I'm solving a problem about $n=180$ firms in a perfect competition market, where demand is given by $D(p)=100-5p$. A firm's cost function is $c(q_i) = 2 \cdot (q_i)^2$. Using $p = c'(q_i)$ and the demand function, I was able to derive the equilibrium quantity $q^* = \frac{1}{2}$ and price $p^* = 2$.

When asked about the profit each firm makes, I calculated a positive result for the individual profit in equilibrium $\pi_i^* = p^*\cdot q_i^* - c(q_i^*) = \frac{1}{2}>0$.

How is this possible? Is there not the assumption that a firm does not make profits in a perfect competition market? I know that there can be short term profits, but we assumed the market is in equilibrium here from the start. I'm new to microeconomics, coming from a mathematical background.

  • 1
    $\begingroup$ Please state the model fully in your post. $\endgroup$ Commented Dec 3, 2020 at 15:58
  • 2
    $\begingroup$ The price is equal to marginal cost. C' is increasing. So when you sell q units at C'(q)=p. You make zero profit on the final traded unit, but positive profit on any other traded unit. $\endgroup$
    – Bayesian
    Commented Dec 3, 2020 at 16:01
  • $\begingroup$ @Bayesian this makes sense! $\endgroup$
    – Friedrich
    Commented Dec 3, 2020 at 16:09

2 Answers 2


Given your parameters there should be profit. There can be profit even in perfect competition if there is less than infinite firms since as pointed out by Bayesian in his +1 comment when price is equal marginal cost there is no profit only on the last unit sold. Here is the full explanation:

The profit function of a firm is given by:

$$\pi = pq_i - aq_i^2$$

So FOC of this problem is given by:

$$\frac{d \pi }{dq_i}= p -2aq_i $$

and hence in the optimum:

$$\frac{p}{2a} = q_i^* $$

Now you can plug the optimum solution above back in to profit function and:

$$\pi = p\frac{p}{2a} - a\left(\frac{p}{2a}\right)^2 = \frac{p^2}{2a} - \frac{p^2}{4a} = \frac{p^2}{4a} = \frac{p^2}{8} \text{ for } a=2$$

Also the supply is given by the sum of optimum quantities across the whole market so:

$$S = \sum^n q_i^* = n(\frac{p}{2a}) = 45p \text{ for } a=2 \text{ and } n =180 $$

where the above assumes all firms are same so that $\sum^n q_i^* = nq^*$

Now equilibrium price will be given by intersection of supply and demand:

$$100−5p = n(\frac{p}{2a}) \implies p^*= \frac{200a}{n+10a}$$

Now the last expression goes to zero as the number of firms increases. However, for $n=180$ and $a=2$ we have:

$$p^*= \frac{400}{200}=2 \implies Q* = \sum^n q_i^* = 90$$

Meaning every firm will produce $q_i^* = 90/180 = 1/2$ and that the profits will be: $1/2$.

However, note the above can be sustained only if you assume no new firms can enter the market. In the long run with free entry - new firms will enter the market up until the $n$ is such that economic profit is zero (remember economic profit is not accounting profit so even at $\pi=0$ people have incentive to be doing business).


The first order condition for profits here (with respect to quantity, since firms are price-takers) is

$$p - 4q_i = 0 \implies q_i^* = \frac 1 4 p,$$

which is the supply function of its firm. So market Supply is

$$nq_i^* = \frac{180}{4} p = 45p,$$

and equilibrium market price

$$45 p = 100 - 5 p \implies p^* = 2$$

and equilibrium market quantity

$$Q* = nq_i^* = 180 \cdot\frac 1 4 (2) = 90$$

Per identical firm we have

$$q^*_i = 0.5$$

The profits are

$$(2)*(0.5) - 2* (0.5)^2 = 0.5$$


So what happens here? Well "perfectly competitive market" among others things means costless and free entry. But the problem above was solved for a fixed number of firms.

At this number of firms, each firm makes an economic profit. This will induce other firms to enter the market, driving equilibrium price down and economic profits to zero.

I guess the OP can understand how one can compute the number of firms required so as to have zero economic profits (hint: it is not finite).

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    $\begingroup$ +1 you beat me to the punch by 1 minute... I should learn to write more concisely :) $\endgroup$
    – 1muflon1
    Commented Dec 3, 2020 at 16:29
  • $\begingroup$ I'll reward elaborateness over timing in this case. Thank you both $\endgroup$
    – Friedrich
    Commented Dec 3, 2020 at 16:31

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