SQA SQA Higher 2019-style practice: A manufacturer makes a closed cylindrical can with a fixed volume of 500 cm^3. The radius is r cm and the height is h cm. Show that the surface area is A(r) = 2π r^2 + 1000r, and hence determine the radius that minimises the surface area.

Volume fixes h: π r^2 h = 500 so h = 500π r^2 (1 mark). Surface area of a closed cylinder is A = 2π r^2 + 2π r h. Substitute h: A = 2π r^2 + 2π r · 500π r^2 = 2π r^2 + 1000r, as required (1 mark for the substitution and simplification). Differentiate, writing 1000r = 1000 r^-1: A'(r) = 4π r - 1000 r^-2 = 4π r - 1000r^2 (1 mark). Set A'(r) = 0: 4π r = 1000r^2, so 4π r^3 = 1000, giving r^3 = 250π and r = [3]250π ≈ 4.30 cm (2 marks). Confirm a minimum: A''(r) = 4π + 2000 r^-3 > 0 for r > 0, so this is a minimum (1 mark). Markers reward the valid surface-area model, correct differentiation of the reciprocal term, solving for r, and a nature check.

ScotlandMathsSyllabus dot point

How do you use differentiation to find the best possible value in a real situation and to describe how quantities change?

Using differentiation to find the optimal value in optimisation problems, the greatest and least values of a function on a closed interval, rates of change, and the motion of a particle through displacement, velocity and acceleration.

A focused answer to the SQA Higher Mathematics applying differential calculus content, covering optimisation problems, the greatest and least values on a closed interval, rates of change, and the displacement, velocity and acceleration of a moving particle.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Optimisation
Greatest and least on a closed interval
Rates and motion
Examples in context
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What this dot point is asking

The SQA wants you to use differentiation to solve optimisation problems, find the greatest and least values of a function on a closed interval, interpret rates of change, and use calculus to describe the motion of a particle through displacement, velocity and acceleration.

Optimisation

Optimisation problems ask for the largest or smallest possible value of some quantity, subject to a constraint. The reliable method has four steps. First, write the quantity you want to optimise (area, volume, cost, distance) as a formula. Second, use the constraint to eliminate every variable except one, so you have a function of a single variable. Third, differentiate and solve $f'(x) = 0$ to locate the stationary points. Fourth, justify the nature of the stationary point with a nature table or the second derivative, and convert the answer back into the quantity the question asked for.

A rectangle has perimeter

20

cm: find the dimensions giving the greatest area

A rectangle has a fixed perimeter of $20$ cm. Find the width and length that maximise the area.

Step 1: Introduce one variable and use the constraint

The perimeter condition links the two dimensions, so we can write one in terms of the other. Let the width be $x$ cm. The perimeter gives $2(\text{length} + \text{width}) = 20$ , so $\text{length} + x = 10$ , meaning the length is $(10 - x)$ cm. Now the area is a function of the single variable $x$ :

A = x(10 - x) = 10x - x^2

Step 2: Differentiate and find the stationary point

To locate the maximum, set the derivative to zero. Differentiating gives $\dfrac{dA}{dx} = 10 - 2x$ . Setting this to zero:

10 - 2x = 0 \implies x = 5

Step 3: Confirm it is a maximum and state the answer

The second derivative is $\dfrac{d^2A}{dx^2} = -2 < 0$ for all $x$ , so $x = 5$ gives a maximum, not a minimum. The dimensions are width $5$ cm and length $10 - 5 = 5$ cm, so the optimal rectangle is a square.

Final answer: The greatest area is $A = 5 \times 5 = 25$ cm $^2$ , achieved when both dimensions equal $5$ cm.

Greatest and least on a closed interval

On a closed interval $[a, b]$ a continuous function attains its greatest and least values either at a stationary point inside the interval or at one of the two endpoints. The procedure is to find every stationary point in $(a, b)$ , evaluate $f$ there, evaluate $f(a)$ and $f(b)$ , and then read off the largest and smallest of all these values. A stationary point that is only a local maximum can still be smaller than an endpoint, which is exactly why the endpoints must always be tested.

Greatest and least value of

f(x) = x^3 - 3x + 1

[-2, 3]

Find the greatest and least values of $f(x) = x^3 - 3x + 1$ on the closed interval $[-2, 3]$ .

Step 1: Find all stationary points inside the interval

Differentiate and set to zero to locate candidates for extreme values:

f'(x) = 3x^2 - 3 = 3(x^2 - 1) = 3(x - 1)(x + 1)

Setting $f'(x) = 0$ gives $x = 1$ and $x = -1$ . Both lie inside $[-2, 3]$ , so both must be checked. On a closed interval the extreme values can occur at stationary points or at the endpoints, so all four candidate points need evaluating.

Step 2: Evaluate $f$ at each candidate

Substitute each $x$ -value into the original function:

f(-1) = (-1)^3 - 3(-1) + 1 = -1 + 3 + 1 = 3

f(1) = 1 - 3 + 1 = -1

f(-2) = -8 + 6 + 1 = -1

f(3) = 27 - 9 + 1 = 19

Step 3: Compare all values and read off the result

The four values are $3$ , $-1$ , $-1$ and $19$ . The largest is $19$ and the smallest is $-1$ .

Final answer: The greatest value is $19$ , attained at the endpoint $x = 3$ . The least value is $-1$ , attained at both the stationary point $x = 1$ and the endpoint $x = -2$ .

Rates and motion

The derivative measures an instantaneous rate of change, so it links naturally to motion in a straight line. If a particle has displacement $s(t)$ at time $t$ , its velocity is the rate of change of displacement and its acceleration is the rate of change of velocity.

Examples in context

A spherical balloon is inflated so that its volume increases. The volume is $V = \dfrac{4}{3}\pi r^3$ , so the rate of change of volume with respect to radius is $\dfrac{dV}{dr} = 4\pi r^2$ , which is exactly the surface area. This says that when the radius is $r$ , each small increase in radius adds a shell of volume equal to the surface area times that increase, a result you can quote directly from differentiating the volume formula.

Try this

Q1. Find the value of $x$ that minimises $y = x^2 - 8x + 3$ . [3 marks]

Cue. $\dfrac{dy}{dx} = 2x - 8 = 0$ , so $x = 4$ ; $\dfrac{d^2y}{dx^2} = 2 > 0$ confirms a minimum.

Q2. A particle has displacement $s = t^3 - 6t^2$ metres. Find its velocity at $t = 4$ . [3 marks]

Cue. $v = \dfrac{ds}{dt} = 3t^2 - 12t$ , so at $t = 4$ , $v = 48 - 48 = 0$ m/s.

Q3. Find the greatest value of $f(x) = 12x - x^3$ on the interval $[0, 3]$ . [4 marks]

Cue. $f'(x) = 12 - 3x^2 = 0$ gives $x = 2$ in range; $f(2) = 24 - 8 = 16$ , $f(0) = 0$ , $f(3) = 36 - 27 = 9$ , so the greatest value is $16$ .

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SQA Higher 20196 marksA manufacturer makes a closed cylindrical can with a fixed volume of

500

^3

. The radius is

r

cm and the height is

h

cm. Show that the surface area is

A(r) = 2\pi r^2 + \dfrac{1000}{r}

, and hence determine the radius that minimises the surface area.

Show worked answer →

Volume fixes $h$ : $\pi r^2 h = 500$ so $h = \dfrac{500}{\pi r^2}$ (1 mark).

Surface area of a closed cylinder is $A = 2\pi r^2 + 2\pi r h$ . Substitute $h$ : $A = 2\pi r^2 + 2\pi r \cdot \dfrac{500}{\pi r^2} = 2\pi r^2 + \dfrac{1000}{r}$ , as required (1 mark for the substitution and simplification).

Differentiate, writing $\dfrac{1000}{r} = 1000 r^{-1}$ : $A'(r) = 4\pi r - 1000 r^{-2} = 4\pi r - \dfrac{1000}{r^2}$ (1 mark).

Set $A'(r) = 0$ : $4\pi r = \dfrac{1000}{r^2}$ , so $4\pi r^3 = 1000$ , giving $r^3 = \dfrac{250}{\pi}$ and $r = \sqrt[3]{\dfrac{250}{\pi}} \approx 4.30$ cm (2 marks).

Confirm a minimum: $A''(r) = 4\pi + 2000 r^{-3} > 0$ for $r > 0$ , so this is a minimum (1 mark). Markers reward the valid surface-area model, correct differentiation of the reciprocal term, solving for $r$ , and a nature check.

SQA Higher 20215 marksA particle moves in a straight line so that its displacement from the origin after

t

seconds is

s(t) = t^3 - 9t^2 + 24t

metres, for

0 \le t \le 6

. Find the times at which the particle is momentarily at rest, and determine its acceleration at each of those times.

Show worked answer →

Velocity is the derivative of displacement: $v(t) = \dfrac{ds}{dt} = 3t^2 - 18t + 24$ (1 mark).

At rest means $v(t) = 0$ : $3t^2 - 18t + 24 = 0$ , divide by $3$ to get $t^2 - 6t + 8 = 0$ , so $(t - 2)(t - 4) = 0$ giving $t = 2$ and $t = 4$ seconds (2 marks), both in $[0, 6]$ .

Acceleration is the derivative of velocity: $a(t) = \dfrac{dv}{dt} = 6t - 18$ (1 mark).

At $t = 2$ : $a = 6(2) - 18 = -6$ m/s $^2$ . At $t = 4$ : $a = 6(4) - 18 = 6$ m/s $^2$ (1 mark). Markers reward $v$ from $s$ , solving the quadratic for both rest times in range, and $a$ from $v$ evaluated at each.

Related dot points

Sources & how we know this

SQA Higher Mathematics Course Specification — SQA (2018)