Short answer: write the quantity you want to maximise or minimise as a function of one variable, use the constraint to remove the other variables, set the derivative equal to zero, then check that your critical point really is a maximum or minimum (and check the endpoints).
The 5-step method
- Draw and label. Name every changing quantity.
- Write the objective. What are you maximising or minimising? (area, cost, volume, distance)
- Use the constraint. Solve it for one variable and substitute, so the objective has a single variable.
- Differentiate and solve f′(x) = 0. Note the allowed domain (lengths can't be negative).
- Justify. Use the second derivative test, a sign chart, or compare values at critical points and endpoints.
Worked example: the fence problem
You have 100 m of fence to enclose a rectangle. What dimensions give the largest area?
- Objective: A = x·y.
- Constraint: 2x + 2y = 100, so y = 50 − x.
- Substitute: A(x) = x(50 − x) = 50x − x², with 0 ≤ x ≤ 50.
- A′(x) = 50 − 2x = 0, so x = 25.
- A″(x) = −2 < 0, so this is a maximum. y = 25.
Answer: a 25 m × 25 m square, area 625 m². The endpoints x = 0 and x = 50 give area 0, which confirms it.
Mistakes that cost marks
- Differentiating before reducing to one variable.
- Forgetting to justify that the point is a maximum or minimum. Most mark schemes give a mark for this.
- Ignoring endpoints on a closed interval, where the true extreme value is sometimes found.
- Answering with x only, when the question asks for the dimensions or the maximum value.
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