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Optimisation with a constraint

June 20, 2011 Leave a comment

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.

Categories: Maths, Optimisation

Optimisation

June 20, 2011 Leave a comment

This is something which you will have come across in A-level, basically optimising a function by setting the derivative (or partial derivative) to 0.

In economics you are optimising things all the time. Firms are looking to optimise revenues or profits, or minimising costs. You will be given demand functions and information on costs etc, and asked to optimise.

For instance, say you have a monopoly firm facing a demand function Q = 200 - 2P, you have fixed costs of 25 and produce at a marginal cost of 5 per unit. You are asked to optimise profits.

From the information here we need to set out our profit function: \pi = TR - TC

First write the demand function in terms of P, the ‘inverse demand’ function: P = 100 - 0.5Q. This helps when we want to get our total revenue function. TR = PQ so here TR = 100Q - 0.5Q^2.

Now lets think of the costs. TC = FC + MC(Q) and we have fixed costs of 25 and marginal cost of 5, so here TC = 25 + 5Q

So now we have an expression for profit, \pi = TR - TC here means \pi = 100Q - 0.5Q^2 - 25 - 5Q so \pi = 95Q - 0.5Q^2 - 25. We optimise profit by setting \frac{d\pi}{dQ}=0, here \frac{d\pi}{dQ}=95-Q so setting it equal to 0 means our profit maximising output will be 95.

You could also have solved that by finding marginal revenue and setting it equal to marginal cost because the monopolist will produce where MR = MC.

We get marginal revenue by differentiating total revenue with respect to output, MR = \frac{d(TR)}{dQ}, here MR = 100 - Q. So if MC is 5 we know that at the production point MR is 5. Putting that into the equation for MR means we have Q of 95.

So our firm is producing an output of 95, which means when you sub that into the original demand function you get a price of 52.5. We can work out the actual value of the profits by going back to our expression for profit, \pi = 95Q - 0.5Q^2 - 25, which gives profits of 4487.5 at the profit-maximising output.

What if you were asked to find the revenue maximising point rather than profit maximising point? You had TR = 100Q - 0.5Q^2 so this time you just set \frac{d(TR)}{dQ}=0. Here \frac{d(TR)}{dQ}=100 - Q so when you set it equal to 0 you get a revenue maximising point of 100. This is of course where the marginal revenue is 0.

If you increase output beyond that point you are getting a negative marginal revenue. Why would this be? Well any firm that faces a downward sloping demand curve will have to decrease its price, if it wants people to buy more of its good. So it has a trade off – it sells more output, but each unit is being sold at a lower price. There comes a point where you lose more by reducing the price on all the units you are selling, than you gain by selling an extra unit, and that’s when you’ve gone past the point where MR = 0. Here, if you can sell 100 units at a price of 50, then you bring in revenue of 5000. But in order to sell 101 you have to reduce the price (this is just reading off the original demand function) to 49.5. This means you get revenue of 4999.5, so your revenue has dropped because you are selling beyond the revenue maximising point.

Categories: Maths, Optimisation