Utility Maximization 066Edd

1. **State the problem:** We want to maximize the utility function $$u = xy + x$$ subject to the budget constraint $$6x + 2y = 110$$. 2. **Set up the Lagrangian:** We introduce a Lagrange multiplier $\lambda$ for the constraint: $$\mathcal{L} = xy + x - \lambda(6x + 2y - 110)$$ 3. **Find the first order conditions (FOCs):** Take partial derivatives and set them equal to zero: $$\frac{\partial \mathcal{L}}{\partial x} = y + 1 - 6\lambda = 0$$ $$\frac{\partial \mathcal{L}}{\partial y} = x - 2\lambda = 0$$ $$\frac{\partial \mathcal{L}}{\partial \lambda} = -(6x + 2y - 110) = 0$$ 4. **Solve the system:** From the second equation: $$x = 2\lambda$$ From the first equation: $$y + 1 = 6\lambda \implies y = 6\lambda - 1$$ Substitute $x$ and $y$ into the budget constraint: $$6(2\lambda) + 2(6\lambda - 1) = 110$$ $$12\lambda + 12\lambda - 2 = 110$$ $$24\lambda = 112$$ $$\lambda = \frac{112}{24} = \frac{14}{3}$$ 5. **Find $x$ and $y$ values:** $$x = 2 \times \frac{14}{3} = \frac{28}{3}$$ $$y = 6 \times \frac{14}{3} - 1 = 28 - 1 = 27$$ 6. **Check the critical values:** $$x = \frac{28}{3} \approx 9.33, \quad y = 27$$ 7. **Use the bordered Hessian to check second order conditions:** The bordered Hessian matrix for two variables with one constraint is: $$H = \begin{bmatrix} 0 & \frac{\partial g}{\partial x} & \frac{\partial g}{\partial y} \\ \frac{\partial g}{\partial x} & \frac{\partial^2 \mathcal{L}}{\partial x^2} & \frac{\partial^2 \mathcal{L}}{\partial x \partial y} \\ \frac{\partial g}{\partial y} & \frac{\partial^2 \mathcal{L}}{\partial y \partial x} & \frac{\partial^2 \mathcal{L}}{\partial y^2} \end{bmatrix}$$ where $g(x,y) = 6x + 2y - 110$. Calculate derivatives: $$\frac{\partial g}{\partial x} = 6, \quad \frac{\partial g}{\partial y} = 2$$ $$\frac{\partial^2 \mathcal{L}}{\partial x^2} = 0$$ $$\frac{\partial^2 \mathcal{L}}{\partial y^2} = 0$$ $$\frac{\partial^2 \mathcal{L}}{\partial x \partial y} = \frac{\partial^2 \mathcal{L}}{\partial y \partial x} = 1$$ So, $$H = \begin{bmatrix} 0 & 6 & 2 \\ 6 & 0 & 1 \\ 2 & 1 & 0 \end{bmatrix}$$ 8. **Calculate determinants for bordered Hessian test:** - The determinant of the full matrix $H$: $$\det(H) = 0 \times (0 \times 0 - 1 \times 1) - 6 \times (6 \times 0 - 1 \times 2) + 2 \times (6 \times 1 - 0 \times 2)$$ $$= 0 - 6 \times (-2) + 2 \times 6 = 12 + 12 = 24 > 0$$ - The leading principal minor (top-left $1 \times 1$ block) is $0$. Since the bordered Hessian determinant is positive and the problem is a maximization with one constraint, the critical point is a maximum. **Final answer:** $$x = \frac{28}{3}, \quad y = 27$$ These values maximize the utility subject to the budget constraint.