I am tagging this as multivariable calculus because it potentially involves taking partial derivatives.
I am working on some mathematical treatment for Cournot duopoly models (not homework, just preparing for a test tomorrow). A simple two player Cournot model is of the form $p=1000-q_1-q_2$ where $p$ stands for price, and $q_i$ is the quantity produced by firm $i$. The payoffs of each firm are given by $pq$ minus some marginal cost times $q$, say $-10q$. In this case, to find best response for firm $i$, I would take its payoff function $\pi_i=(1000-q_i-q_j)q_i-10q_i$ and take its derivative with respect to $q_i$, getting $1000-2q_i-q_j-10$. I would then set it equal to $0$, solving for $q_i$, and getting $q_i=\frac{990-q_j}{2}$, thus getting an expression maximizing $\pi_i$ in terms of $q_j$. Doing the same thing with $q_j$, I get $q_j=\frac{990-q_i}{2}$ since the function is symmetric. Then I set equations equal to each other/substitute one into the other, and get the Nash equilibrium (profit maximization for both), easy.
The issue I am having is dealing with $n$ firms. Assume a>c>0, b>0. Suppose we have total quantity of output $Q=\sum_{i=1}^{n} q_i$ and price be given as $p=a-bQ$. Then let profit of firm $i$ be defined as $\pi_i=p(Q)q_i-cq_i=(a-bQ)q_i-cq_i$. Defining $Q_{-i}$ to denote the sum of quantities of all firms but of firm $i$, we can rewrite the profit function of $i$ as $\pi_i=(a-bq_i-bQ_{-i})q_i-cq_i$ (observe that $q_i+Q_{-i}=Q$). From here on, I can again take a partial derivative with respect to $q_i$, and find the payoff maximization function to be $\pi_i'=a-2bq_i-bQ_{-i}-c$. Rewriting for $q_i$, I get $q_i=\frac{a-bQ_{-i}-c}{2b}$. So, for an arbitrary firm $j$, the maximization function in terms of $Q_{-j}$ will be given by $q_j=\frac{a-bQ_{-j}-c}{2b}$.
The problem I have here is that I can't anymore substitute as easily as I did in the first example with only two firms. The book gives me a hint: Summing the best-response functions over the different players will help. I do not know what to make of it. Could you explain how to find an intersection given $n$ such functions in the form presented above?
EDIT: I might be onto something here, feedback appreciated!
So I decided to follow the books advice and see what comes of it. Summing both sides of $q_i=\frac{a-bQ_{-i}-c}{2b}$ over all the players I get $q_1+\dots +q_n=Q=\frac{na}{2b}-\frac{nc}{2b}-\frac{(n-1)(q_1+\dots + q_n)}{2} \to 2Q=\frac{na}{b}-\frac{nc}{b}-(n-1)(q_1+\dots + q_n)$=$\frac{na}{b}-\frac{nc}{b}-(n-1)(Q) \to (n+1)Q=n\frac{(a-c)}{b} \to Q=\frac{n}{n+1} \frac{(a-c)}{b}$. Thus $q_i=\frac{n}{n+1} \frac{(a-c)}{b} – Q_{-i}$. Hopefully I am at least on the right track, though I can't see where this leads me yet.
Best Answer
As you pointed out, $q_i+Q_{-i}=Q$. Summing over all $i$ yields $$Q+\sum_i Q_{-i}=nQ.$$ Then sum over all $j$ for the best-response functions: \begin{eqnarray} Q&=&\sum_{j}q_j=\frac{n(a-c)}{2b}-\frac 1 2 \sum_{j}Q_{-j}=\frac{n(a-c)}{2b}-\frac 1 2(n-1)Q\\ \Rightarrow Q&=&\frac{n(a-c)}{(n+1)b}. \end{eqnarray} From here, you can solve for $Q$, and then get each $q_i$ because \begin{eqnarray} q_i&=&\frac{a-bQ-c}{2b}+\frac{q_i}2\\ \Rightarrow q_i&=&\frac{a-bQ-c}{b}=\frac{a-c}b-Q\\ &=&\frac{a-c}b-\frac{n(a-c)}{(n+1)b}=\frac{a-c}{b(n+1)}. \end{eqnarray}