Core idea: Optimisation means finding the maximum or minimum value of a quantity.
Examples: • Maximum profit • Minimum cost • Maximum area • Minimum material used
The method: the optimal value always occurs at a stationary point of the model function.
[Diagram: math-stationary-points] - Available in full study mode
Every optimisation problem gives you a formula (or you build one), and asks you to find the input value that makes the output as large or as small as possible.
| Goal | What you find | IB command terms |
|---|---|---|
| Maximum | Largest value of the function | maximise, greatest, most |
| Minimum | Smallest value of the function | minimise, least, cheapest |
Same maths, different context: Whether the question involves profit, area, time, or temperature — the calculus steps are identical.
Only the interpretation changes.
Optimisation method
- Write the model: identify the function to optimise. If not given, build it from the context.
- Differentiate: find f′(x).
- Set f′(x) = 0: solve for the critical x-value(s).
- Classify: confirm a maximum or minimum using a sign diagram or the GDC graph.
- Find the optimal value: substitute x back into f(x). Answer in context with units.
Example — Maximum profit
P(x) = −x² + 8x − 5, where x is units sold (hundreds) and P is profit (thousands of dollars).
Find the maximum profit.
Step by step
- Differentiate.
- Set P′(x) = 0 and solve.
- Classify with a sign diagram: P′(3) = 2 > 0 and P′(5) = −2 < 0, so + then − → maximum.
- Substitute x = 4 to find the maximum profit.
Final answer
Maximum profit = $11 000 when 400 units are sold.
Always state the optimal value: Finding x = 4 is not the final answer.
Substitute into the original function to get the maximum/minimum value itself, and state the units.
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IB exam questions frequently ask you to JUSTIFY whether the stationary point is a maximum or minimum.
Simply saying 'because x = …' earns no marks.
Two AI SL ways to justify: Method 1 — First-derivative sign diagram: test f′ just before and after the point. + then − → maximum; − then + → minimum.
Method 2 — GDC graph: graph the function; a peak is a maximum, a valley is a minimum. (AI SL does not use the second derivative.)
Example — Area problem
A(x) = 20x − 2x² for 0 ≤ x ≤ 10 (area of a rectangular garden).
Find the maximum area and justify it is a maximum.
Step by step
- Differentiate and solve A′(x) = 0.
- Justify with a sign diagram.
- Find the maximum area.
Final answer
Maximum area = 50 m² at x = 5, justified because A′ changes from + to − (a maximum).
Don't forget domain constraints: If the problem gives a restricted domain (e.g. 0 ≤ x ≤ 10), also check the endpoints — the greatest or least value might occur at an endpoint rather than at the stationary point.
Some IB questions give you the formula directly.
Others require you to build it from a word problem.
This is the harder skill — but it follows a consistent pattern.
Building a model
- Identify the quantity to optimise (this is your output variable).
- Identify the variable you control (the input).
- Use any constraints to eliminate extra variables.
- Write the formula for the output in terms of a single variable.
Example — Fencing a garden
Step by step
- Problem — A farmer has 60 m of fencing. He encloses a rectangular garden against a wall (so only 3 sides need fencing). Find the dimensions that maximise area.
- Variables — Length x (parallel to wall) and width w. Constraint: x + 2w = 60 → x = 60 − 2w.
- Area function — A = xw = (60 − 2w)w = 60w − 2w²
- Differentiate — A′(w) = 60 − 4w
- Set = 0 — w = 15 → x = 60 − 30 = 30
- Maximum area — A = 30 × 15 = 450 m² when w = 15 m and length = 30 m.
Constraint elimination: When you have two variables, the constraint (total perimeter, total cost, etc.) lets you write one variable in terms of the other — reducing to a single-variable function you can differentiate.
IB-style question — least metal for a tank
An open-top cylindrical tank must hold 1000 cm³. Find the radius r that uses the least metal (smallest surface area).
Step by step
- Use the fixed volume to write the height in terms of r (this eliminates h).
- Open top → surface area = base + curved side. Substitute h.
- Differentiate, set A′ = 0, and solve for r.
Final answer
r ≈ 6.83 cm (then h ≈ 6.83 cm too — a tank as wide as it is tall).
[Diagram: math-solid-volume] - Available in full study mode