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Profit Maximisation MC= MR Question?

So when MR = MC is supposedly when a firm's profit is maximised.

What i don't understand is how does this actually maximise a firms profit? Surely they are simply just 'breaking even' and that it should actually be MR > MC?

At this point its supposed to be when the difference between total revenue and total cost is the greatest...

Can someone help me out please?
Reply 1
anyone?
Reply 2
Take two situtations
MC<MR.
What this means is that the revenue gained from producing one more unit exceeds the cost required to make this additional unit. Therefore a firm would increase it's production as the increase in revenue per unit is greater than the increase in cost per unit. The firm may end up having a larger cost from producing one more unit of output, but the revenue gained would be greater than the cost incurred hence the firm is more willing to produce more.

MC>MR
If MC is greater than MR then the cost of producing one more additional output exceeds the revenue gained from producing one more additional output. As a result the firm will reduce it's production levels to increase total revenue. While the firm loses some revenue due to the reduction in production, however it would not incur the associated cost at the same time. Hence the firm is more willing to cut back on its production as the reduction in revenue is less than the reduction in price.
Reply 3
Original post by BananaPie
Take two situtations
MC<MR.
What this means is that the revenue gained from producing one more unit exceeds the cost required to make this additional unit. Therefore a firm would increase it's production as the increase in revenue per unit is greater than the increase in cost per unit. The firm may end up having a larger cost from producing one more unit of output, but the revenue gained would be greater than the cost incurred hence the firm is more willing to produce more.

MC>MR
If MC is greater than MR then the cost of producing one more additional output exceeds the revenue gained from producing one more additional output. As a result the firm will reduce it's production levels to increase total revenue. While the firm loses some revenue due to the reduction in production, however it would not incur the associated cost at the same time. Hence the firm is more willing to cut back on its production as the reduction in revenue is less than the reduction in price.


Yes i know this. But it doesn't anser my question.

Why is the profit maximising point when MR = MC and not MR > MC?

I am told that when MR = MC then this is the greatest distance between total revenues and total costs. How does this work?
It's the point at which the cost producing one EXTRA unit equals the revenue gained from one EXTRA unit. Before this point MR was higher than MC. Therefore for every unit before this the firm was making a profit. The firm should therefore make as many units as it can make a profit on if it is a profit maximising firm - this will lead it to produce at the point where the revenue from every extra unit stops exceeding the cost of every extra unit. This would lead it to produce a quantity were producing any less would they are losing potential profit, and producing more would mean they are making a loss on anything produced pass MR=MC.
Original post by lar di da
Yes i know this. But it doesn't anser my question.

Why is the profit maximising point when MR = MC and not MR > MC?

I am told that when MR = MC then this is the greatest distance between total revenues and total costs. How does this work?


Because if MR>MC then there are still more units the firm can produce where it can make more revenue than the cost. Remember MR is not total revenue, it is the extra revenue with each unit.
(edited 11 years ago)
OP, do you do A Level Maths? If so, I can prove it using derivatives :smile:
Original post by SecretDuck
OP, do you do A Level Maths? If so, I can prove it using derivatives :smile:


Can you show me please? Can you also show me the relationship between MR and AR in imperfect competition mathematically?
Reply 8
Original post by LarrikinLibertine
It's the point at which the cost producing one EXTRA unit equals the revenue gained from one EXTRA unit. Before this point MR was higher than MC. Therefore for every unit before this the firm was making a profit. The firm should therefore make as many units as it can make a profit on if it is a profit maximising firm - this will lead it to produce at the point where the revenue from every extra unit stops exceeding the cost of every extra unit. This would lead it to produce a quantity were producing any less would they are losing potential profit, and producing more would mean they are making a loss on anything produced pass MR=MC.


Yes thats it, perfect! thanks. Makes complete sense now
Reply 9
Original post by SecretDuck
OP, do you do A Level Maths? If so, I can prove it using derivatives :smile:


yup i have. do so! it might help me in my course. im actually doing econ at uni lol so itll probably help
Original post by Bushido Brown
Can you show me please? Can you also show me the relationship between MR and AR in imperfect competition mathematically?



Original post by lar di da
yup i have. do so! it might help me in my course. im actually doing econ at uni lol so itll probably help


Total profits = TR - TC

To find the maximum total profit, you need to differentiate the above equation and set it equal to zero. The derivative of a total function is a marginal function - as the marginal function measures the rate of change in TR and TC.

So, we have MR - MC = 0
Which simplifies to MR = MC

In imperfect competition, let's play around with the MR and AR function and see their relationship.
Say p = -q + 200
TR = p * q = -q^2 + 200q
MR = (TR)' = -2q + 200

AR is simply the price function as (p * q / q) equals p. Setting MR equal to AR and solving for q gives us 200 which is both the intercepts for AR and MR. Now, let's find out the relationship between MR and AR using inequalities.
First, we check if MR is ever greater than AR:

-2q + 200 > -q + 200
Which leaves -q > 0
This is a contradiction for all q, q > 0 (also when q = 0). Therefore, MR can never be greater than AR.

Now, we check if MR is ever less than AR:

-2q + 200 < -q + 200
Which leaves -q < 0
Which is always true for all q, q > 0 (also when q = 0).

This says that MR = AR at the same intercept and then MR is below the AR curve. Additionally, check the slopes of the p (or AR) and MR functions. MR's slope is -2 and AR's slope is -1. This means that MR is decreasing at a faster rate than AR.

Hope it helps :smile:
(edited 11 years ago)

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