Why is the MR curve twice as steep as the AR curve in a monopoly's demand curve? Watch

aangel42
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Please help???
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chidona
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(Original post by aangel42)
Please help???
There's no reason why it's necessarily twice as steep (but it is steeper). It's fairly difficult to give an intuitive explanation, but think of the identities:

 MR = \frac{dTR}{dQ}; AR = \frac{TR}{Q}

Total Revenue is typically a Quadratic or higher power (a negative sign is used to give a fairly typical inverted U shape):

 TR = - P.Q^2

Hence:

MR = - 2Q.P
AR = - Q.P

Putting both into the form P = f(Q) (as you would draw it on a graph):

P = - \frac{MR}{2Q}
P = - \frac{AR}{Q}

Thus, the slope of the MR Curve (the top equation) is -1/2 (using dP/dQ), whilst the slope of the AR Curve (the bottom equation) is -1. Hence, in this very simple example you can see how one slope is twice the magnitude of the other.

By the by, the above TR function isn't realistic (it would imply that the optimal production point is at Q = 0), but it's just to show how you can justify the different slopes.
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London Prophet
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Saying TR = PQ^2 is deceptive.

The reason, explained in simple equations is as follows. Both AR and MR cross the y-axis at the same point - that is for the first unit of product both the average and marginal revenue = the price of the first sale.
Call the place where both curves cross the y-axis A (so coordinates (0,A))

Now consider the place where AR crosses the x - axis as B (B,0)
The gradient of AR = -(A/B)
So equation of AR = -(A/B)Q + B (think y = mx + c, but x = Q and y = price)

To get TR, multiply AR by Q, so:

TR = -(A/B)Q^2 + BQ


Marginal revenue is gained by differentiating TR (i.e. the change in total revenue for change in quantity)

so MR = -2(A/B)Q + B

Same position, twice the negative gradient.
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chidona
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(Original post by London Prophet)
Saying TR = PQ^2 is deceptive.
Not at all, it's just a simple example to lead a person through the maths.

All you've done is shift the curve a bit and work backwards - you haven't provided an intuitive explanation either.

Mehhhhh.
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turn and fall
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Its a modelling assumption. We just assume that TR is a quadratic.

Then it is just fairly simple calculus for the proof.
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danny111
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It will be if we assume demand is a straight line.

Then the demand function will be P = a + bQ (like y = mx + c with P on the y-axis and Q on the x-axis)

and hence total revenue will be P*Q = aQ + bQ^2. And marginal revenue will be MR = a + 2bQ which has slope twice of the demand function (2b versus b).

ps and demand line is AR
pps and we have a downward sloping demand curve (i.e. we are not in perfect competition).
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MagicNMedicine
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danny's explanation is the right one

it is twice as steep when we are using a linear demand function
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CerysCole95
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The reason why the MR is twice as steep as the AR (from what I have been taught to remember for exams is...)

It is due to the extra revenue you get from selling one more unit of output and occurs as the price has fallen. The new lower price, however, also applies to all previous units that could have been sold.

Simples, hope those two sentences help
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F000sey
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Easiest way to think of it is say you sell unit 1 for £10. TR is 10 and MR is 10. Then (law of demand) you can sell unit 2 for less e.g. £9, but to be fair to consumer number one you have to lower the price of both units not just the additional one, so TR is now (2 X 9 = 18) while marginal revenue has fallen to 8 (18-10).
AR/Demand falls by £1, MR falls by £2.
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hsdjfh3
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lol
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gmahapatra
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Is this correct?


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Chittesh14
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(Original post by London Prophet)
Saying TR = PQ^2 is deceptive.

The reason, explained in simple equations is as follows. Both AR and MR cross the y-axis at the same point - that is for the first unit of product both the average and marginal revenue = the price of the first sale.
Call the place where both curves cross the y-axis A (so coordinates (0,A))

Now consider the place where AR crosses the x - axis as B (B,0)
The gradient of AR = -(A/B)
So equation of AR = -(A/B)Q + B (think y = mx + c, but x = Q and y = price)

To get TR, multiply AR by Q, so:

TR = -(A/B)Q^2 + BQ


Marginal revenue is gained by differentiating TR (i.e. the change in total revenue for change in quantity)

so MR = -2(A/B)Q + B

Same position, twice the negative gradient.
Shouldn't:
So equation of AR = -(A/B)Q + B (think y = mx + c, but x = Q and y = price) be
So equation of AR = -(A/B)Q + A (think y = mx + c, but x = Q and y = price)? and the same for the rest?
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Huzaifa786
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When <a href=
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Huzaifa786
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When Demand is shown on a graph, it has a peculiar shape that is distinctly known as the demand curve. This demand curve demonstrates the various prices and the demand for the product in respective to it.
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