 1.1% of the 348.7 cm² optimum. In other words, a can that is 10% 'wrong' in radius costs about 1% in metal. Practical criteria - stability on the line, palletising efficiency, label wrap, handling - should therefore dominate the choice, with surface area used only to reject grossly bad shapes.
Surface area is also only a proxy for cost. The ends are blanked from circles of radius r + 4 mm, so the trim scrap scales with r and penalises large radii more than the S formula suggests; the body seam is welded along its full height, so seam time scales with h and penalises tall cans; and very slender cans need thicker wall stock to resist buckling. A better objective would be material cost plus scrap plus seam time, evaluated at two or three candidate shapes rather than optimised analytically. The correct professional move is to state the proxy, quantify its flatness, and then bring in the omitted factors explicitly.