Geometric optimisation

Objective. Use constraints and graphs to compare possible designs.

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Learn: the key idea

Geometric optimisation compares several possible shapes or designs under a shared constraint, such as fixed perimeter or fixed material, to find which gives the best area or volume. Set up an expression for the quantity you want to maximise or minimise in terms of one variable, using the constraint to eliminate the other variable. Sketching a graph of this expression, or comparing values at sensible test points, reveals the optimum. For a rectangle with perimeter 20 m, area = x(10 − x), which is maximised at x = 5, giving a square of area 25 m². Always check the answer makes physical sense within the constraint.

Worked example

A rectangular pen has perimeter 24 m. Write an expression for its area A in terms of width x, and state the maximum possible area.

1 $Length = 12 - x, so A = x(12 - x)$
2 $A is maximised at the midpoint x = 6$
3 $A = 6 \times (12 - 6) = 6 \times 6$
4 $Maximum area = 36 m²$

Answer: 36 m²

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Common mistake. Students often optimise without applying the constraint, leaving two free variables and comparing arbitrary values rather than a properly reduced one-variable function, which gives an unjustified answer.

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