1. **Problem statement:** We have a monopolistic firm producing two goods with inverse demand functions: $$P_1 = 60 - 2Q_1$$ $$P_2 = 80 - Q_2^2$$ and total cost function: $$TC = Q_1^2 + 2Q_1 Q_2 + Q_2^2$$ We want to find the output levels $Q_1$ and $Q_2$ that maximize profit, verify second-order conditions, and calculate maximum profit. 2. **Profit function:** Profit $\pi$ is total revenue minus total cost: $$\pi = P_1 Q_1 + P_2 Q_2 - TC$$ Substitute $P_1$ and $P_2$: $$\pi = (60 - 2Q_1)Q_1 + (80 - Q_2^2)Q_2 - (Q_1^2 + 2Q_1 Q_2 + Q_2^2)$$ 3. **Simplify profit function:** $$\pi = 60Q_1 - 2Q_1^2 + 80Q_2 - Q_2^3 - Q_1^2 - 2Q_1 Q_2 - Q_2^2$$ Combine like terms: $$\pi = 60Q_1 - 3Q_1^2 + 80Q_2 - Q_2^3 - 2Q_1 Q_2 - Q_2^2$$ 4. **First-order conditions (FOCs):** Set partial derivatives of $\pi$ with respect to $Q_1$ and $Q_2$ to zero. $$\frac{\partial \pi}{\partial Q_1} = 60 - 6Q_1 - 2Q_2 = 0$$ $$\frac{\partial \pi}{\partial Q_2} = 80 - 3Q_2^2 - 2Q_1 - 2Q_2 = 0$$ 5. **Solve the system:** From first equation: $$60 - 6Q_1 - 2Q_2 = 0 \implies 6Q_1 = 60 - 2Q_2 \implies Q_1 = \frac{60 - 2Q_2}{6} = 10 - \frac{Q_2}{3}$$ Substitute $Q_1$ into second equation: $$80 - 3Q_2^2 - 2\left(10 - \frac{Q_2}{3}\right) - 2Q_2 = 0$$ Simplify: $$80 - 3Q_2^2 - 20 + \frac{2Q_2}{3} - 2Q_2 = 0$$ $$60 - 3Q_2^2 - \frac{4Q_2}{3} = 0$$ Multiply both sides by 3: $$180 - 9Q_2^2 - 4Q_2 = 0$$ Rearranged: $$9Q_2^2 + 4Q_2 - 180 = 0$$ 6. **Solve quadratic for $Q_2$:** Use quadratic formula: $$Q_2 = \frac{-4 \pm \sqrt{4^2 - 4 \times 9 \times (-180)}}{2 \times 9} = \frac{-4 \pm \sqrt{16 + 6480}}{18} = \frac{-4 \pm \sqrt{6496}}{18}$$ Calculate $\sqrt{6496} \approx 80.6$: $$Q_2 = \frac{-4 \pm 80.6}{18}$$ Two solutions: $$Q_2 = \frac{-4 + 80.6}{18} \approx 4.31$$ $$Q_2 = \frac{-4 - 80.6}{18} \approx -4.69$$ Negative quantity is not feasible, so $Q_2 \approx 4.31$. 7. **Find $Q_1$:** $$Q_1 = 10 - \frac{4.31}{3} = 10 - 1.44 = 8.56$$ 8. **Second-order conditions:** Calculate Hessian matrix of second derivatives: $$\frac{\partial^2 \pi}{\partial Q_1^2} = -6$$ $$\frac{\partial^2 \pi}{\partial Q_2^2} = -6Q_2 - 2 = -6(4.31) - 2 = -25.86 - 2 = -27.86$$ $$\frac{\partial^2 \pi}{\partial Q_1 \partial Q_2} = \frac{\partial^2 \pi}{\partial Q_2 \partial Q_1} = -2$$ Hessian matrix: $$H = \begin{bmatrix} -6 & -2 \\ -2 & -27.86 \end{bmatrix}$$ Check definiteness: - Leading principal minor: $-6 < 0$ - Determinant: $(-6)(-27.86) - (-2)^2 = 167.16 - 4 = 163.16 > 0$ Since first principal minor is negative and determinant positive, Hessian is negative definite, confirming a local maximum. 9. **Calculate maximum profit:** Substitute $Q_1 = 8.56$, $Q_2 = 4.31$ into profit function: $$\pi = 60(8.56) - 3(8.56)^2 + 80(4.31) - (4.31)^3 - 2(8.56)(4.31) - (4.31)^2$$ Calculate stepwise: $$60 \times 8.56 = 513.6$$ $$3 \times 8.56^2 = 3 \times 73.3 = 219.9$$ $$80 \times 4.31 = 344.8$$ $$4.31^3 = 4.31 \times 4.31 \times 4.31 \approx 80.1$$ $$2 \times 8.56 \times 4.31 = 73.7$$ $$4.31^2 = 18.6$$ Sum all: $$\pi = 513.6 - 219.9 + 344.8 - 80.1 - 73.7 - 18.6 = 466.1$$ **Final answers:** - $Q_1 \approx 8.56$ - $Q_2 \approx 4.31$ - Profit is maximized with negative definite Hessian - Maximum profit $\pi_{max} \approx 466.1$