Learn9MA0Paper 1 · Pure Mathematics 1DifferentiationOptimisation and rates of change

Differentiation · Subtopic 4 of 10

Optimisation and Rates of Change

Using derivatives to answer real questions: the largest area, the smallest surface, the cheapest design, and how fast one quantity changes when another one does.

Cambridge 9709 Paper 1Edexcel YMA01 WMA12Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1AQA 7357 Paper 2≈ 50 minRevision 6 · updated 2026-10-27

Section 1 of 14

Learning objectives

Turn a description into a formula in one variable, using a constraint.
Find the maximum or minimum value and check its nature.
Use $\dfrac{dA}{dt}=\dfrac{dA}{dr}\times\dfrac{dr}{dt}$ to link rates of change.
Interpret answers in context, with units.

Pause and think

You have $60$ m of fencing to make a rectangular pen against a long wall (so only three sides need fencing). How would you choose the sides to enclose the most area?

Section 2 of 14

Core idea

Start hereThe biggest box

Take a square of card $12$ cm wide, cut equal squares from each corner, and fold up the sides to make an open box. Cut tiny corners: a flat tray, little volume. Cut huge corners: a tall thin tower, little volume. Somewhere between is the best size. Calculus finds it exactly.

The big idea

Optimisation: write the quantity in terms of one variable, differentiate, set to zero, check it's a max/min. Connected rates of change: link rates with the chain rule, $\dfrac{dA}{dt}=\dfrac{dA}{dr}\times\dfrac{dr}{dt}$ .

1. Optimisation

Cutting bigger corners: volume of the box

x = 0.3

Small corners give a wide, shallow box; big corners a tall, narrow one. The best is in between, where the graph is flat.

Worked exampleThe box

From a $12$ cm square, cut squares of side $x$ cm from the corners and fold. Find the maximum volume.

Goal: $x$ giving the largest $V$ ; then $V$ .
Strategy: Express $V$ in $x$ ; differentiate; solve; check.

$V=x(12-2x)^2$
Base $(12-2x)$ by $(12-2x)$ , height $x$ .
$V=144x-48x^2+4x^3$
Expand.
$\frac{dV}{dx}=144-96x+12x^2=12(x-2)(x-6)$
Differentiate; factorise.
$x=2\ \ (x=6\text{ uses up the whole card})$
Reject the impossible root; $0<x<6$ .
$\frac{d^2V}{dx^2}=-96+24x=-48<0$
At $x=2$ : a maximum.
$V=2\times8^2=128$
Substitute.

AnswerMaximum volume

128

cm³, with

2

cm corners.

✓ Check

x=1.9

1.9\times8.2^2=127.8

;

x=2.1

2.1\times7.8^2=127.8

, both just below

128

✓.

Worked exampleMinimum surface of a can

A closed cylinder holds $250\pi$ cm³. Find the radius that minimises its surface area.

Goal: $r$ and minimum area.
Strategy: Use the volume to remove $h$ .

$\pi r^2h=250\pi\Rightarrow h=\frac{250}{r^2}$
The constraint.
$A=2\pi r^2+2\pi rh=2\pi r^2+\frac{500\pi}r$
Two ends + curved side.
$\frac{dA}{dr}=4\pi r-\frac{500\pi}{r^2}=0\Rightarrow r^3=125\Rightarrow r=5$
Differentiate; solve.
$\frac{d^2A}{dr^2}=4\pi+\frac{1000\pi}{r^3}>0$
Minimum.

Answer

r=5

cm,

h=10

cm,

A=150\pi

cm².

✓ CheckHeight = diameter: a known optimal can shape ✓.

Optimisation method

Drawing diagram…

The constraint step is where most marks are won or lost.

2. Connected rates of change

\frac{dA}{dt}=\frac{dA}{dr}\times\frac{dr}{dt}

$\frac{d}{dt}$: rate of change with time ("how fast").
$\frac{dA}{dr}$: from the geometry formula linking $A$ and $r$ .

Worked exampleRipple

A ripple's radius grows at $3$ cm/s. How fast is its area growing when $r=10$ ?

Goal: $\frac{dA}{dt}$ at $r=10$ .
Strategy: $A=\pi r^2$ ; chain rule.

$\frac{dA}{dr}=2\pi r$
Differentiate.
$\frac{dA}{dt}=2\pi r\times3=6\pi r$
Chain.
$r=10:\ 60\pi\approx188$
Substitute.

Answer

60\pi\approx188

cm²/s.

✓ CheckUnits: cm²/s ✓.

Worked exampleInflating a balloon

A sphere's volume increases at $100$ cm³/s. How fast is $r$ increasing when $r=5$ ?

Goal: $\frac{dr}{dt}$ .
Strategy: Rearrange the chain rule.

$V=\tfrac43\pi r^3\Rightarrow\frac{dV}{dr}=4\pi r^2=100\pi$
At $r=5$ .
$\frac{dr}{dt}=\frac{dV/dt}{dV/dr}=\frac{100}{100\pi}=\frac1\pi$
Divide.

Answer

\frac1\pi\approx0.318

cm/s.

✓ CheckAs

r

grows,

\frac{dr}{dt}

shrinks for the same inflow (more surface to cover) ✓.

Pause and think

A farmer fences three sides of a rectangle against a wall with 40 m. Maximum area?

Summary

Optimise: one variable, differentiate, solve, check.
Rates: multiply derivatives along the chain.

Optimise: Make as large or small as possible.
Rate of change: A derivative with respect to time.

Open box from a 12 cm square: V = x(12 − 2x)²

Setting

\dfrac{dV}{dx}=0

gives

x=2

(and

x=6

, which gives no box). The second derivative is negative at

x=2

, so the largest volume is

128\text{ cm}^3

HistoryKepler's wine barrels. In 1613, Johannes Kepler bought wine for his wedding and was surprised by how the merchant measured the volume with a single rod. He went on to study which barrel shapes hold the most wine for a given measurement, and found that Austrian barrels were very close to the best possible shape. His 1615 book on it was an early step towards calculus.

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