Modelling with Differentiation inc. Optimisation (AQA GCSE Further Maths)

Revision Note

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Written by: Jamie Wood

Reviewed by: Dan Finlay

Modelling with Differentiation inc. Optimisation

How can I use differentiation to solve modelling questions?

Derivatives can be calculated for any variables – not just y and x
In every case the derivative is a formula giving the rate of change of one variable with respect to the other variable
For example if $A = 4 {πr}^{2}$ then $\frac{d A}{d r} = 8 πr$
- $\frac{d A}{d r}$ is the rate of change of $A$ with respect to $r$

Differentiation can be used to find maximum and minimum points of a function (see Stationary Points & Turning Points)
Therefore it can be used to solve maximisation and minimisation problems in modelling questions
- For example you may want to
  - Maximise the volume of a container
  - Minimise the amount of fuel used

Model Diff Illustr 2, A Level & AS Maths: Pure revision notes

Examiner Tips and Tricks

Exam questions on this topic will often be divided into two parts:
- First a 'Show that...' part where you derive a given formula from the information in the question
- And then a 'Find...' part where you use differentiation to answer a question about the formula
Even if you can't answer the first part you can still use the formula to answer the second part

Worked Example

A cuboid has length $4 x$ cm, width $x$ cm, and height $(\frac{3}{x} - 5)$ cm.

(a) $\begin{matrix} \end{matrix}$ Show that the volume, $V$ cm³ is given by $V = 12 x - 20 x^{2}$ .

The volume of a cuboid is " $V = l e n g t h \times w i d t h \times h e i g h t$ "

$∴ V = 4 x \times x \times (\frac{3}{x} - 5)$

Expand and simplify

$\begin{array}{rcl} V & = & 4 x^{2} (\frac{3}{x} - 5) \\ V & = & \frac{12 x^{2}}{x} - 20 x^{2} \end{array}$

$∴ V = 12 x - 20 x^{2}$

(b) Find the maximum volume of the cuboid.

Differentiate V with respect to x

$\frac{d V}{d x} = 12 - 40 x$

At the maximum volume, $\frac{d V}{d x} = 0$

$∴ 12 - 40 x = 0$

Solve for x

$\begin{array}{rcl} 40 x & = & 12 \\ x & = & \frac{12}{40} = 0.3 \end{array}$

So the value of x, at the maximum volume is 0.3
Find the maximum volume by substituting x = 0.3 in to the formula for V

$V = 12 (0.3) - 20 {(0.3)}^{2} = 1.8$

The maximum volume of the cuboid is 1.8 cm³

(c) Prove that your answer is a maximum value.

Using the second derivative is usually the easiest way to find the nature of a stationary point

$\frac{d^{2} y}{d x^{2}} = - 40$ < 0

The value of the second derivative (at $x = 0.3$ ) is negative, therefore V = 1.8 cm³ is a maximum volume

Last updated: 5 October 2024

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