Minimizing Cost of Cylindrical Can

Calculus 1 · Axiom Academy

EXAMPLE Minimizing Cost of a Cylindrical Can Optimization with a constraint: cheapest material for a can of fixed volume A closed cylindrical can must hold . The metal for the top and bottom costs 0.02 per cm ^2 , and the metal for the curved side costs 0.01 per cm ^2 . Find the radius r and height h that minimize the total cost of the metal. A taller, thinner can has more (cheap) side and less (expensive) lid — the trade-off we are optimizing. Nice work — you optimized a real manufacturing problem end to end. The moves that made it work: Two pieces of surface: a cylinder has two circular ends (combined area ) and a curved side (area ). Cost-weight each area: when materials cost different amounts, multiply each area by its own price before adding. Kill the extra variable: the volume constraint lets you replace h with , leaving a one-variable function C(r) . Optimize: gives ; setting it to zero yields . Result: and — and notice h=4r exactly at the optimum, with confirming it is a minimum. This "minimize material for a fixed volume" pattern is the heart of packaging and engineering design.

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