# Why marginal revenue must equal marginal cost?

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?

• Both quantities yield the same profit. Therefore if q=5 maximizes profit so does q=4. May 16 '19 at 13:41
• That's the point. Both quantities yield the same profit, yet q=5 is considered to be optimal (MR=MC). Why? May 16 '19 at 15:07
• 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.) May 16 '19 at 15:57
• 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. May 16 '19 at 18:06

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.