Stationary points and optimisation · Applications of differentiation

In General Maths you found stationary points by setting $\frac{dy}{dx} = 0$ (see calculus). Further Maths asks you to classify them, then to use the same idea to find the best value of something: the largest area, the smallest surface, the greatest height.

Stationary points and their nature

At a stationary point the tangent is flat: $\frac{dy}{dx} = 0$ . The second derivative tells you which kind it is:

$\dfrac{d^2y}{dx^2} < 0$ : the gradient is falling through 0, so it is a maximum;
$\dfrac{d^2y}{dx^2} > 0$ : the gradient is rising through 0, so it is a minimum.

Maximum and minimumFlat tangents: the gradient goes + 0 − at a maximum and − 0 + at a minimum

Worked example · WAEC 2013

WAEC 2013 · Paper 2 · Q12 (a)

Find the maximum and minimum points of the curve $y = 2x^3 - 3x^2 - 12x + 4$ .

Where is the gradient zero?
- ${\frac{dy}{dx} = 6x^2 - 6x - 12}$ .
- Take out 6: ${6(x^2 - x - 2) = 6(x - 2)(x + 1)}$ .
- So ${\frac{dy}{dx} = 0}$ at ${x = 2}$ and ${x = -1}$ .
Think first. Differentiate, then factorise.
Which is which?
- ${\frac{d^2y}{dx^2} = 12x - 6}$ .
- At ${x = -1}$ : ${-18 < 0}$ , so a maximum.
- At ${x = 2}$ : ${18 > 0}$ , so a minimum.
Think first. Find the second derivative and its sign at each point.
The points
- ${x = -1}$ : ${y = -2 - 3 + 12 + 4 = 11}$ . The maximum point is ${(-1, 11)}$ .
- ${x = 2}$ : ${y = 16 - 12 - 24 + 4 = -16}$ . The minimum point is ${(2, -16)}$ .
Think first. Put each x into the curve, not into dy/dx.

More: stationary points

Increasing and decreasing functions

A function is increasing where its graph rises from left to right, so its gradient is positive. It is decreasing where the graph falls, so its gradient is negative:

f'(x) > 0: \text{ increasing} \qquad f'(x) < 0: \text{ decreasing}

Rising and fallingf′(x) > 0 where the curve rises, f′(x) < 0 where it falls; the turning points separate the two

To find the interval, differentiate and solve the inequality. When $f'(x)$ is a quadratic, find its roots first. With a positive $x^2$ term it is negative between the roots and positive outside them (see quadratic inequalities).

Worked example

Find the range of values of

x

for which

f(x) = x^3 - 6x^2 + 9x + 1

is decreasing.

The gradient
- ${f'(x) = 3x^2 - 12x + 9}$ .
- Take out 3: ${f'(x) = 3(x^2 - 4x + 3)}$ .
- Factorise: ${f'(x) = 3(x - 1)(x - 3)}$ .
Think first. Differentiate, then factorise.
Where is it zero?
- ${f'(x) = 0}$ at ${x = 1}$ and ${x = 3}$ .
Think first. These are the turning points.
Where is it negative?
- ${f'(x)}$ has a positive $x^2$ term, so it is negative between the roots.
- Test ${x = 2}$ : ${3(1)(-1) = -3 < 0}$ . It agrees.
- So $f$ is decreasing for ${1 < x < 3}$ .
Think first. Positive x² term: negative between the roots or outside them?

More: increasing and decreasing

WAEC 2020 · Paper 1 · Q29In what interval is the function $f : x \to 2x - x^2$ increasing?

Sketching a curve

A sketch shows the shape and the key points; it is not a plotted graph, so you don’t need a table of values. Find these, then join them with a smooth curve:

\begin{aligned} x = 0 &: \text{ the } y\text{-intercept} \\ y = 0 &: \text{ the } x\text{-intercepts} \\ \tfrac{dy}{dx} = 0 &: \text{ the turning points} \end{aligned}

Sketching a cubicPut x = 0 and y = 0 for the intercepts, find the turning points, then show where the ends go

Intercepts. Put $x = 0$ for where it cuts the $y$ -axis. Put $y = 0$ and factorise for where it meets the $x$ -axis. A squared factor such as $x^2$ means the curve touches the axis there instead of crossing it.
Turning points. Solve $\frac{dy}{dx} = 0$ and use $\frac{d^2y}{dx^2}$ to say which is a maximum and which a minimum.
The ends. For large $|x|$ the highest power wins. For a cubic with a positive $x^3$ term the curve comes up from the bottom left and goes off to the top right; with a negative $x^3$ term it is the other way round.
Join the points with a smooth curve, and label each point with its coordinates.

Worked example · WAEC 2014

WAEC 2014 · Paper 2 · Q11 (b)

Sketch the curve $y = 9x - x^3$ .

Calculate the total area bounded by the $x$ -axis and the curve $y = 9x - x^3$ .

Where it cuts the axes
- ${x = 0}$ gives ${y = 0}$ : the curve passes through the origin.
- ${y = 0}$ : ${x(9 - x^2) = 0}$ .
- Difference of two squares: ${x(3 - x)(3 + x) = 0}$ .
- So it crosses the $x$ -axis at ${x = -3}$ , ${0}$ and ${3}$ .
Think first. Put x = 0, then put y = 0 and factorise.
The turning points
- ${\frac{dy}{dx} = 9 - 3x^2 = 0}$ , so ${x^2 = 3}$ .
- So ${x = \sqrt3}$ or ${x = -\sqrt3}$ .
- ${\frac{d^2y}{dx^2} = -6x}$ .
- At ${x = \sqrt3}$ it is negative: a maximum.
- There ${y = 9\sqrt3 - 3\sqrt3 = 6\sqrt3 \approx 10.4}$ .
- At ${x = -\sqrt3}$ it is positive: a minimum, ${y = -6\sqrt3 \approx -10.4}$ .
Think first. Set dy/dx = 0, then use the second derivative.
The ends
- For large $|x|$ the ${-x^3}$ term wins.
- Large negative $x$ : $y$ is large and positive, so the curve starts at the top left.
- Large positive $x$ : $y$ is large and negative, so it ends at the bottom right.
Think first. Which term wins when x is large?
The sketch
- From the top left, down through ${(-3, 0)}$ to the minimum ${(-\sqrt3, -6\sqrt3)}$ .
- Up through the origin to the maximum ${(\sqrt3, 6\sqrt3)}$ .
- Down through ${(3, 0)}$ and off to the bottom right.
Think first. Join the points in order from left to right.
(ii) The total area
- ${\int_0^3 (9x - x^3)\,dx = \left[\frac{9x^2}{2} - \frac{x^4}{4}\right]_0^3}$ .
- ${= \frac{81}{2} - \frac{81}{4} = \frac{81}{4}}$ .
- The part from $-3$ to $0$ is the same size below the axis: ${\frac{81}{4}}$ .
- Total: ${\frac{81}{4} + \frac{81}{4} = 40.5}$ square units (see areas below the axis).
Think first. The curve is below the axis on one side of 0. Integrate each part separately.

More: sketching curves

WAEC 2009 · Paper 2 · Q10Using the same axes, sketch the curves $y = 6 - x - x^2$ and $y = 3x^2 - 2x + 3$ .

Optimisation: the best value

“Largest” and “smallest” questions are stationary-point questions in disguise. The quantity usually depends on two letters, linked by a fact the question gives (a fixed volume, a fixed length of fence). Use the fact to write the quantity in one letter first:

1. write the quantity in one letter, using the given fact2. differentiate and set it = 03. solve, then check it is a max or min4. answer the question asked

A best-value problemOne letter, differentiate, set to zero, check, answer

Slide $x$ and look for the flat tangent:

Find the best xPick a problem, slide x

352area (m²)28dA/dx = 60 − 4x

x = 8

Two sides of x m and one of (60 − 2x) m. So A = x(60 − 2x) = 60x − 2x². The gradient is positive, so increasing x improves it. Solve 60 − 4x = 0 to jump straight to x = 15.

Worked example · WAEC 2022

WAEC 2022 · Paper 2 · Q10

A solid rectangular block has a base which measures $3x\text{ cm}$ by $2x\text{ cm}$ . The height of the block is $y\text{ cm}$ and its volume is $72\text{ cm}^3$ .

Express $y$ in terms of $x$ .

Find an expression for the total surface area of the block in terms of $x$ only;

Find the value of $x$ for which the total surface area has a stationary value (2 d.p.).

Use the volume
- ${3x \times 2x \times y = 72}$ , so ${6x^2y = 72}$ .
- So ${y = \frac{12}{x^2}}$ .
Think first. Volume = length × width × height. Make y the subject.
The surface area
- Top and bottom: ${2 \times 3x \times 2x = 12x^2}$ .
- Two faces $3x$ by $y$ : ${6xy}$ . Two faces $2x$ by $y$ : ${4xy}$ .
- So ${A = 12x^2 + 10xy}$ .
- Substitute ${y = \frac{12}{x^2}}$ : ${A = 12x^2 + \frac{120}{x}}$ .
Think first. Six faces, in three pairs. What is the area of each pair?
Set dA/dx = 0
- ${\frac{dA}{dx} = 24x - \frac{120}{x^2}}$ .
- Set it to 0: ${24x = \frac{120}{x^2}}$ , so ${24x^3 = 120}$ .
- So ${x^3 = 5}$ and ${x = \sqrt[3]5 \approx 1.71}$ .
- Check: ${\frac{d^2A}{dx^2} = 24 + \frac{240}{x^3} > 0}$ , so this is the smallest surface area.
Think first. Differentiate 12x² + 120x⁻¹.

Greatest height

For an object thrown upwards, the height $h$ depends on the time $t$ . At the top it is momentarily at rest: its velocity $\frac{dh}{dt}$ is 0. So the greatest height is a maximum-point question.

Worked example · WAEC 2016

WAEC 2016 · Paper 2 · Q7

A ball is thrown vertically upwards from ground level and its height after $t$ seconds is $(15.4t - 4.9t^2)\text{ m}$ . Find, correct to one decimal place, the:

time it takes to reach the maximum height;

maximum height it attains.

When is it at rest?
- ${\frac{dh}{dt} = 15.4 - 9.8t = 0}$ .
- So ${t = \frac{15.4}{9.8} = \frac{11}{7} \approx 1.6}$ s.
Think first. Differentiate h and set it to 0.
The greatest height
- ${15.4 \times \frac{11}{7} = 24.2}$ .
- ${4.9 \times \left(\frac{11}{7}\right)^2 = 4.9 \times \frac{121}{49} = 12.1}$ .
- So ${h = 24.2 - 12.1 = 12.1}$ m.
Think first. Put t = 11/7 into the height, not the velocity.

More: motion

Your turn

WAEC 2018 · Paper 2 · Q8

A body is thrown vertically upwards. Its height $h$ metres at time $t$ seconds is given by $h = \left(12t - \frac25t^2\right)\text{ m}$ . Find:

(a)
the time at which it is momentarily at rest;
(b)
the maximum height reached by the body.

Worked solution (try it first)

(a)

The body is momentarily at rest when its velocity is zero: $\dfrac{dh}{dt} = 12 - \dfrac45t = 0$ .
So $\dfrac45t = 12$ and $t = 15$ s.

(b)

Put $t = 15$ into the height: $h = 12(15) - \frac25(15)^2$
$= 180 - 90$
$= 90$ m.

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