A law is passed requiring a monopolistic soft-drink manufacturer to separate the production department and the marketing department. The marketing department chooses the price $P > 0$ to charge for a bottle of the firm's soft drink and the production department chooses the level of output $Q > 0$. The two departments are forbidden to discuss their decisions with each other and, therefore, move simultaneously. Managers in both departments want to maximize the firm's profit.
$$\pi = PS - (Q^2) / 2$$
where $S$ denotes the firm's sales. Sales can not exceed the firm's output, nor can they exceed the market demand. Unsold output is thrown away. This means $S = \min\{Q, D(P)\}$ where market demand is
$$D(P) = \begin{cases}6 - P & {\rm if}\; P < 6\\ 0 & {\rm if} \; P \geq 6\end{cases}$$
Find all Nash equilibria.
I'm unsure how to approach this question. Can anyone explain to me in detail?
For $D(P)\lt Q$, the first term in $\pi$ doesn't depend on $Q$, and the second term has a minimum at $Q=0$, which is inadmissible, so there are no equilibria in this sector.
For $D(P)\gt Q$, we have $S=Q$, and thus $\pi=PQ-Q^2/2$. The derivative with respect to $P$ is zero only for $Q=0$, which is again inadmissible, so we don't get an equilibrium from this case, either.
That leaves $D(P)=Q$. For $P\ge6$, this means $Q=0$, which is inadmissible. For $P\lt6$, it means $\pi=P\min(Q,6-P)-Q^2/2$. Raising $Q$ lowers the second term while leaving the first term unchanged, and thus lowers the profit. Lowering $Q$ lowers the first term with derivative $P$ and raises the second term with derivative $-Q$, and thus lowers the profit if $P\ge Q$. Lowering $P$ lowers the first term, whereas increasing $P$ lets the first term change as $P(6-P)$, with derivative $6-2P$, which is non-positive for $P\ge3$. Thus we have equilibria for any $3\le P\lt6$ and $Q=6-P$.