# Exercise 6.1 Theory of IO (Jean Tirole's book)

I am trying to solve exercise 6.1 of the famous book from Jean Tirole, The Theory of Industrial of Organization. I am stuck on a part he deems "straightforward":

"Suppose there are two firms, with unit costs $$c_1 < c_2$$. Let $$p^m(c)$$ denote the monopoly price for unit cost $$c$$; it maximizes $$(p-c)D(p)$$. If the two firms could get together and sign a contract, they would let firm 1 produce everything and charge $$p^m(c_1)$$. This would maximize industry profits. The "pie" could then be divided between the two firms through an arbitrary lump-sum transfer from firm 1 to firm 2. But suppose that it is illegal for firms to overtly agree and use side payments. We can determine the set of industry allocations that are efficient for the two firms, constrained by the fact that side payments are prohibited [...]. To this purpose, we fix a profit target $$\bar\Pi_2$$ for firm 2 in the interval $$[0, \Pi^m(c_2)]$$ and look for profit-sharing agreements (without transfers) in which both firms charge the same price $$p$$. They choose market shares $$s_1$$ and $$s_2$$ such that $$s_1 + s_2 = 1$$. The interpretation of these market shares is that firm $$i$$ produces exactly $$q_i = s_iD(p)$$; if $$s_i<1/2$$, the consumers who go to firm $$i$$ and are rationed buy from firm $$j$$. Given the profit target $$\bar\Pi_2$$ for firm 2, the efficient allocation is a choice of price $$p$$ and market shares $$s_1$$ and $$s_2$$ so as to solve:

\begin{align} \max_{p, s_1, s_2} & (p-c_1)s_1 D(p) \\ s.t. & (p-c_2)s_2 D(p) \geq \bar\Pi_2 \\ & s_1 + s_2 = 1 \end{align}

Suppose that the profit function (p-c)D(p) is concave for all $$p$$ and all $$c$$.

(i) After substituting $$s_1$$, obtain the first-order conditions Show that

$$p^m(c_1) \leq p \leq p^m(c_2)$$

Show that the objective function is concave (I have done this, idea is to realize from the FOCs that $$D(p)+(p-c_1)D(p) \leq 0$$ and that $$D(p)+ (p-c_2)D(p) \geq 0$$, and then you prove the result by contradiction by showing that it can't neither be that $$p^* < p^m(c_1)$$ nor that $$p > p^m(c_2)$$, then for the second point, it's just tedious algebra).

(ii) Show that

$$(p-c_1)D'(p) + D(p) + \frac{(c_2 - c_1)\bar\Pi_2}{(p-c_2)^2} = 0$$

This is the one I can't solve... Any help?

I reached this point:

$$(p-c_1)D'(p) + D(p) + \frac{(p-c_1)\bar\Pi_2}{(p-c_2)^2}\frac{(p-c_2)D'(p) + D(p)}{D(p)-\bar\Pi_2/((p-c_2)D(p))} = 0$$

For notational convenience, I'll write $$D$$ instead of $$D(p)$$.
We have the problem: $$\max_{s_2, p} (p - c_1)(1 - s_2) D \text{ s.t. } (p - c_2) s_2 D \ge \overline{\Pi}_2.$$ The first order conditions give (where $$\lambda$$ is the Lagrange multiplier): \begin{align*} &(1 - s_2) D + (p - c_1)(1 - s_2)D' - \lambda s_2 D - \lambda (p - c_2)s_2 D' = 0,\\ &-(p - c_1) D - \lambda (p - c_2) D = 0,\\ &(p - c_2) s_2 D = \overline{\Pi}_2. \end{align*} The second condition tells us that $$\lambda = - \dfrac{p - c_1}{p - c_2}$$. Substitute into the first condition.
$$(1 - s_2)D + (p - c_1)(1 - s_2) D' + \frac{(p - c_1)}{(p - c_2)} s_2 D + (p - c_1)s_2 D' = 0.$$ Gathering terms: $$\frac{D}{p - c_2}\left((1 - s_2)(p - c_2) + s_2(p - c_1)\right) + D'(p - c_1) = 0.$$
The third first order condition gives: $$s_2 = \frac{\overline{\Pi}_2}{(p - c_2)D},$$ So: $$(1 - s_2)(p - c_2) + s_2(p - c_1) = (p - c_2) - \frac{\overline{\Pi}_2}{D} + \frac{\overline{\Pi}_2 (p - c_1)}{D(p - c_2)}$$ Using this, we get the condition: \begin{align*} 0 &= \frac{D}{p - c_2}\left((p - c_2) - \frac{\overline{\Pi}_2}{D} + \frac{\overline{\Pi}_2 (p - c_1)}{D(p - c_2)}\right) + D'(p - c_1),\\ &= D - \frac{\overline{\Pi}_2}{p - c_2} + \frac{\overline{\Pi}_2(p - c_1)}{(p - c_2)^2} + D'(p - c_1),\\ &= D + D'(p - c_1) + \overline{\Pi}_2 \frac{c_2 - c_1}{(p - c_2)^2}. \end{align*}