Optimisation with a stated domain

Translate the constraint first. If a rectangle has fixed perimeter, its second side can be written in terms of the first. This gives an area function and a domain of positive side lengths. Differentiate to find stationary candidates, then check that they belong to that domain.

Finding a stationary point is only the first step. Use the derivative sign, second derivative or an explicit comparison, and consider endpoints when they belong to the domain. In an open or unbounded domain an optimum may fail to be attained. Return to the context: an excluded zero length or a negative dimension is not a usable solution.

Worked example

A rectangle has perimeter 24 cm. Find its greatest possible area.

1. If one side is x, the other is 12 − x, with 0 < x < 12.
2. A = 12x − x² and A′ = 12 − 2x, so the stationary value is x = 6.
3. A″ = −2, and A = 36 − (x − 6)² proves this is the global maximum.

Answer: 36 cm², attained by a 6 cm square

Revise first: Differentiating powers, Completing the square.