1
$\begingroup$
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?

$\endgroup$
4
  • $\begingroup$ Both quantities yield the same profit. Therefore if q=5 maximizes profit so does q=4. $\endgroup$
    – Bayesian
    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? $\endgroup$ 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.) $\endgroup$ 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
    May 16, 2019 at 18:06

2 Answers 2

0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.