Home > Maths, Optimisation > Optimisation with a constraint

## Optimisation with a constraint

In A level maths you will be used to solving standard optimisation problems like $y = -3x^2 + 18x - 4$ find the value of x that gives you the maximum value of y. You just differentiate it and set it equal to zero, ie $\frac{dy}{dx} = -6x + 18$ so when $0 = -6x + 18$, x = 3 so y = 23.

In economics you do a lot of optimisation but a lot of the time you are working within a budget constraint.

For example suppose a firm has a production function, $Q = 2K^{0.3}L^{0.7}$

This shows you how the output is related to the inputs of capital (K) and labour (L), so if you have 10 units of capital and 12 units of labour you will get $Q = 2(10^{0.3})(12^{0.7})=22.723$.

Now if we wanted to ‘optimise’ our output, and we had an unlimited budget, then we could just keep buying more capital and hiring more labour and our output would go up and up and we could reach any level of output we wanted. However in the real world, there will be a cost for capital and a cost for labour.

Suppose a unit of capital costs 2 and a unit of labour costs 3, and we have a total budget of 30. There are lots of ways we could use that budget, we could afford 3 units of capital and 8 units of labour and have an output of 11.921, or we could afford 6 of each and have output of 12 etc. How do we know what is the ‘optimum’ way to spend our budget to maximise output given that constraint? This is a very ‘real world’ problem that you can use mathematics to solve.

You solve this using something the Lagrange multiplier which is something you probably won’t have been taught at A-level, but all you need to know is partial differentiation and you can do it.

I won’t go into the full mathematical reasoning behind why it works here, just show you how to do it.

Basically you have to optimise an original function: $f(x,y)$ subject to a constraint:$\phi(x,y)=M$

So you set up a new function: $g(x,y,\lambda)=f(x,y)+\lambda(M-\phi(x,y))$ and find the three partial derivatives and set them equal to zero, ie $\frac{\partial g}{\partial x} = 0, \frac{\partial g}{\partial y} = 0, \frac{\partial g}{\partial \lambda} = 0$. Then you solve the simultaneous equations and you have your values for x and y.

It will make more sense if we look at it in terms of this example above:

Objective function is: $Q = 2K^{0.3}L^{0.7}$
Constraint is $2K + 3L = 30$

So we set up our new function: $g(K,L,\lambda)=Q = 2K^{0.3}L^{0.7}+\lambda(30-(2K + 3L))$

Now we do our three partial derivatives and set them equal to 0:

$\frac{\partial g}{\partial K} = 0.6K^{-0.7}L^{0.7}-2\lambda = 0$ (1)
$\frac{\partial g}{\partial L} = 1.4K^{0.3}L^{-0.3}-3\lambda = 0$ (2)
$\frac{\partial g}{\partial \lambda} =30 - 2K - 3L = 0$ (3)

Now we can solve our simultaneous equations…
First put equation (1) and equation (2) in terms of $\lambda$
$0.3K^{-0.7}L^{0.7}=\lambda$
$\frac{1.4}{3}K^{0.3}L^{-0.3}=\lambda$

So $0.3K^{-0.7}L^{0.7}=\frac{1.4}{3}K^{0.3}L^{-0.3}$

We can multiply both sides of this equation by $K^{0.7}$ and we get $0.3L^{0.7}=\frac{1.4}{3}KL^{-0.3}$

Now we can multiply both sides of this equation by $L^{0.3}$ and we get $0.3L=\frac{1.4}{3}K$ so $L=\frac{1.4}{0.9}K$

Now we can substitute this into equation (3) to get it all in terms of K, so $30 - 2K - 3(\frac{1.4}{0.9}K) = 0$
So $30 - 2K - 3(\frac{1.4}{0.9}K) = 0 \Rightarrow 30 = 2K + 3(\frac{1.4}{0.9}K)\Rightarrow 30 = \frac{20}{3}K$ so $K = 4.5$

We can now put this back into our expression for L to get $L=\frac{1.4}{0.9}4.5$ so $L = 7$

So our optimum choice is to spend our budget of 30 on 4.5 units of capital and 7 units of labour, this will give us a total output Q of 12.262. No other combination, within our budget constraint, would reach a higher value of output.

A nice practical application of partial differentiation to solving an economic problem.

Advertisements
Categories: Maths, Optimisation
1. No comments yet.
1. No trackbacks yet.