Example
Zach and Yann compete in the market for coffee. They face a demand `P = 1400 - 2Q`, and both have the same marginal cost equal to 200.
Zach's best response is `Q_Z = 300 - \frac{Q_Y}{2}`. Yann's best response is `Q_Y = 300 - \frac{Q_Z}{2}`.
$$ \begin{align*} Q_Z &= 300 - \frac{Q_Y}{2} \\ Q_Z &= 300 - \frac{300 - \frac{Q_Z}{2}}{2} \\ Q_Z &= 300 - \frac{300}{2} + \frac{Q_Z}{4} \\ Q_Z - \frac{Q_Z}{4} &= 300 - 150 \\ \frac{3}{4} Q_Z &= 150 \\ Q_Z &= 200 \end{align*} $$Finally, plug `Q_Z = 200` into Yann's best response
$$ \begin{align*} Q_Y &= 300 - \frac{Q_Z}{2} \\ Q_Y &= 300 - \frac{200}{2} \\ Q_Y &= 300 - 100 \\ Q_Y &= 200 \end{align*} $$The Cournot equilibrium is (200, 200).
Question
The inverse demand on the market for coffee is `P = 19008 - 3 ( Q_Z + Q_Y )`.
Zach faces marginal costs equal to `576`.
Yann faces marginal costs equal to `1800`.
What is the Cournot Equilibrium?
In equilibrium, `Q_Z = 2184.0` and `Q_Y = 1776.0`.
Zach's revenue:
$$ \begin{align*} R \left( Q_Z \right) &= P Q_Z \\ &= \left( 19008 - 3 ( Q_Z + Q_Y ) \right) Q_Z \\ &= \left( 19008 - 3 Q_Z - 3 Q_Y \right) Q_Z \\ &= 19008 Q_Z - 3 Q_Z^2 - 3 Q_Y Q_Z \end{align*} $$
Zach's marginal revenue:
$$ \begin{align*} MR (Q_Z) \begin{align*} MR ( Q_Z ) &= 19008 - 3 Q_Y - 2 \times 3 Q_Z \\ &= 19008 - 3 Q_Y - 6 Q_Z \end{align*} \end{align*} $$Zach's marginal cost:
$$ \begin{align*} MC (Q_Z) = 576 \end{align*} $$Zach maximizes profit when `MR \left( Q_Z \right) = MC \left( Q_Z \right)`
\begin{align*} MC ( Q_Z ) &= MR ( Q_Z ) \\ 576 &= 19008 - 3 Q_Y - 6 Q_Z \\ 6 Q_Z &= 19008 - 576 - 3 Q_Y \\ Q_Z &= \frac{19008 - 576}{6} - \frac{Q_Y}{2} \\ Q_Z &= 3072.0 - \frac{Q_Y}{2} \end{align*}Yann's revenue:
$$ \begin{align*} R \left( Q_Y \right) &= P Q_Y \\ &= \left( 19008 - 3 ( Q_Z + Q_Y ) \right) Q_Y \\ &= \left( 19008 - 3 Q_Y - 3 Q_Z \right) Q_Y \\ &= 19008 Q_Y - 3 Q_Y^2 - 3 Q_Z Q_Y \end{align*} $$ Yann's marginal revenue: $$ \begin{align*} MR (Q_Y) \begin{align*} MR ( Q_Y ) &= 19008 - 3 Q_Z - 2 \times 3 Q_Y \\ &= 19008 - 3 Q_Z - 6 Q_Y \end{align*} \end{align*} $$ Yann's marginal cost: $$ \begin{align*} MC (Q_Y) = 1800 \end{align*} $$ Yann's best response solves \begin{align*} MC ( Q_Y ) &= MR ( Q_Y ) \\ 1800 &= 19008 - 3 Q_Y - 6 Q_Z \\ 6 Q_Y &= 19008 - 1800 - 3 Q_Z \\ Q_Y &= \frac{19008 - 1800}{6} - \frac{Q_Z}{2} \\ Q_Y &= 2868.0 - \frac{Q_Z}{2} \end{align*}
Plug `Q_Y = 2868.0 - \frac{Q_Z}{2}` into Zach's best response:
$$ $$ \begin{align*} Q_Z &= 3072.0 - \frac{Q_Y}{2} \\ Q_Z &= 3072.0 - \frac{ 2868.0 - \frac{Q_Z}{2} }{ 2 } \\ Q_Z &= 3072.0 - \frac{ 2868.0 }{ 2 } + \frac{Q_Z}{4} \\ Q_Z - \frac{Q_Z}{4} &= 3072.0 - \frac{ 2868.0 }{ 2 } \\ \frac{3}{4} Q_Z &= 1638.0 \\ Q_Z &= \frac{4}{3} 1638.0 \\ Q_Z &= 2184.0 \end{align*} $$ $$So Zach's quantity is `Q_Z = 2184.0`. Plug that into Yann's best response function:
$$ $$ \begin{align*} Q_{ Y } &= 2868.0 - \frac{Q_Z}{2} \\ &= 2868.0 - \frac{ 2184.0 }{2} \\ &= 2868.0 - 1092.0 \\ &= 1776.0 \end{align*} $$ $$