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Optimisation

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.

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Categories: Maths, Optimisation
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