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$P(Q)$ represents a market where demand $Q$ is related to price $P$ by $$P(Q) = Q^{-\frac{1}{2}}$$ In this market there are $m$ identical producers, say firm 1, 2, up to $m$ which can produce any non-negative quantity say $q$ at costs $$q^2$$

We make the profit function: $$\Pi=Q^{-\frac{1}{2}}\cdot q -q^2$$ For firm $q_i$: $$\Pi=q_i\cdot\sum q_{-i} - q_i^2$$ Deriving and set to $0$: $$q_i=\frac{2m-1}{4m}$$ And $$q*=\frac{1}{m-1}$$

Is this ok? I just don't know anymore, have been working for hours on this......

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    You should start by calculating the optimal quantity of one firm as a function of the quantity of the other firms. That is, you should find best response functions.2012-11-25
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    There are $m$ firms. If $Q_{-i}$ is the aggregate quantity chose by producers different from $i$, then $i$ maximizes $P(Q_{-i}+q)-q^2$ with respect to $q$. Doing this for all levels of $Q_{-i}$ gives you the best response function of $i$.2012-11-25
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    Sorry, you have to maximize $P(Q_{-i}+q)q-q^2$ with respect to $q$ (revenue minus cost).2012-11-26
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    Profit is revenue (price times quantity) minus cost. The usual assumption (which Cournot made too) is that firms try to maximize profits.2012-11-26
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    In principle, you have to find $m$ quantities $q_1,\ldots,q_m$ such that $q_i$ is a best response to $Q_{-i}=\sum_{j\neq i}q_j$. It might be simpler to first look for a symmetric equilibrium, a $q$ that is a best response to $(n-1)q$. If this works out, everyone playing $q$ is an equilibrium.2012-11-26
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    In a symmetric equilibrium in which everyone produces $q$, we have $Q_{-i}=(m-1)q$. If you have the profit maximizing $q^*$ as a function of $Q_{-i}$, a symmetric equilibrium is simply a solution to $q=q^*((n-1)q)$.2012-11-26
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    Okay, let's go t chat.2012-11-26
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    Bob do you have the solution? Because i have the same problem. thanks,2012-11-26

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