1. **Problem statement:** Two firms produce quantities $q_1$ and $q_2$ facing a market demand $P = 100 - 1.5Q$ where $Q = q_1 + q_2$. Their cost functions are $C_1 = 100q_1$ and $C_2 = 80 + 0.5q_2^2$. We want to find the profit of each firm when they collude to maximize joint profit. 2. **Formula and approach:** Under collusion, firms maximize total profit $\Pi = \Pi_1 + \Pi_2$ where $$\Pi_1 = Pq_1 - C_1 = (100 - 1.5Q)q_1 - 100q_1$$ $$\Pi_2 = Pq_2 - C_2 = (100 - 1.5Q)q_2 - (80 + 0.5q_2^2)$$ Total profit: $$\Pi = (100 - 1.5(q_1 + q_2))(q_1 + q_2) - 100q_1 - 80 - 0.5q_2^2$$ 3. **Rewrite total profit:** $$\Pi = (100 - 1.5Q)Q - 100q_1 - 80 - 0.5q_2^2$$ where $Q = q_1 + q_2$. 4. **Maximize total profit:** Take partial derivatives w.r.t. $q_1$ and $q_2$ and set to zero. $$\frac{\partial \Pi}{\partial q_1} = \frac{\partial}{\partial q_1}[(100 - 1.5Q)Q - 100q_1 - 80 - 0.5q_2^2] = 0$$ $$\frac{\partial \Pi}{\partial q_2} = \frac{\partial}{\partial q_2}[(100 - 1.5Q)Q - 100q_1 - 80 - 0.5q_2^2] = 0$$ 5. **Calculate derivatives:** Since $Q = q_1 + q_2$, $$\frac{\partial}{\partial q_1}[(100 - 1.5Q)Q] = (100 - 1.5Q) \cdot 1 + Q \cdot (-1.5) = 100 - 3Q$$ So, $$\frac{\partial \Pi}{\partial q_1} = 100 - 3Q - 100 = -3Q = 0 \implies Q = 0$$ This suggests a boundary solution, so check carefully. For $q_2$: $$\frac{\partial}{\partial q_2}[(100 - 1.5Q)Q] = 100 - 3Q$$ Then, $$\frac{\partial \Pi}{\partial q_2} = 100 - 3Q - q_2 = 0$$ 6. **Re-express derivatives carefully:** Actually, the derivative of $(100 - 1.5Q)Q$ w.r.t. $q_i$ is: $$\frac{\partial}{\partial q_i}[(100 - 1.5Q)Q] = (100 - 1.5Q) \cdot 1 + Q \cdot (-1.5) = 100 - 3Q$$ So for $q_1$: $$\frac{\partial \Pi}{\partial q_1} = 100 - 3Q - 100 = -3Q = 0 \implies Q=0$$ For $q_2$: $$\frac{\partial \Pi}{\partial q_2} = 100 - 3Q - q_2 = 0$$ 7. **This inconsistency suggests re-deriving carefully:** Rewrite total profit: $$\Pi = (100 - 1.5(q_1 + q_2))(q_1 + q_2) - 100q_1 - 80 - 0.5q_2^2$$ Expand: $$\Pi = 100(q_1 + q_2) - 1.5(q_1 + q_2)^2 - 100q_1 - 80 - 0.5q_2^2$$ Simplify: $$\Pi = 100q_1 + 100q_2 - 1.5(q_1^2 + 2q_1q_2 + q_2^2) - 100q_1 - 80 - 0.5q_2^2$$ $$\Pi = 100q_2 - 1.5q_1^2 - 3q_1q_2 - 1.5q_2^2 - 80 - 0.5q_2^2$$ $$\Pi = 100q_2 - 1.5q_1^2 - 3q_1q_2 - 2q_2^2 - 80$$ 8. **Partial derivatives:** $$\frac{\partial \Pi}{\partial q_1} = -3q_1 - 3q_2 = 0$$ $$\frac{\partial \Pi}{\partial q_2} = 100 - 3q_1 - 4q_2 = 0$$ 9. **Solve system:** From first: $$-3q_1 - 3q_2 = 0 \implies q_1 = -q_2$$ Substitute into second: $$100 - 3(-q_2) - 4q_2 = 0 \implies 100 + 3q_2 - 4q_2 = 0 \implies 100 - q_2 = 0 \implies q_2 = 100$$ Then, $$q_1 = -100$$ Negative output is not feasible, so check constraints or re-examine cost functions. 10. **Check cost functions:** Given $C_1 = 100q_1$ (linear) and $C_2 = 80 + 0.5q_2^2$ (quadratic), $q_1$ must be non-negative. 11. **Reconsider derivative sign:** In step 8, derivative w.r.t. $q_1$ is: $$\frac{\partial \Pi}{\partial q_1} = 100 - 3q_1 - 3q_2 - 100 = -3q_1 - 3q_2$$ But we missed the $100q_1$ term in expansion. Let's re-derive carefully: Original profit: $$\Pi = (100 - 1.5Q)Q - 100q_1 - 80 - 0.5q_2^2$$ Expand: $$\Pi = 100Q - 1.5Q^2 - 100q_1 - 80 - 0.5q_2^2$$ Recall $Q = q_1 + q_2$: $$\Pi = 100(q_1 + q_2) - 1.5(q_1 + q_2)^2 - 100q_1 - 80 - 0.5q_2^2$$ Simplify: $$\Pi = 100q_1 + 100q_2 - 1.5(q_1^2 + 2q_1q_2 + q_2^2) - 100q_1 - 80 - 0.5q_2^2$$ $$\Pi = 100q_2 - 1.5q_1^2 - 3q_1q_2 - 1.5q_2^2 - 80 - 0.5q_2^2$$ $$\Pi = 100q_2 - 1.5q_1^2 - 3q_1q_2 - 2q_2^2 - 80$$ 12. **Partial derivatives:** $$\frac{\partial \Pi}{\partial q_1} = -3q_1 - 3q_2 = 0$$ $$\frac{\partial \Pi}{\partial q_2} = 100 - 3q_1 - 4q_2 = 0$$ 13. **Solve system:** From first: $$-3q_1 - 3q_2 = 0 \implies q_1 = -q_2$$ Substitute into second: $$100 - 3(-q_2) - 4q_2 = 0 \implies 100 + 3q_2 - 4q_2 = 0 \implies 100 - q_2 = 0 \implies q_2 = 100$$ Then, $$q_1 = -100$$ Again negative $q_1$ is infeasible. 14. **Conclusion:** The negative $q_1$ suggests the cost function $C_1 = 100q_1$ is too high to produce positive output under collusion. The firm 1 produces zero. Set $q_1 = 0$, then maximize profit w.r.t. $q_2$: $$\Pi = 100q_2 - 2q_2^2 - 80$$ Derivative: $$\frac{d\Pi}{dq_2} = 100 - 4q_2 = 0 \implies q_2 = 25$$ 15. **Calculate profits:** Price: $$P = 100 - 1.5(0 + 25) = 100 - 37.5 = 62.5$$ Firm 1 profit: $$\Pi_1 = Pq_1 - C_1 = 62.5 \times 0 - 100 \times 0 = 0$$ Firm 2 profit: $$\Pi_2 = Pq_2 - C_2 = 62.5 \times 25 - (80 + 0.5 \times 25^2) = 1562.5 - (80 + 312.5) = 1562.5 - 392.5 = 1170$$ **Final answer:** $$q_1 = 0, q_2 = 25$$ $$\Pi_1 = 0, \Pi_2 = 1170$$