In an oligopoly a few firms each know that their own output moves the price. Cournot's model treats quantities as the choice and looks for a Nash equilibrium. Each firm solves a small version of Monopoly with Calculus: a monopoly over the demand left over by its rival.
🔑 Key idea
With inverse demand $P = a - b(q_1 + q_2)$ and marginal costs $c_1, c_2$, firm 1's best response to $q_2$ is $$q_1 = \frac{a - c_1 - bq_2}{2b}.$$ The Cournot equilibrium solves both best responses at once: $q_1 = \dfrac{a - 2c_1 + c_2}{3b}$. With equal costs each firm makes $\dfrac{a - c}{3b}$, and total output lies between the monopoly and competitive levels.
💡 Why it works
Firm 1 takes $q_2$ as fixed and maximizes $(a - bq_1 - bq_2)q_1 - c_1q_1$. Its marginal revenue is $a - 2bq_1 - bq_2$: like a monopoly facing the residual demand $a - bq_2 - bq_1$. Setting it equal to $c_1$ gives the best response. Each firm's best response falls as its rival produces more. The equilibrium is where the two best-response lines cross. Each firm ignores the fact that its extra output also lowers its rival's profit, so together they produce more than a monopoly (or a cartel) would.
Best responses for $P = 100 - Q$ and $c = 10$, with $q_1$ across and $q_2$ up. Firm 1's: $q_1 = 45 - q_2/2$, drawn as $q_2 = 90 - 2q_1$. Firm 2's: $q_2 = 45 - q_1/2$. They cross at (30, 30).
✏️ Worked example: Symmetric duopoly
$P = 100 - Q$, $c = 10$ for both firms.
Best response: $q_1 = \dfrac{90 - q_2}{2}$. By symmetry $q_1 = q_2 = q$, so $q = \dfrac{90 - q}{2}$ and $q = 30$.
The low-cost firm produces more and earns more, but the high-cost firm still produces: unlike Bertrand competition, it isn't driven out.
Answer: $q_1 = 32$, $q_2 = 26$.
⚠️ Common mistake
Maximizing each firm's profit as if it were a monopoly, leaving the rival's output out of the price. That gives each firm the monopoly quantity (45 above), a total of 90 and a price of 10: neither firm's choice is a best response to the other's. Check: plug your $q_2$ into firm 1's best response; you must get back your $q_1$.