Consider a case of differentiated products:
$q_i = 12 – pi + bp_j$ where b > 0
Assume no costs so that:
Profit for firm i = $q_i (p_i, p_j)\cdot p_i$
Want to maximise the profit function: $[12 – p_i + bp_j] * p_i$
Assuming firms are symmetric: $p_i = p_j = p$
Max: $[12 – p + bp] \cdot p$
Differentiated: $12 + (2b – 2) \cdot p = 0$
$p = \frac{12}{ 2 – 2b}$
However I'm supposed to have $p = \frac{12} b$
Best Answer
Usually you use both profit functions separately,
$$\Pi_i= [12 - p_i + bp_j] \cdot p_i$$
Differentiating w.r.t $p_i$ and setting the derivative equal to $0$.
$$\frac{\partial \Pi_i}{\partial p_i}=12-2\cdot p_i+bp_j=0 $$
$$\Rightarrow p_i=6+\frac{b}2\cdot p_j\quad (*)$$
Two options for the next steps:
1) You can use the symmetry argument, which means that $p_i=p_j$. We use the reaction function of firm $i$.
$$p_i=6+\frac{b}2\cdot p_i\quad$$
$$p_i\cdot \left(1-\frac{b}2\right)=6\Rightarrow \boxed{p_i=\frac{12}{2-b}}$$
2) You can use the reaction function of firm j: $$ p_j=6+\frac{b}2\cdot p_i$$ Combined with $(*)$ you have a small equation system. The solution is the same as in $1)$.