Modelling with Differentiation

Exam code: 4037

Written by: Ashika|Reviewed by: Caroline Carroll|Updated 2 July 2026

Modelling with Differentiationvideo

Modelling with differentiation

How is differentiation used in modelling questions?

Derivatives can be calculated for any variables – not just y and x
The derivative is a formula giving the rate of change of one variable with respect to the other variable
- For example if $A = 4 π r^{2}$ then $\frac{d A}{d r} = 8 π r$
- $\frac{d A}{d r}$ is the rate of change of $A$ with respect to $r$
The phrase 'increasing at a rate of' means the rate of change of one variable with respect to time
- $\frac{d}{d t}$

Differentiation can be used to find maximum and minimum points of a function
- In modelling, this is called optimisation
- Second derivative tests help to determine is the point is a maximum or minimum

Optimisation

What is optimisation?

In general, optimisation is finding the best way to do something
In mathematics, optimisation is finding the maximum or minimum output of a function
- For example, finding the maximum possible profit or minimum costs

Differentiation can be used to solve optimisation problems in modelling questions
- For example you may want to
  - Maximise the volume of a container
  - Minimise the amount of fuel used

Example of using differentiation to maximise a function

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

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

Modelling with Differentiationvideo

Modelling with differentiation

How is differentiation used in modelling questions?

Optimisation

Optimisation

What is optimisation?

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