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.

A-Level Deep Dive: Optimisation Problems

Spec mapping

AQA 7357 specification, Paper 1 — Pure Mathematics, Section G: Differentiation. AQA explicitly requires candidates to "apply differentiation to find gradients, tangents and normals, maxima and minima and stationary points," and to "identify where functions are increasing or decreasing." Optimisation sits at the synoptic apex of this section: it pulls together stationary-point location ( $dy/dx = 0$ ), classification ( $d^2y/dx^2$ sign), boundary analysis on closed intervals, and modelling judgement (AO3). Optimisation is examined in Paper 1 (calculus questions) and routinely surfaces in Paper 3's mechanics or statistics modelling contexts. The AQA formula booklet does not list standard volumes, surface areas, or perimeters — these geometric formulae must be memorised. Expect a marks split of roughly AO1 35%, AO2 25%, AO3 40% on a full optimisation question.

Worked example with full mark scheme

Question (8 marks):

A rectangular sheet of card measures 30 cm by 20 cm. Equal squares of side $x$ cm are cut from each corner, and the resulting flaps are folded up to form an open-top box.

(a) Show that the volume $V$ cm $^3$ of the box is given by $V = 4x^3 - 100x^2 + 600x$ . (2)

(b) Find the value of $x$ that maximises $V$ , justifying that this value gives a maximum. (5)

Solution with mark scheme:

(a) Step 1 — express the box dimensions.

After cutting and folding, the base measures $(30 - 2x)$ by $(20 - 2x)$ cm and the height is $x$ cm.

$V = x(30 - 2x)(20 - 2x)$

M1 — correct dimensions written down (length, width, height in terms of $x$ ). Common error: students write the base as $(30 - x)(20 - x)$ , forgetting that two squares are removed along each edge.

Step 2 — expand.

$V = x(600 - 60x - 40x + 4x^2) = x(4x^2 - 100x + 600) = 4x^3 - 100x^2 + 600x$

A1 — correct expansion to the printed form. (Show-that questions demand visible algebra; just stating the answer scores zero.)

(b) Step 1 — differentiate.

$\dfrac{dV}{dx} = 12x^2 - 200x + 600$

M1 — correct application of $\frac{d}{dx}(ax^n) = nax^{n-1}$ term by term.

Step 2 — set $dV/dx = 0$ .

$12x^2 - 200x + 600 = 0 \implies 3x^2 - 50x + 150 = 0$

Using the quadratic formula:

$x = \dfrac{50 \pm \sqrt{2500 - 1800}}{6} = \dfrac{50 \pm \sqrt{700}}{6} = \dfrac{50 \pm 10\sqrt{7}}{6} = \dfrac{25 \pm 5\sqrt{7}}{3}$

M1 — solving the quadratic correctly (formula or completed-square).

Numerically: $x \approx 12.74$ or $x \approx 3.92$ .

Step 3 — reject the invalid root.

The domain is $0 < x < 10$ (since the shorter side is 20 cm and $2x < 20$ ). Therefore $x \approx 12.74$ is rejected and $x = \dfrac{25 - 5\sqrt{7}}{3} \approx 3.92$ cm.

A1 — correct value with the rejection of the spurious root explicitly justified by the physical domain.

Step 4 — classify using the second derivative.

$\dfrac{d^2V}{dx^2} = 24x - 200$

At $x \approx 3.92$ : $24(3.92) - 200 = 94.08 - 200 = -105.92 < 0$ .

Since the second derivative is negative, this stationary point is a maximum.

M1 A1 — correct second derivative evaluated at the stationary point with sign-based conclusion.

$V \approx 4(60.24) - 100(15.37) + 600(3.92) \approx 240.96 - 1536.6 + 2352 \approx 1056$

B1 — $V \approx 1056$ cm $^3$ (3 s.f.).

Total: 8 marks (M4 A3 B1).

Specimen question modelled on the AQA 7357 Paper 1 format

Question (6 marks): A closed cylindrical can of radius $r$ cm and height $h$ cm has a fixed volume of $500$ cm $^3$ .

(a) Show that the surface area $S$ cm $^2$ of the can satisfies $S = 2\pi r^2 + \dfrac{1000}{r}$ . (2)

(b) Find the value of $r$ that minimises $S$ , giving your answer in exact form. Justify that $S$ is minimised. (4)

Mark scheme decomposition by AO:

(a)

M1 (AO3.1a) — using $V = \pi r^2 h = 500$ to express $h = \dfrac{500}{\pi r^2}$ .
A1 (AO1.1b) — substituting into $S = 2\pi r^2 + 2\pi r h$ to obtain $S = 2\pi r^2 + 2\pi r \cdot \dfrac{500}{\pi r^2} = 2\pi r^2 + \dfrac{1000}{r}$ .

(b)

M1 (AO1.1b) — $\dfrac{dS}{dr} = 4\pi r - \dfrac{1000}{r^2}$ .
M1 (AO1.1b) — setting $dS/dr = 0$ : $4\pi r^3 = 1000$ , so $r^3 = \dfrac{250}{\pi}$ , giving $r = \sqrt[3]{\dfrac{250}{\pi}}$ .
A1 (AO2.5) — answer in exact form, e.g. $r = \left(\dfrac{250}{\pi}\right)^{1/3}$ .
B1 (AO2.4) — $\dfrac{d^2S}{dr^2} = 4\pi + \dfrac{2000}{r^3} > 0$ for all $r > 0$ , so $S$ is minimised.

Total: 6 marks split AO1 = 3, AO2 = 2, AO3 = 1. This is an AO3-rich modelling question — the constraint elimination in (a) is exactly the modelling step, and the second-derivative justification in (b) is rewarded for reasoning, not just procedure.

Synoptic links

Connects to:

Section G — Stationary points and classification: every optimisation problem reduces to locating stationary points of a single-variable function. The classification step ( $d^2y/dx^2 > 0$ minimum, $< 0$ maximum, $= 0$ inconclusive) is examined identically whether the context is an abstract polynomial or a real-world model.
Section G — Second derivative as concavity: the sign of $d^2y/dx^2$ tells you concavity. A function concave down at a stationary point has a local maximum; concave up gives a local minimum. This intuition extends seamlessly to inflection-point analysis.
Section A and B — Geometric formulae: optimisation routinely demands recall of $V = \pi r^2 h$ for cylinders, $V = \tfrac{4}{3}\pi r^3$ for spheres, $S = 2\pi r^2 + 2\pi r h$ for closed cylinders, $A = \tfrac{1}{2}ab\sin C$ for triangles. None are in the AQA formula booklet — memorise them.
Section O — Modelling and assumption-evaluation: AO3 marks reward candidates who comment on whether the optimum is physically reasonable (does $x$ lie in the valid domain? would the box still be rigid?) and identify modelling assumptions (zero card thickness, perfectly rigid folds).
Calculus of variations (university Year 2): the A-Level technique "differentiate, set to zero, classify" is the finite-dimensional shadow of the Euler–Lagrange equation, which optimises functionals over function spaces — used in Lagrangian mechanics, geodesics, and minimal surfaces.

Mark-scheme literacy

Optimisation questions on 7357 typically split AO marks across all three objectives:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	30–40%	Differentiating correctly, solving the resulting equation, evaluating the function at the critical point
AO2 (reasoning / interpretation)	20–30%	Justifying the maximum/minimum (second derivative sign or first-derivative sign change), rejecting invalid roots, presenting exact form when requested
AO3 (problem-solving)	30–40%	Setting up the function in one variable from a geometric or physical description, identifying the constraint, eliminating the constrained variable, commenting on physical reasonableness

Examiner-rewarded phrasing: "since $d^2V/dx^2 < 0$ at $x = \dots$ , this is a maximum"; " $x \approx 12.74$ is rejected because $x < 10$ is required for the box to exist"; "the model assumes zero card thickness, which is reasonable for thin card but not for corrugated board". Phrases that lose marks: "this is a max because the graph looks like a max" (no calculus reasoning); leaving multiple roots without rejecting invalid ones; failing to comment on the physical domain.

A specific AQA pattern to watch: questions that ask candidates to "comment on the validity of the model" expect at least one substantive sentence identifying a real-world limitation (thickness, rigidity, friction, idealised shapes). One-word answers like "valid" score zero.

Grade-band model answers

3-mark question

Question: A function is defined by $f(x) = x^3 - 12x + 5$ . Find the $x$ -coordinates of the stationary points and classify each.

Grade C response (~180 words):

$f'(x) = 3x^2 - 12$ . Setting $f'(x) = 0$ : $3x^2 = 12$ , so $x^2 = 4$ , giving $x = \pm 2$ .

$f''(x) = 6x$ . At $x = 2$ : $f''(2) = 12 > 0$ , so minimum. At $x = -2$ : $f''(-2) = -12 < 0$ , so maximum.

Examiner commentary: Full marks (3/3). The candidate correctly differentiates, finds both stationary points, and classifies each using the second-derivative test. The reasoning chain (compute $f''$ , evaluate at each critical point, conclude based on sign) is the standard A-Level template and is presented cleanly. This is the minimum acceptable answer for a procedural classification question — anything less detailed risks losing the AO2 mark for justification.

Grade A response (~210 words):*

Differentiate: $f'(x) = 3x^2 - 12$ . Set $f'(x) = 0$ : $3x^2 - 12 = 0 \implies x^2 = 4 \implies x = \pm 2$ .

Second derivative: $f''(x) = 6x$ .

At $x = 2$ : $f''(2) = 12 > 0$ , so $f$ has a local minimum at $x = 2$ .

At $x = -2$ : $f''(-2) = -12 < 0$ , so $f$ has a local maximum at $x = -2$ .

Examiner commentary: Full marks (3/3). The candidate uses the conditional connectives ("so", "implies") that signal logical flow, names the second-derivative test implicitly by computing $f''$ before invoking sign-based classification, and uses the precise terminology "local minimum" and "local maximum" rather than the looser "min" and "max". On a longer question, this kind of presentation discipline accumulates AO2 marks consistently.

6-mark question

Question: A farmer has 200 metres of fencing and wants to enclose a rectangular field. Find the dimensions that maximise the enclosed area, and state the maximum area, justifying that it is a maximum.

Grade B response (~250 words):

Let length = $x$ and width = $y$ . Perimeter: $2x + 2y = 200$ , so $y = 100 - x$ .

Area: $A = xy = x(100 - x) = 100x - x^2$ .

$dA/dx = 100 - 2x$ . Setting to zero: $100 - 2x = 0 \implies x = 50$ .

So $y = 50$ and $A = 50 \times 50 = 2500$ m $^2$ .

$d^2A/dx^2 = -2 < 0$ , so maximum.

Examiner commentary: Method correct, secured 5/6. The candidate sets up the constraint, eliminates $y$ , differentiates and solves, and applies the second-derivative test. The answer is correct (50 m by 50 m, 2500 m $^2$ ). They lose the final AO3 mark because no comment is made on the physical reasonableness — for instance, that the optimum is a square (geometrically intuitive), or that the model assumes the fencing has zero thickness and the field has straight sides on flat ground. AQA's AO3 marks for modelling-context optimisation routinely include "comment on reasonableness", and silent answers forfeit them.

Grade A response (~310 words):*

Let the rectangle have length $x$ m and width $y$ m, with $x, y > 0$ .

The perimeter constraint: $2x + 2y = 200 \implies y = 100 - x$ , valid for $0 < x < 100$ .

The area to maximise:

$A(x) = x(100 - x) = 100x - x^2$

Differentiate:

$\dfrac{dA}{dx} = 100 - 2x$

Set $dA/dx = 0$ : $x = 50$ . Therefore $y = 50$ m and $A = 2500$ m $^2$ .

Classify: $\dfrac{d^2A}{dx^2} = -2 < 0$ for all $x$ , so the stationary point is a maximum (and indeed the global maximum on the open interval since $A$ is a downward parabola).

The optimum field is a 50 m by 50 m square with maximum enclosed area 2500 m $^2$ .

Comment on reasonableness: the result that the optimal rectangle is a square is the classical isoperimetric result — for a fixed perimeter, the rectangle of maximum area is a square. The model assumes flat ground, straight-line fencing, and that all 200 m of fencing is used on the perimeter (no gates or overlaps). These are reasonable for a planning estimate but a real farmer would account for terrain irregularities and gate widths.

Examiner commentary: Full marks (6/6). Beyond the procedural correctness, the candidate writes the domain constraint explicitly ( $0 < x < 100$ ), uses display-math notation, justifies the maximum as global on the domain, names the isoperimetric result connecting the answer to a wider mathematical principle, and produces a substantive paragraph on modelling assumptions. Every AO3 mark is locked in by these touches.

9-mark question

Question: An open-top box is to be made from a square sheet of metal of side 24 cm by cutting equal squares of side $x$ cm from each corner and folding up the sides.

(a) Show that the volume $V$ cm $^3$ is given by $V = x(24 - 2x)^2$ . (2)

(b) Find the exact value of $x$ that maximises $V$ , justifying your answer. (5)

Grade A response (~370 words):*

(a) After cutting and folding, the base is a square of side $(24 - 2x)$ cm and the height is $x$ cm. Therefore $V = x(24 - 2x)^2$ , valid for $0 < x < 12$ . (M1 dimensions, A1 form.)

(b) Expand: $V = x(576 - 96x + 4x^2) = 4x^3 - 96x^2 + 576x$ .

Differentiate:

$\dfrac{dV}{dx} = 12x^2 - 192x + 576 = 12(x^2 - 16x + 48)$

Set $dV/dx = 0$ : $x^2 - 16x + 48 = 0$ . Using the formula:

$x = \dfrac{16 \pm \sqrt{256 - 192}}{2} = \dfrac{16 \pm 8}{2} = 12 \text{ or } 4$

Reject $x = 12$ (boundary; gives $V = 0$ ). So $x = 4$ .

Classify: $\dfrac{d^2V}{dx^2} = 24x - 192$ . At $x = 4$ : $24(4) - 192 = -96 < 0$ , so maximum.

Maximum volume: $V = 4(24 - 8)^2 = 4 \times 256 = 1024$ cm $^3$ .

(c) The model assumes the metal has zero thickness; in practice the cut squares and folded sides would not meet exactly along the edges, slightly reducing the achievable volume. It also assumes perfectly rigid folds at exactly 90°, with no material lost to cutting. For thin sheet metal these assumptions are reasonable; for thick metal or for a craft requiring sealed seams, they break down.

Examiner commentary: Full marks (9/9). Part (a) is concise and explicit. Part (b) factors the quadratic cleanly, explicitly rejects the boundary root with reasoning, applies the second-derivative test with computed values, and reports the maximum volume — all five marks secured. Part (c) gives a substantive comment identifying two distinct modelling limitations (zero thickness, perfect 90° folds) and notes when each matters. The phrasing "for thin sheet metal these assumptions are reasonable; for thick metal ... they break down" is exactly the conditional reasoning AQA AO3 mark schemes reward. Total: 9/9.

A-Level-depth misconceptions

The errors that distinguish A from A* on optimisation questions:

Forgetting to classify. Setting $dy/dx = 0$ finds all stationary points, including minima, maxima, and inflections. Without the second-derivative test (or first-derivative sign analysis), you cannot claim a maximum. Examiners deduct the classification mark routinely.
Ignoring boundary behaviour on a closed domain. Optimising on $[a, b]$ requires checking $f(a)$ , $f(b)$ , and interior stationary points. A function can attain its maximum at a boundary even when an interior stationary point exists.
Failing to eliminate the constraint. Optimisation always reduces to one variable. Students who differentiate $V = \pi r^2 h$ partially with respect to $r$ and $h$ separately are doing multivariable calculus (university content) and will not obtain the right answer at A-Level. Use the constraint to eliminate one variable first.
Confusing local and global maxima. $d^2y/dx^2 < 0$ guarantees a local maximum. On a bounded interval, the global maximum may be at an endpoint. A* candidates note when they are claiming local versus global.
Sign error in the second derivative. A common slip: differentiating $dV/dx = 12x^2 - 200x + 600$ to get $d^2V/dx^2 = 24x - 200$ is straightforward, but evaluating the sign at the stationary point requires substitution — students sometimes evaluate at $x = 0$ instead, getting the wrong sign and the wrong classification.
Inconclusive second-derivative test. When $d^2y/dx^2 = 0$ at a stationary point, the test is inconclusive — the point may be a maximum, minimum, or point of inflection. Fall back to the first-derivative sign test.
Domain error on physical problems. $x$ in a box problem must satisfy $0 < x < L/2$ where $L$ is the shorter side. Roots outside this interval must be rejected with explicit reasoning.

Common errors and mark-loss patterns on optimisation

Three patterns repeatedly cost candidates marks on Paper 1 optimisation questions. They are all about reasoning discipline, not technique.

Silent classification. Candidates locate the stationary point correctly and report the answer without checking whether it is a maximum or minimum. The classification mark (typically 1 mark via second-derivative test, sometimes 2 marks if "justify" is asked) is forfeited even when the stationary point happens to be the right type.
No comment on physical reasonableness. Modelling-context questions on AQA Paper 3 (and increasingly Paper 1) include explicit "comment" or "evaluate the model" prompts. One-word answers ("valid", "reasonable") score zero. Substantive commentary identifying a specific assumption and its limitation is required.
Domain neglect. Quadratic stationary-point equations often produce two roots, only one of which lies in the physical domain. Failing to reject the invalid root explicitly — with a phrase like "rejected because $x > 10$ would mean the side length is negative" — typically loses the A1 reasoning mark even when the final numerical answer is correct.

This pattern is endemic to Paper 1 optimisation questions: candidates know the calculus, lose marks on reasoning.

Going further — university and academic signposting

Optimisation at A-Level points directly toward several undergraduate trajectories:

Multivariable calculus (Year 1): real optimisation problems involve multiple constrained variables. The Lagrange multiplier method extends single-variable optimisation to functions of several variables under one or more equality constraints, with the optimum characterised by $\nabla f = \lambda \nabla g$ .
Convex optimisation (Year 2/3): when the objective function is convex and the feasible region is convex, every local minimum is a global minimum and efficient algorithms (gradient descent, interior-point methods) are guaranteed to converge. This is the foundation of modern machine learning, operations research, and signal processing.
Calculus of variations (Year 2/3): optimising over function spaces (rather than points) leads to the Euler–Lagrange equation. Applications include the brachistochrone (curve of fastest descent), geodesics on curved surfaces, and Lagrangian mechanics — where the path a particle takes minimises the action integral.
KKT conditions (Year 3): the Karush–Kuhn–Tucker conditions generalise Lagrange multipliers to inequality constraints, giving necessary conditions for optimality in nonlinear programming. This is the theoretical backbone of constrained optimisation in economics, engineering, and AI.

Oxbridge interview prompt: "Find the rectangle of maximum area inscribed in a circle of radius 1. Now find the rectangle of maximum perimeter. Why are these different problems geometrically?"

Additional A* practice: cylinder of fixed surface area

A common A* trap on 7357 is to swap the constraint and objective: instead of fixing volume and minimising surface area (the standard "optimal can" problem), fix surface area and maximise volume.

Worked example: A closed cylindrical container has total surface area $600\pi$ cm $^2$ . Find the radius that maximises the volume, and state the maximum volume in exact form.

The surface area: $2\pi r^2 + 2\pi r h = 600\pi \implies r^2 + rh = 300 \implies h = \dfrac{300 - r^2}{r}$ .

The volume: $V = \pi r^2 h = \pi r^2 \cdot \dfrac{300 - r^2}{r} = \pi r(300 - r^2) = 300\pi r - \pi r^3$ .

Differentiate: $\dfrac{dV}{dr} = 300\pi - 3\pi r^2$ . Set to zero: $r^2 = 100 \implies r = 10$ (rejecting $r = -10$ ).

Classify: $\dfrac{d^2V}{dr^2} = -6\pi r$ . At $r = 10$ : $-60\pi < 0$ , so maximum.

Then $h = \dfrac{300 - 100}{10} = 20$ cm, and $V = \pi(100)(20) = 2000\pi$ cm $^3$ .

Why A candidates spot this immediately:* at the optimum, $h = 2r$ — the optimal closed cylinder has height equal to its diameter. This is the same proportion that minimises surface area for fixed volume, illustrating the duality between these two optimisation problems. Recognising recurring geometric optima (square for fenced rectangle, equilateral triangle for fixed-perimeter triangle, $h = 2r$ for closed cylinder) lets you sanity-check answers.

A subtlety: the constraint $r^2 + rh = 300$ requires $r < \sqrt{300} \approx 17.3$ for $h > 0$ . Always check that the optimum lies in the physical domain — if it doesn't, the maximum is attained at a boundary instead.

AQA A-Level alignment footer

This content is aligned with the AQA A-Level Mathematics (7357) specification, Paper 1 — Pure Mathematics, Section G: Differentiation. For the most accurate and up-to-date information, please refer to the official AQA specification document.

Visual summary

graph TD
    A["Real-world problem:<br/>maximise or minimise<br/>a quantity"] --> B["Identify objective<br/>and constraint"]
    B --> C["Express objective<br/>in two variables<br/>(e.g. V in r and h)"]
    C --> D["Use constraint to<br/>eliminate one variable<br/>(e.g. h in terms of r)"]
    D --> E["Objective in<br/>one variable:<br/>f(x)"]
    E --> F["Differentiate:<br/>compute df/dx"]
    F --> G["Set df/dx = 0<br/>and solve"]
    G --> H{"Roots in<br/>physical domain?"}
    H -->|"No"| I["Reject root<br/>with reasoning"]
    H -->|"Yes"| J["Compute d-squared-f/dx-squared<br/>at the stationary point"]
    J --> K{"Sign of second derivative?"}
    K -->|"Negative"| L["Local maximum"]
    K -->|"Positive"| M["Local minimum"]
    K -->|"Zero"| N["Inconclusive:<br/>use first-derivative<br/>sign test"]
    L --> O["Evaluate f at optimum<br/>and at boundaries"]
    M --> O
    N --> O
    O --> P["Comment on physical<br/>reasonableness<br/>and modelling assumptions"]

    style L fill:#27ae60,color:#fff
    style M fill:#3498db,color:#fff
    style P fill:#e67e22,color:#fff

Optimisation Problems

Optimisation Problems

What Is Optimisation?

Setting Up Equations from Word Problems

Example 1: Maximising Area of a Rectangle

Example 2: Minimising Surface Area of a Cylinder

Verifying the Nature of a Turning Point

Method 1: Second Derivative Test

Method 2: Sign of the First Derivative

Example 3: Box Problem (Classic A-Level)

Example 4: Profit Maximisation

Common Pitfalls in Optimisation

Summary

A-Level Deep Dive: Optimisation Problems

Spec mapping

Worked example with full mark scheme

Specimen question modelled on the AQA 7357 Paper 1 format

Synoptic links

Mark-scheme literacy

Grade-band model answers

3-mark question

6-mark question

9-mark question

A-Level-depth misconceptions

Common errors and mark-loss patterns on optimisation

Going further — university and academic signposting

Additional A* practice: cylinder of fixed surface area

AQA A-Level alignment footer

Visual summary

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