Optimisation Problems

This lesson covers optimisation — the process of using differentiation to find the maximum or minimum value of a quantity in a real-world context. Optimisation problems are a staple of A-Level Mathematics and appear frequently on AQA papers. They test your ability to translate a written description into algebra, differentiate, solve, and interpret your result.

What Is Optimisation?

In mathematics, optimisation means finding the value of a variable that makes a given quantity as large as possible (maximum) or as small as possible (minimum). In A-Level problems, you are typically asked to maximise an area, volume, or profit, or to minimise a cost, surface area, or length.

The general strategy is:

Identify the quantity to be optimised (e.g. area, volume, cost).
Write it as a function of a single variable using given constraints.
Differentiate and set the derivative equal to zero.
Solve to find the critical value(s).
Verify using the second derivative (or other method) that you have a maximum or minimum.
Answer the question — give the optimised value, not just the value of x.

Exam Tip: A very common mistake is to stop after finding x. The question usually asks for the maximum area, minimum cost, etc. — so substitute back to find the actual optimised value.

Setting Up Equations from Word Problems

The hardest part of optimisation is often the first step: turning the words into algebra. Here are some common setups.

Example 1: Maximising Area of a Rectangle

A farmer has 120 metres of fencing. She wants to create a rectangular pen against an existing wall (so only three sides need fencing). Find the dimensions that maximise the area.

Step 1: Let the width perpendicular to the wall be x metres. Then the length parallel to the wall is (120 − 2x) metres (since two widths and one length use up the 120 m of fencing).

Step 2: The area is:

A = x(120 − 2x) = 120x − 2x²

Step 3: Differentiate:

dA/dx = 120 − 4x

Step 4: Set dA/dx = 0:

120 − 4x = 0
4x = 120
x = 30

Step 5: Check the second derivative:

d²A/dx² = −4

Since d²A/dx² < 0, this is a maximum.

Step 6: The maximum area is A = 30(120 − 60) = 30 × 60 = 1800 m².

The dimensions are 30 m by 60 m.

Example 2: Minimising Surface Area of a Cylinder

A closed cylindrical can must hold 500 cm³ of liquid. Find the radius that minimises the total surface area.

Step 1: Let the radius be r cm and the height be h cm. The volume constraint gives:

V = πr²h = 500

So:

h = 500/(πr²)

Step 2: The total surface area is:

S = 2πr² + 2πrh

Substitute for h:

S = 2πr² + 2πr × 500/(πr²)
S = 2πr² + 1000/r

Step 3: Differentiate:

dS/dr = 4πr − 1000/r²

Step 4: Set dS/dr = 0:

4πr = 1000/r²
4πr³ = 1000
r³ = 250/π
r = (250/π)^(1/3) ≈ 4.30 cm

Step 5: Check the second derivative:

d²S/dr² = 4π + 2000/r³

Since both terms are positive for r > 0, d²S/dr² > 0, confirming a minimum.

Step 6: The height is h = 500/(π × 4.30²) ≈ 8.60 cm (which equals 2r, a well-known result for optimal cylinders).

The minimum surface area is approximately S = 2π(4.30)² + 1000/4.30 ≈ 349 cm².

Verifying the Nature of a Turning Point

At A-Level, you must always verify whether your stationary point is a maximum or minimum. The two standard methods are:

Method 1: Second Derivative Test

If d²y/dx² > 0, the point is a local minimum.
If d²y/dx² < 0, the point is a local maximum.
If d²y/dx² = 0, the test is inconclusive — use another method.

Method 2: Sign of the First Derivative

Check the sign of dy/dx on either side of the stationary point:

If dy/dx changes from positive to negative → maximum.
If dy/dx changes from negative to positive → minimum.

Exam Tip: The second derivative test is quicker and is preferred unless the second derivative is zero or very difficult to compute.

Example 3: Box Problem (Classic A-Level)

A rectangular sheet of card measures 24 cm by 15 cm. Squares of side x cm are cut from each corner and the sides folded up to make an open box. Find the value of x that maximises the volume.

Setting up:

After cutting and folding, the box has:

Length = (24 − 2x) cm
Width = (15 − 2x) cm
Height = x cm

The volume is:

V = x(24 − 2x)(15 − 2x)

Expand:

V = x(360 − 48x − 30x + 4x²)
V = x(360 − 78x + 4x²)
V = 360x − 78x² + 4x³

Differentiate:

dV/dx = 360 − 156x + 12x²

Set dV/dx = 0:

12x² − 156x + 360 = 0
x² − 13x + 30 = 0
(x − 3)(x − 10) = 0
x = 3 or x = 10

Since the width is 15 cm, we need 15 − 2x > 0, so x < 7.5. Therefore x = 10 is invalid.

Check x = 3 using the second derivative:

d²V/dx² = −156 + 24x

At x = 3: d²V/dx² = −156 + 72 = −84 < 0 → maximum.

Maximum volume = 3(24 − 6)(15 − 6) = 3 × 18 × 9 = 486 cm³.

Example 4: Profit Maximisation

A company sells widgets at a price of p pounds each. The number sold per week is n = 800 − 20p. Each widget costs £5 to produce, and the company has fixed weekly costs of £1000.

Find the price that maximises weekly profit.

Setting up:

Revenue = np = (800 − 20p)p = 800p − 20p²

Cost = 5n + 1000 = 5(800 − 20p) + 1000 = 4000 − 100p + 1000 = 5000 − 100p

Profit = Revenue − Cost:

P = 800p − 20p² − (5000 − 100p)
P = 800p − 20p² − 5000 + 100p
P = −20p² + 900p − 5000

Differentiate:

dP/dp = −40p + 900

Set dP/dp = 0:

−40p + 900 = 0
p = 22.50

Check: d²P/dp² = −40 < 0 → maximum.

Maximum profit = −20(22.5)² + 900(22.5) − 5000 = −10125 + 20250 − 5000 = £5125 per week.

Common Pitfalls in Optimisation

Not eliminating a variable: If your function has two unknowns (e.g. r and h), use the constraint to write one in terms of the other before differentiating.
Ignoring domain restrictions: Physical constraints (lengths must be positive, x < certain value) may eliminate one of your solutions.
Forgetting to verify max/min: Always use the second derivative test or another method.
Stopping too early: Find the actual maximum/minimum value, not just the value of the variable.

Summary

Optimisation problems require you to form an expression for the quantity to be maximised or minimised, then use differentiation.
Use constraints to reduce the expression to a single variable.
Set the first derivative equal to zero to find stationary points.
Use the second derivative test to verify the nature of the turning point.
Always check that your solution is physically valid (positive lengths, within domain).
Give your final answer in the context of the question.

Exam Tip: Optimisation questions are worth a lot of marks. Lay out your working clearly: state what you are maximising/minimising, show the constraint, show the substitution, differentiate, solve, verify, and state the answer in context. Every step earns marks even if you make an arithmetic slip later.