A-level · Differentiation

Stationary points (maxima and minima)Exam-style questions

Finding where a curve turns, and using it to find the greatest or least value of something. What a stationary point is and why dy/dx = 0 finds it; full coordinates and which roots to keep; telling a maximum from a minimum with the second derivative; what to do when the second derivative is zero; where a curve is increasing or decreasing; and optimisation questions in context.

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25 questions · A-level level · mark scheme and worked solution on each · no sign-up to try

DifficultyAll Level 3 Level 4 Level 5

1
A rectangular enclosure of area $200\ \text{m}^2$ is built against a straight wall, so only three sides need fencing. The side parallel to the wall has length $x$ m.
(a) Show that the total length of fencing is $L = x + \dfrac{400}{x}$ . (3 marks)
(b) Find the value of $x$ that minimises the fencing used, and the minimum length. (4 marks)
Level 5 · 7 marksTry it
2
A rectangle has its base on the $x$ -axis and its two upper vertices on the curve $y = 12 - x^2$ (with $y \ge 0$ ), symmetric about the $y$ -axis. The upper-right vertex is at $(x, 12 - x^2)$ .
(a) Show that the area of the rectangle is $A = 24x - 2x^3$ . (2 marks)
(b) Find the value of $x$ that maximises the area, and the maximum area. (5 marks)
Level 5 · 7 marksTry it
3
The diagram shows the curve with equation $y = (x + 1)(x - 3)^{2}$ . The curve has a maximum point $P$ and a minimum point $Q$ , and $Q$ lies on the $x$ -axis. The shaded region is bounded by the curve and the $x$ -axis.
(a) State the $x$ -coordinate of $Q$ .

(1 mark)
(b) Calculate the $x$ -coordinate of $P$ .

(4 marks)
(c) Find the area of the shaded region.

(4 marks)
(d) Find the least value of the gradient of the curve.

(4 marks)
Level 5 · 13 marksTry it
4
A company makes open-topped containers for storing grain.

A container is modelled so that
- the base of the container is a circle of radius $r$ cm
- the curved surface of the container is perpendicular to the base
- the container has no lid
- the total area of the base and the curved surface is $300\pi$ cm $^2$
The volume of the container is $V$ $V$ cm $^3$ $^{3}$ .
(a) Show that

$V = \frac{\pi}{2}\left(300r - r^3\right)$

making your method clear.

(4 marks)

Using algebraic differentiation,
(b) find the value of $r$ $r$ that makes $V$ $V$ stationary.

(4 marks)
(c) Prove, by further differentiation, that $V$ $V$ is a maximum at this value of $r$ $r$ .

(2 marks)
(d) Hence find the maximum volume of the container, giving your answer in terms of $\pi$ $π$ .

(1 mark)
Level 5 · 11 marksTry it
5
Food cans are made in the shape of a closed cylinder of radius $r$ cm. The capacity of each can is $250\pi$ cm $^3$ . The metal used for the curved surface costs $0.2$ pence per cm $^2$ , and the metal used for the top and the base costs $0.6$ pence per cm $^2$ .
(a) Show that the total cost, $C$ pence, of the metal for one can is given by $C = 1.2\pi r^2 + 100\pi r^{-1}$ .

(4 marks)
(b) Use calculus to determine the minimum cost of the metal for one can, giving your answer correct to the nearest penny. You should justify that it is a minimum.

(6 marks)
(c) Suggest one reason why the cans might not be made with the radius found in part (b).

(1 mark)
Level 5 · 11 marksTry it

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