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I'm doing an introductory economics course, having never done economics before. In our topics, we covered monopoly and the principle that the marginal revenue slope is twice that of the demand slope. I have also seen it stated elsewhere that the y-intercept of the two is also the same. However, if I calculate the marginal revenue equation and if I graph it, I get a different y-intercept to that of the demand slope (see below). As a result, I am always 0.5 units away from the given solution.

If the demand equation is as follows: $P = k + aQ$, then $MR = Q\times(k+aQ) - (Q-1)(k+a(Q-1))$. This simplifies to $MR = (k-a) + 2aQ$. (Note, a is normally negative, so $-a$ would essentially be adding to $k$).

Can anyone explain to me what I'm doing wrong?

The demand curve of one example was given as: $P = 120 - 2Q$

Here are my calculations: $MR = Q \times (120 - 2Q) - (Q - 1) \times (120 - 2(Q - 1))$

$MR = 120 Q - 2Q^2 - (Q - 1) \times (122 - 2Q)$

$MR = 120Q - 2Q^2 - (122Q - 2Q^2 - 122 + 2Q)$

$MR = 120Q - 2Q^2 - 124Q + 2Q^2 + 122$

$MR = 122 - 4Q$

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Your marginal revenue is not calculated correctly. Marginal revenue $(MR)$ is the derivate of total revenue which is equal price times quantity $TR=PQ$.

In your case $TR$ should be:

$$TR=(k+aQ)Q \implies MR = \frac{dTR}{dQ} = k+2aQ$$

If the demand is given as: $P = 120−2Q$ then:

$$TR= (120-2Q)Q \implies MR = \frac{dTR}{dQ} = 120-4Q $$

Also made a graph for you by simulating the $MR$ (red) and demand (blue) in R

enter image description here

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  • $\begingroup$ That makes sense, thank you! So if I was creating a graph, it would be incorrect to use change in total revenue as my marginal revenue curve? $\endgroup$ – E P Oct 7 '20 at 19:52
  • $\begingroup$ @EstellePretorius you are welcome (if you think my answer solved your issue consider accepting it). Also, with regards to your question you should consider the change in total revenue - but you have to consider the instantaneous rate of change (i.e. derivative). There are also derivative analogues for discrete functions but in this case the demand was specified as a continuous function $P= 120-Q$. $\endgroup$ – 1muflon1♦ Oct 7 '20 at 19:55
  • $\begingroup$ Perfect, thanks a million!! $\endgroup$ – E P Oct 7 '20 at 19:57

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