Optimisation

This lesson covers optimisation — using differentiation to find maximum and minimum values in practical contexts — as required by the Edexcel A-Level Mathematics specification (9MA0). You need to be able to set up equations from word problems, differentiate, find stationary points, and verify their nature.

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What Is Optimisation?

Optimisation means finding the maximum or minimum value of a quantity. In A-Level Mathematics, this typically involves:

1. Setting up a function to be optimised (e.g., area, volume, cost, distance).
2. Writing the function in terms of one variable (using constraints to eliminate others).
3. Differentiating and setting dy/dx = 0 to find stationary points.
4. Using the second derivative (or other method) to verify the nature.
5. Answering the question in context.

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Step-by-Step Strategy

Step	Action
1	Read the problem carefully. Draw a diagram if appropriate.
2	Define variables and write down what you need to maximise or minimise.
3	Write the function in terms of one variable (use constraints to eliminate others).
4	Differentiate and set the derivative equal to zero.
5	Solve for the variable.
6	Verify it is a maximum or minimum (using d²y/dx² or context).
7	Answer the original question (give the maximum/minimum value, not just the x-value).

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Worked Example 1: Maximising Area

A farmer has 100 m of fencing to enclose a rectangular pen against an existing wall. The wall forms one side of the rectangle. Find the dimensions that maximise the area.

Setting up: Let the width (perpendicular to wall) = x metres. Let the length (parallel to wall) = y metres.

Fencing constraint: 2x + y = 100, so y = 100 - 2x.

Area: A = xy = x(100 - 2x) = 100x - 2x²

Differentiating: dA/dx = 100 - 4x

Set dA/dx = 0: 100 - 4x = 0 → x = 25 m Then y = 100 - 2(25) = 50 m

Verify: d²A/dx² = -4 < 0 → Maximum. ✓

Maximum area: A = 25 × 50 = 1250 m²

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Worked Example 2: Minimising Material

An open-topped box has a square base of side x cm and height h cm. Its volume is 500 cm³. Find the dimensions that minimise the surface area.

Setting up: Volume: V = x²h = 500, so h = 500/x².

Surface area (no top): S = x² + 4xh = x² + 4x(500/x²) = x² + 2000/x

Differentiating: dS/dx = 2x - 2000/x² = 2x - 2000x⁻²

Set dS/dx = 0: 2x = 2000/x² 2x³ = 2000 x³ = 1000 x = 10 cm

Then h = 500/100 = 5 cm

Verify: d²S/dx² = 2 + 4000/x³ = 2 + 4000/1000 = 2 + 4 = 6 > 0 → Minimum. ✓

Minimum surface area: S = 100 + 2000/10 = 100 + 200 = 300 cm²

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Worked Example 3: Maximum Volume of a Box

A square sheet of cardboard has side length 20 cm. Squares of side x cm are cut from each corner and the sides are folded up to make an open box. Find x for maximum volume.

Setting up: Base: (20 - 2x) × (20 - 2x), height: x

Volume: V = x(20 - 2x)² = x(400 - 80x + 4x²) = 4x³ - 80x² + 400x

Domain: 0 < x < 10 (the side must remain positive).

Differentiating: dV/dx = 12x² - 160x + 400 = 4(3x² - 40x + 100) = 4(3x - 10)(x - 10)

Set dV/dx = 0: (3x - 10)(x - 10) = 0 x = 10/3 or x = 10

Since x = 10 gives V = 0 (no box), the solution is x = 10/3 ≈ 3.33 cm.

Verify: d²V/dx² = 24x - 160 At x = 10/3: d²V/dx² = 24(10/3) - 160 = 80 - 160 = -80 < 0 → Maximum. ✓

Maximum volume: V = (10/3)(20 - 20/3)² = (10/3)(40/3)² = (10/3)(1600/9) = 16000/27 ≈ 592.6 cm³

Exam Tip: Always state the domain of the variable — this helps you reject invalid solutions and ensures you are optimising over the correct range.

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Worked Example 4: Shortest Distance

Find the point on the curve y = √x that is closest to the point (3, 0).

Setting up: Let the point on the curve be (x, √x) where x ≥ 0.

Distance squared: D = (x - 3)² + (√x)² = (x - 3)² + x = x² - 6x + 9 + x = x² - 5x + 9

(We minimise D rather than √D — the minimum of D gives the minimum of √D.)

Differentiating: dD/dx = 2x - 5

Set dD/dx = 0: x = 5/2 = 2.5

Verify: d²D/dx² = 2 > 0 → Minimum. ✓

Closest point: (5/2, √(5/2)) = (2.5, √(2.5))

Minimum distance = √D = √(6.25 - 12.5 + 9) = √(2.75) = √(11/4) = √11 / 2 ≈ 1.66 units

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Worked Example 5: Maximising Revenue

A cinema charges £p per ticket and sells (200 - 10p) tickets. Find the price that maximises revenue.

Setting up: Revenue: R = p(200 - 10p) = 200p - 10p²

Domain: 0 < p < 20 (tickets sold must be positive).

Differentiating: dR/dp = 200 - 20p

Set dR/dp = 0: p = £10

Verify: d²R/dp² = -20 < 0 → Maximum. ✓

Maximum revenue: R = 10(200 - 100) = 10 × 100 = £1000

Tickets sold at this price: 200 - 10(10) = 100 tickets

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Worked Example 6: Optimising with Exponentials

The profit P (in thousands of pounds) from selling x items is modelled by P = 50xe⁻⁰·¹ˣ. Find the number of items that maximises profit.

Solution: dP/dx = 50e⁻⁰·¹ˣ + 50x(-0.1)e⁻⁰·¹ˣ = 50e⁻⁰·¹ˣ(1 - 0.1x) = 50e⁻⁰·¹ˣ(1 - x/10)

Set dP/dx = 0: Since e⁻⁰·¹ˣ > 0, we need 1 - x/10 = 0 → x = 10 items

Verify: d²P/dx² = 50(-0.1)e⁻⁰·¹ˣ(1 - 0.1x) + 50e⁻⁰·¹ˣ(-0.1)

At x = 10: d²P/dx² = 50(-0.1)e⁻¹(0) + 50e⁻¹(-0.1) = -5e⁻¹ < 0 → Maximum ✓

Maximum profit: P = 50(10)e⁻¹ = 500/e ≈ £183,900 (or £183.9 thousand)

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Worked Example 7: Cylinder in a Sphere

A right circular cylinder is inscribed in a sphere of radius R. Show that the maximum volume of the cylinder is (4πR³)/(3√3).

Setting up: Let the cylinder have radius r and height 2h. By Pythagoras (with the sphere radius): r² + h² = R², so r² = R² - h²

Volume: V = πr²(2h) = 2πh(R² - h²) = 2π(R²h - h³)

Differentiating: dV/dh = 2π(R² - 3h²)

Set dV/dh = 0: R² - 3h² = 0 → h² = R²/3 → h = R/√3

Then r² = R² - R²/3 = 2R²/3

V = 2π × (R/√3) × (2R²/3) = 4πR³/(3√3) ∎

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Common Mistakes

Mistake	Correction
Giving the value of x, not the maximum/minimum value	The question usually asks for the optimal VALUE (area, volume, etc.), not just the x-value
Not stating the domain	Always define the range of valid x values
Forgetting to verify max/min	Use the second derivative test or check the context
Trying to optimise with two variables	Use the constraint to eliminate one variable first
Not drawing a diagram	A diagram helps you set up the correct equations

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Summary

• Optimisation uses differentiation to find maximum or minimum values in context.
• The standard method: set up a function of one variable, differentiate, set derivative to zero, solve, verify, and answer in context.
• Use constraints (e.g., perimeter, volume) to eliminate extra variables.
• Always verify the nature of the stationary point (second derivative test or boundary analysis).
• State the domain of your variable and answer the question that was asked.

A-Level Deep Dive: Optimisation

Spec mapping

Edexcel 9MA0 specification section 9 — Differentiation, sub-strand 9.6 covers differentiation to find points of inflection, maxima and minima, and identify these as appropriate. Apply differentiation to gradients, tangents and normals, and stationary points (refer to the official specification document for exact wording). Optimisation problems sit at the apex of section 9 because they synthesise every prior idea — finding $\frac{dy}{dx}$dxdy​, classifying stationary points, and (decisively for AO3 marks) constructing the function to be optimised from a verbal context. Optimisation is examined in 9MA0-01 Pure Mathematics (typical 6–10 mark Q at the back of Paper 1) and 9MA0-03 Statistics and Mechanics, where mechanics modelling questions ask candidates to maximise range of a projectile or minimise time of flight. The Edexcel formula booklet does not list standard optimisation results — geometric formulas (volume of cylinder, surface area of cone) must be reconstructed from first principles or memorised.

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Worked example with full mark scheme

Question (8 marks): A rectangular sheet of metal measures $24$24 cm by $15$15 cm. Equal squares of side $x$x cm are cut from each corner and the sides are folded up to form an open-top box.

(a) Show that the volume $V$V cm$^3$3 of the box is given by $V = x(24 - 2x)(15 - 2x)$V=x(24−2x)(15−2x). (2)

(b) Find the value of $x$x that maximises $V$V, and verify it gives a maximum. Give the maximum volume to 3 significant figures. (6)

Solution with mark scheme:

(a) Step 1 — identify the dimensions of the folded box.

After cutting squares of side $x$x from each corner and folding, the base measures $(24 - 2x)$(24−2x) by $(15 - 2x)$(15−2x) and the height is $x$x.

B1 — correct identification of the three dimensions, with the height equal to the side of the cut square. Common slip: writing the base as $(24 - x)$(24−x) by $(15 - x)$(15−x) — forgetting that two squares are removed along each edge.

B1 — assembling the volume $V = x(24 - 2x)(15 - 2x)$V=x(24−2x)(15−2x) as required (printed answer; both B marks awarded only with full justification of the dimensions).

(b) Step 1 — expand to a polynomial in $x$x.

$V = x(24 - 2x)(15 - 2x) = x(360 - 48x - 30x + 4x^2) = x(360 - 78x + 4x^2)$V=x(24−2x)(15−2x)=x(360−48x−30x+4x2)=x(360−78x+4x2)

$V = 4x^3 - 78x^2 + 360x$V=4x3−78x2+360x

M1 — multiplying out and expressing $V$V as a polynomial. Either keeping the factorised form or expanding is acceptable, but differentiating the factorised form via product rule is messier; the expansion is the safer route.

Step 2 — differentiate.

$\frac{dV}{dx} = 12x^2 - 156x + 360$dxdV​=12x2−156x+360

M1 — applying the power rule term-by-term.

Step 3 — set $\frac{dV}{dx} = 0$dxdV​=0 and solve.

$12x^2 - 156x + 360 = 0 \implies x^2 - 13x + 30 = 0 \implies (x - 3)(x - 10) = 0$12x2−156x+360=0⟹x2−13x+30=0⟹(x−3)(x−10)=0

So $x = 3$x=3 or $x = 10$x=10.

M1 — solving the quadratic correctly (factorisation, formula, or completing the square all acceptable).

Step 4 — reject the inadmissible root.

Since $15 - 2x > 0$15−2x>0 for the box to exist, we need $x < 7.5$x<7.5. Therefore $x = 10$x=10 is rejected.

A1 — explicit rejection of $x = 10$x=10 with a domain-based reason. Candidates who simply ignore the second root lose this mark; candidates who reject it without justification lose it as well.

Step 5 — classify the stationary point.

$\frac{d^2V}{dx^2} = 24x - 156$dx2d2V​=24x−156

At $x = 3$x=3: $\frac{d^2V}{dx^2} = 72 - 156 = -84 < 0$dx2d2V​=72−156=−84<0, so this is a maximum.

M1 — second-derivative test (or sign-change test on $\frac{dV}{dx}$dxdV​) applied correctly with a clear conclusion.

Step 6 — compute the maximum volume.

$V(3) = 3(24 - 6)(15 - 6) = 3 \cdot 18 \cdot 9 = 486$V(3)=3(24−6)(15−6)=3⋅18⋅9=486 cm$^3$3 (already 3 s.f.).

A1 — correct numerical maximum with units.

Total: 8 marks (B1 B1 M1 M1 M1 A1 M1 A1).

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Specimen question modelled on the Edexcel 9MA0 Paper 1 format

Question (6 marks): A closed cylindrical tank has volume $1000$1000 cm$^3$3. The cost per cm$^2$2 of the curved surface is twice the cost per cm$^2$2 of the two circular ends. Find the radius $r$r that minimises the total material cost, giving your answer to 3 significant figures.

Mark scheme decomposition by AO:

• B1 (AO3.1a) — modelling step: setting up the cost function. With unit cost $k$k for the ends and $2k$2k for the curved surface, total cost $C = k \cdot 2\pi r^2 + 2k \cdot 2\pi r h$C=k⋅2πr2+2k⋅2πrh, or any equivalent. The AO3 mark recognises that the candidate correctly identifies which surface gets the higher unit cost.
• M1 (AO3.1b) — using the constraint $\pi r^2 h = 1000$πr2h=1000 to eliminate $h$h: $h = \dfrac{1000}{\pi r^2}$h=πr21000​.
• M1 (AO1.1b) — substituting to obtain a function of one variable: $C(r) = 2k\pi r^2 + 4k\pi r \cdot \dfrac{1000}{\pi r^2} = 2k\pi r^2 + \dfrac{4000k}{r}$C(r)=2kπr2+4kπr⋅πr21000​=2kπr2+r4000k​.
• M1 (AO1.1b) — differentiating: $\dfrac{dC}{dr} = 4k\pi r - \dfrac{4000k}{r^2}$drdC​=4kπr−r24000k​.
• A1 (AO1.1b) — solving $\dfrac{dC}{dr} = 0$drdC​=0: $4\pi r^3 = 4000$4πr3=4000, so $r^3 = \dfrac{1000}{\pi}$r3=π1000​, giving $r \approx 6.83$r≈6.83 cm.
• A1 (AO2.4) — verifying the minimum (second derivative $4k\pi + \dfrac{8000k}{r^3} > 0$4kπ+r38000k​>0 for $r > 0$r>0, so any stationary point is a minimum) and commenting that the constant $k$k does not affect the optimal $r$r.

Total: 6 marks split AO1 = 3, AO2 = 1, AO3 = 2. This is an AO3-heavy question — the modelling marks reward translating "twice the cost per cm$^2$2" into a correctly weighted objective. Candidates who treat all surfaces as equal cost (a common slip) lose the B1 and propagate an incorrect optimum.

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Synoptic links

Connects to:

1. Section 9 — Stationary points (sub-strand 9.6): every optimisation question is a stationary-point question wearing a context. The procedural core — set $\frac{dy}{dx} = 0$dxdy​=0, solve, classify — is identical. What changes is the front-end work of building the function from a constraint.

2. Section 9 — Second derivative test: the classification step uses $\frac{d^2y}{dx^2}$dx2d2y​. A negative value at a stationary point indicates a local maximum; a positive value indicates a local minimum. If the second derivative is zero, a sign-change test on $\frac{dy}{dx}$dxdy​ is required.