1. **Problem Statement:** Find the values of $x$ and $y$ that maximize the utility function $U(x,y) = x^y$ subject to the budget constraint $3x + 4y = 72$ using the Lagrange multiplier method. 2. **Set up the Lagrangian:** $$\mathcal{L}(x,y,\lambda) = x^y + \lambda (72 - 3x - 4y)$$ 3. **Find partial derivatives and set to zero:** $$\frac{\partial \mathcal{L}}{\partial x} = y x^{y-1} - 3\lambda = 0$$ $$\frac{\partial \mathcal{L}}{\partial y} = x^y \ln(x) - 4\lambda = 0$$ $$\frac{\partial \mathcal{L}}{\partial \lambda} = 72 - 3x - 4y = 0$$ 4. **From the first two equations:** $$y x^{y-1} = 3\lambda \quad (1)$$ $$x^y \ln(x) = 4\lambda \quad (2)$$ Divide (2) by (1): $$\frac{x^y \ln(x)}{y x^{y-1}} = \frac{4\lambda}{3\lambda} \Rightarrow \frac{x \ln(x)}{y} = \frac{4}{3}$$ So, $$y = \frac{3 x \ln(x)}{4}$$ 5. **Substitute $y$ into budget constraint:** $$3x + 4y = 72 \Rightarrow 3x + 4 \times \frac{3 x \ln(x)}{4} = 72$$ Simplify: $$3x + 3x \ln(x) = 72$$ $$3x (1 + \ln(x)) = 72$$ $$x (1 + \ln(x)) = 24$$ 6. **Solve for $x$ numerically:** Try $x=5$: $$5 (1 + \ln(5)) \approx 5 (1 + 1.609) = 5 \times 2.609 = 13.045 < 24$$ Try $x=10$: $$10 (1 + \ln(10)) = 10 (1 + 2.3026) = 10 \times 3.3026 = 33.026 > 24$$ Try $x=7$: $$7 (1 + \ln(7)) = 7 (1 + 1.9459) = 7 \times 2.9459 = 20.621 < 24$$ Try $x=8$: $$8 (1 + \ln(8)) = 8 (1 + 2.0794) = 8 \times 3.0794 = 24.635 > 24$$ Try $x=7.8$: $$7.8 (1 + \ln(7.8)) = 7.8 (1 + 2.0541) = 7.8 \times 3.0541 = 23.82 < 24$$ Try $x=7.85$: $$7.85 (1 + \ln(7.85)) = 7.85 (1 + 2.061) = 7.85 \times 3.061 = 24.01 \approx 24$$ So, $x^* \approx 7.85$ 7. **Find $y^*$:** $$y^* = \frac{3 x^* \ln(x^*)}{4} = \frac{3 \times 7.85 \times 2.061}{4} = \frac{48.5}{4} = 12.13$$ 8. **Find $\lambda^*$:** From equation (1): $$\lambda = \frac{y x^{y-1}}{3} = \frac{12.13 \times 7.85^{12.13 -1}}{3}$$ Calculate $7.85^{11.13}$ approximately: $$\ln(7.85^{11.13}) = 11.13 \times \ln(7.85) = 11.13 \times 2.061 = 22.93$$ So, $$7.85^{11.13} = e^{22.93} \approx 9.3 \times 10^9$$ Then, $$\lambda^* \approx \frac{12.13 \times 9.3 \times 10^9}{3} = 3.76 \times 10^{10}$$ 9. **Economic interpretation of $\lambda$:** The Lagrange multiplier $\lambda$ represents the marginal utility of income, i.e., how much the maximum utility increases if the consumer's income increases by one unit. 10. **Estimate change in utility if income increases from 72 to 73:** $$\Delta U \approx \lambda^* \times \Delta I = 3.76 \times 10^{10} \times 1 = 3.76 \times 10^{10}$$ **Final answers:** $$x^* \approx 7.85, \quad y^* \approx 12.13, \quad \lambda^* \approx 3.76 \times 10^{10}$$