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Maximisation Maths

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Required Knowledge: Differentiation

We can use differentiation to find the rate of change which has useful applications in economics. For example to find something that is marginal (one more) we can differentiate. If you are not familiar with differentiating please see this lesson here.

So to find the marginal cost when total cost = 6q2 + 7q + 5 we would differentiate to find dTC/dq = 12q + 7, therefore MC = 12q + 7. If asked to find the marginal cost when quantity = 5, then we would differentiate the total costs and substitute q = 5.
This is the same as if we wanted to find marginal profit or marginal revenue.

Total Revenue = Price*Quantity
For example, find the total revenue when p=200-2q and q=2. P=196, Q=2, therefore TR = 196*2 = 392.

We know that profit maximisation occurs when the different between total revenue and total costs is at its greatest. This is the same as the output level when marginal cost equals marginal revenue (MR=MC).
If given marginal revenue and marginal cost to find maximum profit (denoted by ∏) we would set them equal.

Example 1
For example MR = 70-10q and MC = 25q, find the maximum profit. 
70-10q = 25q
70 = 35q
Therefore the output level at which profit is maximised is 2. We can calculate the profits at this level by substituting q into ∏=TR-TC.

Example 2
p = 10 - 4q
TC = 15+q2
Calculate the output required to maximise profit and find the profit level at this point.
TR = (10-4q)*q = 10q-4q2
TC = 15+q2
∏= TR-TC = (10q-4q2)-(15+q2) = 10q-15-5q2

Maximum profit occurs at MR = MC
MR = dTR/dq = 10-8q
MC = dTC/dq = 2q
10-8q = 2q
10 = 10q
q = 1
Therefore profit is maximised at an output level of 1.
To find the profit level at this point we will substitute q=1 back into (10q-15-5q2) which is TR-TC
10-15-5 = -10.
Therefore the maximum profit the firm can make is -10 at an output level of 1. 

NB: As highlighted in this example it is important to remember that a firm doesn't always make a profit even if it is trying to maximise profits, although the profit in the example above is negative, this is the best outcome, if it had produced an output level that wasn't 1 it would have made even less money.

Alternatively firms may not be trying to maximise profit and may instead be trying to maximise total revenue. To calculate the quantity they should produce in this case we would find dTR/dq and set this equal to zero (mathematically we are looking for a turning point, as this would be the maximum or the minimum, we do this by differentiating and setting the differential to 0). We can then re-arrange the equation to calculate q.
We may also be asked to prove that this is a maximum and not a minimum, to do this find the second differential (d2y/dx2) and substitute q in, if the answer is positive then we have a minimum, and if the answer is negative then we have a maximum.

For example, TR=200q-0.5q2 find the output level needed in order to maximise TR and prove that this is a maximum.
dTR/dq = 200-q
200 - q = 0
q = 200
d2TR/dq2 = -1; Negative therefore our point must be a maximum
Hence an output of 200 units should be produced in order to maximise total revenue and we have proven that this will maximise it, and not minimise it!

1. Find an equation for the marginal cost when TC = 4q3 + 7q2 + 2
2. Find the marginal cost when TC = 12q2 + 14q + 2 when q=3
3. Find the marginal revenue when TR = 6q2 + 5 and q=1 and an equation for marginal revenue when TR=6q2 + 5
4. Find the total revenue, in terms of q, when p=6q2 + 9q + 7
5. Find the marginal revenue, in terms of q, when p=5q + 6
6. Calculate the maximum profit and the quantity produced when p = 300-3q and TC = 15+2q2.
7. Find the quantity needed to maximise total revenue when TR = 100q-5q2 and hence prove this is a maximum

Page last updated on 20/10/13