Price  Quantity   TR      TC   Profit   MR    MC
6         0       0       3     -3     -     -
6         1       6       5      1     6     2
6         2       12      8      4     6     3
6         3       18      12     6     6     4
6         4       24      17     7     6     5
6         5       30      23     7     6     6

Consider table above. If my understanding is correct, optimal quantity equals 5, because at this point MR = MC. But after producing 5th widget, profit does not increase. Then why is producing 5 widgets considered to be profit maximizing quantity, but not 4?

  • $\begingroup$ Both quantities yield the same profit. Therefore if q=5 maximizes profit so does q=4. $\endgroup$
    – Bayesian
    Commented May 16, 2019 at 13:41
  • $\begingroup$ That's the point. Both quantities yield the same profit, yet q=5 is considered to be optimal (MR=MC). Why? $\endgroup$ Commented May 16, 2019 at 15:07
  • 3
    $\begingroup$ The reason is that you are working with discrete quantities. Therefore your MR and MC are approximations. If you use continuous functions profit is maximized at the point where MR=MC. (In this case you'd get an even higher profit if you'd produce 4.5 units.) $\endgroup$ Commented May 16, 2019 at 15:57
  • 1
    $\begingroup$ My point was that q=4 is optimal as well and no one would argue with this. If your profit is differentiable then by definition "profit = revenue - cost" is maximized at some quantity with "marginal revenue - marginal cost =0". This is where the equality comes from. You only provided a table - we don't even know how costs look like for, e.g., q=4.5. $\endgroup$
    – Bayesian
    Commented May 16, 2019 at 18:06

2 Answers 2


The condition $MC=MR$ comes from studying the case where you can produce any quantity, for example, $4.0, 4.1, 4.41, 4.987$ or any fraction of production. In that case (under some general conditions), $MC=MR$ will give you the unique quantity that maximizes profits.

In contrast, in your table, you can only produce whole units, for example, $3, 4, 5$, etc. In that case, $MC=MR$ is an approximate condition to find the optimal profits and usually, it gives you the unique solution. Your example is special in the sense that there are 2 optimal quantities and the $MC=MR$ condition only gives you one such solution, but you can actually choose either.


Literally from the definition of Marginal Cost; it is the cost of producing one additional unit.

As your marginal revenue remains constant, pretty clear that you are going to make the maximum profits as the curves intersect at MC = MR. In the specific case that you consider, YES, you can choose either of 4 units or 5 units to be produced but, as Maarten said, your profits will definitely rise if you consider Optimal Production -2 units VS -1 unit in a general scenario.

This is the simple the reason why we say that MC = MR equilibrium is the optimal point to maximize profits. Doesn't mean you can't reach a similar state of profits before it can happen, just that MC = MR will for sure give you a maximal profit.

Also, emphasis on the point Discrete Quantities. Under a continuous distribution, your profits will keep increasing as long as the next extra unit produced will sell for more than it's production cost; i.e. till MC tends to the value of MR.


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