← Calculus methods

Optimisation: maximising and minimising

Always the hardest part of at least one question, every year — a real design or space constraint, reduced to one variable, then optimised with calculus.

The four-step method

  • 1. Write the quantity to optimise (area, volume, surface area) as a formula — it will usually involve TWO variables at first.
  • 2. Use the given CONSTRAINT (a fixed perimeter, a fixed volume, a fixed length of material) to write one variable in terms of the other.
  • 3. Substitute to get the quantity as a function of ONE variable, then differentiate and set the derivative to zero.
  • 4. Solve for that variable, then find the other one and the optimised quantity — and PROVE it is a maximum or minimum (see the next section).
Worked example Maximising an area under a material constraint A logo shape has a total perimeter of 80 cm, made of two straight sides (each length h) and two semicircular ends (each radius r), where the perimeter is 2h + 2πr = 80. Find the values of h and r that maximise the enclosed area A = 2rh + πr². From the constraint: 2h + 2πr = 80, so h = 40 − πr. Substitute into the area formula: A = 2r(40 − πr) + πr² = 80r − 2πr² + πr² = 80r − πr². Differentiate: dA/dr = 80 − 2πr. Set dA/dr = 0: 80 − 2πr = 0, so r = 40/π ≈ 12.73 cm. Find h: h = 40 − π(12.73) ≈ 0.02 cm (essentially zero — the perimeter is almost entirely used by the semicircular ends at the maximum). Substitute back to find the maximum area: A = 80(12.73) − π(12.73)² ≈ 509.3 cm². ✓ r ≈ 12.7 cm, and the maximum area ≈ 509 cm² — note how the constraint equation is what lets a two-variable problem become a one-variable calculus problem at all. ✕

Forgetting to eliminate a variable before differentiating

Differentiating a formula that still contains TWO variables (e.g. A = 2rh + πr² with both r and h present) produces a meaningless partial expression — calculus in this standard is always single-variable. The constraint equation MUST be used to substitute one variable out completely before dA/dr = 0 means anything. If your derivative still has two letters representing lengths or quantities in it, you have skipped the elimination step.

Practise just this