Topics

Optimisation

Covers AS91578

What optimisation asks

An optimisation problem asks for the largest or smallest value something can take — maximum volume, minimum cost, largest area for a fixed perimeter.
The mathematics is just stationary points. The difficulty is building the equation, and that is exactly where the marks are.
The 2025 report: "Modelling for optimisation problems continues to be a skill demonstrated by high-performing candidates. Candidates need a lot more practice at understanding how to deal with creating appropriate models."

The method

Draw a diagram and label every unknown with a letter.
Write the objective — the quantity to be maximised or minimised — as an equation.
Write the constraint — the fixed condition given in the problem.
Use the constraint to eliminate a variable, so the objective depends on one variable only.
Differentiate the objective and set it to zero.
Solve, and reject any answer that makes no physical sense.
Answer the question that was asked — often the maximum value, sometimes the dimensions that achieve it.

Why the constraint matters

The objective usually starts with two variables — a box's volume depends on both length and height.
You cannot differentiate a two-variable expression with the tools in this standard.
The constraint is what removes the second variable. Finding it is half the modelling.
Worked through — a rectangle with perimeter 40 m, maximise the area:
- Objective: $A = xy$
- Constraint: $2x + 2y = 40$ , so $y = 20 - x$
- Substitute: $A = x(20 - x) = 20x - x^2$
- Differentiate: $\dfrac{dA}{dx} = 20 - 2x$
- Set to zero: $x = 10$ , so $y = 10$ and $A = 100$ m²
- The optimal rectangle is a square — a result worth remembering as a sanity check.

Formulae you are expected to know

The Level 2 specification says candidates "should be familiar with common 2D and 3D shapes and their associated formulae", and the same expectation carries to Level 3:
- Rectangle: $A = xy$ , $P = 2x + 2y$
- Circle: $A = \pi r^2$ , $C = 2\pi r$
- Cylinder: $V = \pi r^2 h$ , curved surface $= 2\pi rh$ , total surface $= 2\pi r^2 + 2\pi rh$
- Cone: $V = \frac{1}{3}\pi r^2 h$ , curved surface $= \pi r l$
- Sphere: $V = \frac{4}{3}\pi r^3$ , surface $= 4\pi r^2$
- Cuboid: $V = xyz$ , surface $= 2(xy + yz + xz)$

Rejecting invalid solutions

Physical quantities cannot be negative. A length of $-3$ m is not an answer.
State the rejection explicitly: " $x = -3$ is rejected because a length must be positive."
Also check the domain. If a piece of wire 20 cm long is cut, then $0 < x < 20$ , and any solution outside that range is invalid.

Justifying the maximum or minimum

The 2026 specification is explicit: at Achieved and Merit the paper will not require you to prove which you have found. It will say something like "you may assume your solution is a minimum".
At Excellence, or when the question asks, justify it by:
- the second derivative — negative for a maximum, positive for a minimum, or
- testing gradients either side, or
- an argument from context — "the area must be zero at both ends of the range, so the single stationary point in between is the maximum".

Optimisation on a closed interval

If the variable is restricted, the extreme value may occur at an endpoint rather than at a stationary point.
Evaluate the objective at every stationary point in range and at both endpoints, then compare the values.

Tips and Tricks

Draw the diagram and label it. Almost every optimisation mark that is lost is lost before any calculus happens, in the translation from words to equations. A labelled sketch makes the constraint visible.

Reduce to one variable before differentiating. If your objective still contains two letters, you have not used the constraint yet. Differentiating anyway is the most common structural error.

The report notes Merit candidates "demonstrated the procedure to set up models for optimisation questions, in terms of one variable, before correctly differentiating it". That phrase — in terms of one variable — is the whole technique.

Answer the actual question. If it asks for the maximum volume, give the volume, not the value of $r$ that produces it. If it asks for the dimensions, give all of them. Read the last line of the question again before you stop.

Reject impossible answers in writing. A negative length or a radius larger than the available material must be discarded with a stated reason, not silently dropped.

At Achieved and Merit, do not waste time proving the nature if the paper has told you to assume it — but do check whether the question is asking you to justify, because at Excellence it often is.

Worked ExampleMinimising the surface area of a can

A cylindrical can is to hold 500 cm³ of liquid. Find the radius that minimises the total surface area, and the minimum surface area, correct to 3 significant figures.

Step 1 — Draw and label

A closed cylinder with radius $r$ and height $h$ , both in centimetres. Two unknowns, so we will need a constraint.

Step 2 — Write the objective

The quantity to minimise is the total surface area — two circular ends plus the curved side:

$A = 2\pi r^2 + 2\pi rh$

This contains two variables, so it cannot be differentiated yet.

Step 3 — Write the constraint

The volume is fixed at 500 cm³:

$\pi r^2 h = 500$

Step 4 — Eliminate $h$

Rearrange the constraint to make $h$ the subject:

$h = \frac{500}{\pi r^2}$

Substitute into the objective:

$A = 2\pi r^2 + 2\pi r\left(\frac{500}{\pi r^2}\right)$

Simplify — the $\pi$ cancels and one $r$ cancels:

$A = 2\pi r^2 + \frac{1000}{r}$

Step 5 — Differentiate

Convert to index form first:

$A = 2\pi r^2 + 1000r^{-1}$

$\frac{dA}{dr} = 4\pi r - 1000r^{-2} = 4\pi r - \frac{1000}{r^2}$

Step 6 — Set the derivative to zero

State the condition explicitly before rearranging:

$\frac{dA}{dr} = 0 \quad \Rightarrow \quad 4\pi r - \frac{1000}{r^2} = 0$

$4\pi r = \frac{1000}{r^2}$

Multiply both sides by $r^2$ :

$4\pi r^3 = 1000$

$r^3 = \frac{1000}{4\pi} = \frac{250}{\pi}$

Step 7 — Solve

$r = \sqrt[3]{\frac{250}{\pi}} = \sqrt[3]{79.577...} = 4.3012...$

$r \approx 4.30 \text{ cm}$

Only one real cube root exists, and it is positive, so there is nothing to reject.

Step 8 — Confirm it is a minimum

$\frac{d^2A}{dr^2} = 4\pi + 2000r^{-3} = 4\pi + \frac{2000}{r^3}$

For any $r > 0$ both terms are positive, so $\frac{d^2A}{dr^2} > 0$ throughout the physically meaningful domain. The stationary point is therefore a minimum, and since it is the only one it is the global minimum.

Step 9 — Find the minimum surface area

Substitute the unrounded value $r = 4.3012...$ into the area expression:

$A = 2\pi(4.3012...)^2 + \frac{1000}{4.3012...}$

$= 116.24... + 232.49... = 348.73...$

$A \approx 349 \text{ cm}^2$

Step 10 — Sense-check

The corresponding height is $h = \dfrac{500}{\pi(4.3012)^2} = 8.602...$ cm, which is exactly $2r$ . The optimal can is as tall as it is wide — a standard result, and a good check that the working is right ✓

Optimisation

Covers AS91578

What optimisation asks

An optimisation problem asks for the largest or smallest value something can take — maximum volume, minimum cost, largest area for a fixed perimeter.
The mathematics is just stationary points. The difficulty is building the equation, and that is exactly where the marks are.
The 2025 report: "Modelling for optimisation problems continues to be a skill demonstrated by high-performing candidates. Candidates need a lot more practice at understanding how to deal with creating appropriate models."

The method

Draw a diagram and label every unknown with a letter.
Write the objective — the quantity to be maximised or minimised — as an equation.
Write the constraint — the fixed condition given in the problem.
Use the constraint to eliminate a variable, so the objective depends on one variable only.
Differentiate the objective and set it to zero.
Solve, and reject any answer that makes no physical sense.
Answer the question that was asked — often the maximum value, sometimes the dimensions that achieve it.

Why the constraint matters

The objective usually starts with two variables — a box's volume depends on both length and height.
You cannot differentiate a two-variable expression with the tools in this standard.
The constraint is what removes the second variable. Finding it is half the modelling.
Worked through — a rectangle with perimeter 40 m, maximise the area:
- Objective: $A = xy$
- Constraint: $2x + 2y = 40$ , so $y = 20 - x$
- Substitute: $A = x(20 - x) = 20x - x^2$
- Differentiate: $\dfrac{dA}{dx} = 20 - 2x$
- Set to zero: $x = 10$ , so $y = 10$ and $A = 100$ m²
- The optimal rectangle is a square — a result worth remembering as a sanity check.

Formulae you are expected to know

The Level 2 specification says candidates "should be familiar with common 2D and 3D shapes and their associated formulae", and the same expectation carries to Level 3:
- Rectangle: $A = xy$ , $P = 2x + 2y$
- Circle: $A = \pi r^2$ , $C = 2\pi r$
- Cylinder: $V = \pi r^2 h$ , curved surface $= 2\pi rh$ , total surface $= 2\pi r^2 + 2\pi rh$
- Cone: $V = \frac{1}{3}\pi r^2 h$ , curved surface $= \pi r l$
- Sphere: $V = \frac{4}{3}\pi r^3$ , surface $= 4\pi r^2$
- Cuboid: $V = xyz$ , surface $= 2(xy + yz + xz)$

Rejecting invalid solutions

Physical quantities cannot be negative. A length of $-3$ m is not an answer.
State the rejection explicitly: " $x = -3$ is rejected because a length must be positive."
Also check the domain. If a piece of wire 20 cm long is cut, then $0 < x < 20$ , and any solution outside that range is invalid.

Justifying the maximum or minimum

The 2026 specification is explicit: at Achieved and Merit the paper will not require you to prove which you have found. It will say something like "you may assume your solution is a minimum".
At Excellence, or when the question asks, justify it by:
- the second derivative — negative for a maximum, positive for a minimum, or
- testing gradients either side, or
- an argument from context — "the area must be zero at both ends of the range, so the single stationary point in between is the maximum".

Optimisation on a closed interval

If the variable is restricted, the extreme value may occur at an endpoint rather than at a stationary point.
Evaluate the objective at every stationary point in range and at both endpoints, then compare the values.

Tips and Tricks

Reduce to one variable before differentiating. If your objective still contains two letters, you have not used the constraint yet. Differentiating anyway is the most common structural error.

Reject impossible answers in writing. A negative length or a radius larger than the available material must be discarded with a stated reason, not silently dropped.

Worked ExampleMinimising the surface area of a can

A cylindrical can is to hold 500 cm³ of liquid. Find the radius that minimises the total surface area, and the minimum surface area, correct to 3 significant figures.

Step 1 — Draw and label

A closed cylinder with radius $r$ and height $h$ , both in centimetres. Two unknowns, so we will need a constraint.

Step 2 — Write the objective

The quantity to minimise is the total surface area — two circular ends plus the curved side:

$A = 2\pi r^2 + 2\pi rh$

This contains two variables, so it cannot be differentiated yet.

Step 3 — Write the constraint

The volume is fixed at 500 cm³:

$\pi r^2 h = 500$

Step 4 — Eliminate $h$

Rearrange the constraint to make $h$ the subject:

$h = \frac{500}{\pi r^2}$

Substitute into the objective:

$A = 2\pi r^2 + 2\pi r\left(\frac{500}{\pi r^2}\right)$

Simplify — the $\pi$ cancels and one $r$ cancels:

$A = 2\pi r^2 + \frac{1000}{r}$

Step 5 — Differentiate

Convert to index form first:

$A = 2\pi r^2 + 1000r^{-1}$

$\frac{dA}{dr} = 4\pi r - 1000r^{-2} = 4\pi r - \frac{1000}{r^2}$

Step 6 — Set the derivative to zero

State the condition explicitly before rearranging:

$\frac{dA}{dr} = 0 \quad \Rightarrow \quad 4\pi r - \frac{1000}{r^2} = 0$

$4\pi r = \frac{1000}{r^2}$

Multiply both sides by $r^2$ :

$4\pi r^3 = 1000$

$r^3 = \frac{1000}{4\pi} = \frac{250}{\pi}$

Step 7 — Solve

$r = \sqrt[3]{\frac{250}{\pi}} = \sqrt[3]{79.577...} = 4.3012...$

$r \approx 4.30 \text{ cm}$

Only one real cube root exists, and it is positive, so there is nothing to reject.

Step 8 — Confirm it is a minimum

$\frac{d^2A}{dr^2} = 4\pi + 2000r^{-3} = 4\pi + \frac{2000}{r^3}$

Step 9 — Find the minimum surface area

Substitute the unrounded value $r = 4.3012...$ into the area expression:

$A = 2\pi(4.3012...)^2 + \frac{1000}{4.3012...}$

$= 116.24... + 232.49... = 348.73...$

$A \approx 349 \text{ cm}^2$

Optimisation

What optimisation asks

The method

Why the constraint matters

Formulae you are expected to know

Rejecting invalid solutions

Justifying the maximum or minimum

Optimisation on a closed interval

Step 1 — Draw and label

Step 2 — Write the objective

Step 3 — Write the constraint

Step 4 — Eliminate hhh

Step 5 — Differentiate

Step 6 — Set the derivative to zero

Step 7 — Solve

Step 8 — Confirm it is a minimum

Step 9 — Find the minimum surface area

Step 10 — Sense-check

Optimisation

What optimisation asks

The method

Why the constraint matters

Formulae you are expected to know

Rejecting invalid solutions

Justifying the maximum or minimum

Optimisation on a closed interval

Step 1 — Draw and label

Step 2 — Write the objective

Step 3 — Write the constraint

Step 4 — Eliminate hhh

Step 5 — Differentiate

Step 6 — Set the derivative to zero

Step 7 — Solve

Step 8 — Confirm it is a minimum

Step 9 — Find the minimum surface area

Step 10 — Sense-check

Step 4 — Eliminate $h$

Step 4 — Eliminate $h$