Many Economics courses at university require you to have studied A-level Mathematics. You may wonder what kind of maths an Economics course could involve. Some applications of A-Level Maths concepts in Economics courses at university include:
- Statistics– interpreting data to better understand the world around us is key in economics. in Econometrics makes use of topics such as the normal distribution and hypothesis testing.
- Differentiation– a lot of economics considers marginal changes and optimal conditions, for instance in microeconomics (like the example below).
- Algebra- many models in economics are specified mathematically, such as the Solow Growth Model in Macroeconomics.
- Graphs- to visualise models, graphs are often used by microeconomists and macroeconomists.
The following example will demonstrate the use of differentiation, a key topic in A-level maths, in a standard economic model.
Example
Employed individuals in the UK work 6 hours per day, on average1. When this figure is aggregated across all members of a population the resulting figure is the country’s total labour supply. Modelling how labour supply decisions are made at an individual level is valuable to economic analysis. For example, labour supply models are used in welfare and tax policy analysis, the output of such models can have economy–wide consequences.
The following model will illustrate a simple labour supply decision.
Individuals seek to maximise their happiness (or utility), denoted U, by consuming goods and services, c , and taking leisure time, L.
This can be represented by a function such as:
\[U(c,L)=\log(c)+\log(L)\]
The amount of leisure time taken is determined by the number of hours in a day minus the hours worked, h:
\[L=24-h \text{ (constraint)}\]
Consumption is determined by the income from employment, which is calculated by multiplying the number of hours worked, h, by the wage, w:
\[c=wh \text{ (constraint)}\]
Substituting both constraints into U:
\[U(h)=\log(wh)+\log(24−h)\]
Differentiating U with respect to h:
\[U'(h)=\frac{1}{h}+\frac{1}{h-24}\]
The first-order condition for an optimisation problem is that this first derivative is equal to zero. Setting this equal to zero:
\[0=\frac{1}{h}+\frac{1}{h−24} →h=12\]
to prove this is a maximum, the second-order condition, that U”(h)<0 must be true.
\[U”(h)=−\frac{1}{h^2}−\frac{1}{(h−24)^2}\]
\[−\frac{1}{h^2}−\frac{1}{(h−24)^2}<0\]
Therefore, the point where U'(h)=0 is a maximum.
In this simple model, an individual’s utility is maximised when labour supply is 12 hours.
As with all mathematical models, this model offers a framework to simplify to a real-world problem. It is not necessarily an accurate model, but nevertheless can be useful for the study of labour supply problems.
This model has necessitated the use of both the first and second derivative, including the derivative of logarithms. This is just one of many applications of A-level Maths in university level Economics.
