# Stackelberg Equilibrium

In a Stackelberg model, a firm enter the market first. Its quantity anticipates the quantity that the other firm will produce.

## Example

The inverse demand for coffee beans is $$P = 1400 - 2 Q_Z - 2 Q_Y$$

Zach chooses his quantity Q_Z first, then Yann follows.

## Yann's Best Response Function

When Yann enters the market, he chooses the best response to Q_Z.

Yann's revenue is R_Y = PQ = \left(1400 - 2 Q_Z - 2 Q_Y\right) Q_Y.

Yann's marginal revenue is MR_Y = 1400 - 2Q_Z - 4Q_Y.

Yann's marginal cost is MC_Y = 200.

Yann maximizes profit when \begin{align*} MC_Y &= MR_Y \\ 200 &= 1400 - 2Q_Z - 4Q_Y \\ 4Q_Y &= 1200 - 2Q_Z \\ Q_Y &= 300 - \frac{Q_Z}{2} \end{align*}

## Zach's profit maximization problem

Zach's maximizes $$\Pi_Z = \left(1400 - 2 Q_Z - 2 Q_Y\right) Q_Z -200 Q_Z$$

Plugging Q_Y = 300 - \frac{Q_Z}{2} gives \begin{align*} \Pi_Z &= \left(1400 - 2 Q_Z - 600 + Q_Z \right) Q_Z - 200 Q_Z \\ &= \left( 800 - Q_Z \right) Q_Z - 200 Q_Z \\ \end{align*}

Zach's profit is composed of a revenue R_Z = \left( 800 - Q_Z \right) Q_Z and a cost C_Z= 200 Q_Z.

When Zach maximizes profit \begin{align*} MC_Z &= MR_Z \\ 200 &= 800 - 2 Q_Z \\ 2 Q_Z &= 600 \\ Q_Z &= 300 \end{align*}

Zach produces 300 coffee beans. Yann produces Q_Y = 300 - \frac{300}{2} = 150 coffee beans.

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