1. **Problem Statement:** We have two firms in a duopoly with demand functions: $$q_1 = 100 - 2p_1 + p_2$$ $$q_2 = 100 - 2p_2 + p_1$$ and constant marginal costs. We want to find the Bertrand equilibrium prices. 2. **Bertrand Equilibrium Concept:** In Bertrand competition, firms choose prices to maximize profits given the other firm's price. The equilibrium occurs where neither firm can increase profit by changing its price. 3. **Profit Functions:** Profit for firm 1: $$\pi_1 = (p_1 - c_1) q_1 = (p_1 - c_1)(100 - 2p_1 + p_2)$$ Profit for firm 2: $$\pi_2 = (p_2 - c_2) q_2 = (p_2 - c_2)(100 - 2p_2 + p_1)$$ ### (a) Both firms have marginal cost $c_1 = c_2 = 10$ 4. **First-order conditions:** Take derivative of profits w.r.t own price and set to zero. For firm 1: $$\frac{\partial \pi_1}{\partial p_1} = (100 - 2p_1 + p_2) + (p_1 - 10)(-2) = 0$$ Simplify: $$100 - 2p_1 + p_2 - 2p_1 + 20 = 0$$ $$120 + p_2 - 4p_1 = 0$$ $$4p_1 = 120 + p_2$$ $$p_1 = \frac{120 + p_2}{4}$$ For firm 2: $$\frac{\partial \pi_2}{\partial p_2} = (100 - 2p_2 + p_1) + (p_2 - 10)(-2) = 0$$ Simplify: $$100 - 2p_2 + p_1 - 2p_2 + 20 = 0$$ $$120 + p_1 - 4p_2 = 0$$ $$4p_2 = 120 + p_1$$ $$p_2 = \frac{120 + p_1}{4}$$ 5. **Solve the system:** Substitute $p_2$ into $p_1$: $$p_1 = \frac{120 + \frac{120 + p_1}{4}}{4} = \frac{120 + 30 + \frac{p_1}{4}}{4} = \frac{150 + \frac{p_1}{4}}{4}$$ Multiply both sides by 4: $$4p_1 = 150 + \frac{p_1}{4}$$ Multiply both sides by 4 again: $$16p_1 = 600 + p_1$$ $$16p_1 - p_1 = 600$$ $$15p_1 = 600$$ $$p_1 = 40$$ Then: $$p_2 = \frac{120 + 40}{4} = \frac{160}{4} = 40$$ 6. **Equilibrium prices:** $$p_1^* = 40, \quad p_2^* = 40$$ ### (b) Firm 1's marginal cost $c_1 = 30$, Firm 2's marginal cost $c_2 = 10$ 7. **First-order conditions:** For firm 1: $$\frac{\partial \pi_1}{\partial p_1} = (100 - 2p_1 + p_2) + (p_1 - 30)(-2) = 0$$ Simplify: $$100 - 2p_1 + p_2 - 2p_1 + 60 = 0$$ $$160 + p_2 - 4p_1 = 0$$ $$4p_1 = 160 + p_2$$ $$p_1 = \frac{160 + p_2}{4}$$ For firm 2: $$\frac{\partial \pi_2}{\partial p_2} = (100 - 2p_2 + p_1) + (p_2 - 10)(-2) = 0$$ Simplify: $$100 - 2p_2 + p_1 - 2p_2 + 20 = 0$$ $$120 + p_1 - 4p_2 = 0$$ $$4p_2 = 120 + p_1$$ $$p_2 = \frac{120 + p_1}{4}$$ 8. **Solve the system:** Substitute $p_2$ into $p_1$: $$p_1 = \frac{160 + \frac{120 + p_1}{4}}{4} = \frac{160 + 30 + \frac{p_1}{4}}{4} = \frac{190 + \frac{p_1}{4}}{4}$$ Multiply both sides by 4: $$4p_1 = 190 + \frac{p_1}{4}$$ Multiply both sides by 4: $$16p_1 = 760 + p_1$$ $$16p_1 - p_1 = 760$$ $$15p_1 = 760$$ $$p_1 = \frac{760}{15} = 50.67$$ Then: $$p_2 = \frac{120 + 50.67}{4} = \frac{170.67}{4} = 42.67$$ 9. **Equilibrium prices:** $$p_1^* = 50.67, \quad p_2^* = 42.67$$ **Final answers:** (a) $p_1 = 40$, $p_2 = 40$ (b) $p_1 = 50.67$, $p_2 = 42.67$