Key Idea: Optimisation is the real-world application of stationary points: finding the maximum profit, minimum cost, shortest time, or greatest area given a constraint. The technique is identical to Topic 5.6 — differentiate and set f'(x) = 0 — but now you must first build the function from a word problem, and interpret your answer with units in context.
✅ The 5-step optimisation method
Example: A farmer has 100 m of fence to enclose a rectangular field with one side along a wall (no fence needed there). Maximise the area. Let x = width of the field. Then length = 100 − 2x. Area A = x(100 − 2x) = 100x − 2x² A'(x) = 100 − 4x = 0 → x = 25 m A''(x) = −4 < 0 → this is a maximum ✓ Length = 100 − 50 = 50 m. Maximum area = 25 × 50 = 1250 m²
Always check the domain after finding the critical x. An x that gives a negative length or area is outside the valid domain. If f''(x) = 0 at the critical point, test values either side of x to confirm max or min.
Paper 1 and 2: Show every step — setting up the function, differentiating, setting f'(x) = 0, verifying nature, and concluding in context. Marks are awarded at each stage. A correct final answer with no working earns few marks in extended questions. Context is essential: Don't just say 'x = 25' — say 'the width should be 25 m to achieve the maximum area of 1250 m²'.
IB-style question [5 marks]
A manufacturer's profit, in dollars, from selling x items is modelled by P(x) = 60x − x² − 200. Find the number of items that maximises the profit, and the maximum profit.
Step by step:
Maximum profit occurs where the derivative is zero.
Substitute x = 30 to find the maximum profit.
Final answer:
30 items gives the maximum profit of $700.