Model, differentiate, set to 0, classify, answer: Optimisation finds a maximum or minimum in a real context. The recipe: (1) write the quantity to optimise; (2) use a constraint to get it in one variable; (3) differentiate, set f'(x) = 0; (4) classify; (5) answer the question (often the actual max/min value).
One variable is the goal: You can only differentiate a function of one variable — use the constraint to eliminate the second variable first.
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Use the constraint to substitute: Write the target quantity (area, cost, volume…), then use the given constraint (a fixed perimeter, total length, etc.) to express it with a single variable.
IB-style question — set up the area
A rectangular pen uses a wall as one side and 40 m of fencing for the other three sides. With width x, write the area A in terms of x.
Step by step
- Constraint: two widths + one length = 40, so length = 40 − 2x.
- Area = width × length.
Final answer
A = 40x − 2x².
Read the constraint carefully: A wall side means only three sides are fenced — getting the constraint right is half the problem.
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Differentiate, solve f'(x) = 0, classify: Differentiate the one-variable model, solve f'(x) = 0 for the optimal x, classify it (second-derivative test), then answer — usually the maximum/minimum value.
IB-style question — maximum area
Maximise the pen area A = 40x − 2x² (from the 40 m of fencing). Find the dimensions and the maximum area.
Step by step
- A'(x) = 40 − 4x = 0 ⇒ x = 10. A''(x) = −4 < 0 ⇒ maximum.
- Length = 40 − 20 = 20; area = 10 × 20.
Final answer
Width 10 m, length 20 m; maximum area 200 m².
Answer the actual question: After finding x, give what's asked — the dimensions, the maximum value, or both.
IB-style question — optimise a transcendental function
A function is f(x) = 2x eˣ.
Find the x-coordinate of its stationary point and show that it is a minimum.
Step by step
- Differentiate with the product rule.
- Set f′ = 0. Since eˣ ≠ 0, only the bracket can be zero.
- Second derivative test: f″(x) = 2eˣ(2 + x); at x = −1 it is positive ⇒ minimum.
Final answer
Minimum at x = −1 (since f″(−1) > 0).
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Models like T = ax + b/x: Many cost/material problems give a model like T = ax + b/x. Rewrite b/x as bx⁻¹, differentiate, set T'(x) = 0, and solve (a positive value of x).
IB-style question — reciprocal model
The cost of a container is T = x + 36/x (x > 0). Find the value of x that minimises T, and the minimum cost.
Step by step
- T = x + 36x⁻¹, so T' = 1 − 36x⁻² = 0 ⇒ x² = 36 ⇒ x = 6.
- T''(x) = 72x⁻³ > 0 ⇒ minimum; T(6) = 6 + 6.
Final answer
x = 6 gives the minimum cost T = 12.
IB-style question — least material for a closed can
A closed cylindrical can has volume 250π cm³ and radius r cm.
(a) Show that its surface area is S = 2πr² + 500π/r.
(b) Find the radius that minimises the surface area.
Step by step
- (a) Use the volume to write the height in one variable (this set-up is the Topic 3.1 part).
- Substitute into the closed-cylinder surface area S = 2πr² + 2πrh.
- (b) Differentiate and set S′ = 0.
- Solve, and note S″ = 4π + 1000π/r³ > 0, so it is a minimum.
Final answer
(a) S = 2πr² + 500π/r. (b) r = 5 cm gives the least material.
Keep the positive root: x² = 36 gives x = ±6, but a length/quantity must be positive, so x = 6.