1. Problem 1: Maximize production given budget constraints. 1.1 The production function is $$N(x,y) = 10 x^{0.6} y^{0.4}$$ where $x$ is labor units and $y$ is capital units. 1.2 The budget constraint for part (a) is given by the total cost of labor and capital: $$30x + 60y = 300000$$. 1.3 Form the Lagrangian function: $$\mathcal{L}(x,y,\lambda) = 10 x^{0.6} y^{0.4} - \lambda (30x + 60y - 300000)$$. 1.4 Take partial derivatives and set them to zero: $$\frac{\partial \mathcal{L}}{\partial x} = 6 x^{-0.4} y^{0.4} - 30 \lambda = 0$$ $$\frac{\partial \mathcal{L}}{\partial y} = 4 x^{0.6} y^{-0.6} - 60 \lambda = 0$$ $$\frac{\partial \mathcal{L}}{\partial \lambda} = 30x + 60y - 300000 = 0$$ 1.5 From the first two equations, express $\lambda$: $$\lambda = \frac{6 x^{-0.4} y^{0.4}}{30} = \frac{4 x^{0.6} y^{-0.6}}{60}$$ Simplify: $$\frac{6 x^{-0.4} y^{0.4}}{30} = \frac{4 x^{0.6} y^{-0.6}}{60}$$ Multiply both sides by denominators: $$6 x^{-0.4} y^{0.4} \times 60 = 4 x^{0.6} y^{-0.6} \times 30$$ $$360 x^{-0.4} y^{0.4} = 120 x^{0.6} y^{-0.6}$$ Divide by 120: $$3 x^{-0.4} y^{0.4} = x^{0.6} y^{-0.6}$$ Rewrite powers: $$3 = x^{0.6 + 0.4} y^{-0.6 - 0.4} = x^{1.0} y^{-1.0} = \frac{x}{y}$$ So, $$\frac{x}{y} = 3 \Rightarrow x = 3y$$ 1.6 Substitute into budget constraint: $$30(3y) + 60y = 300000$$ $$90y + 60y = 300000$$ $$150y = 300000$$ $$y = 2000$$ Therefore, $$x = 3 \times 2000 = 6000$$ 1.7 Calculate maximum production: $$N(6000, 2000) = 10 \times 6000^{0.6} \times 2000^{0.4}$$ Calculate powers: $$6000^{0.6} = e^{0.6 \ln 6000}, 2000^{0.4} = e^{0.4 \ln 2000}$$ Using approximate logs: $\ln 6000 \approx 8.6995$, $\ln 2000 \approx 7.6009$ $$6000^{0.6} \approx e^{5.2197} \approx 185.59$$ $$2000^{0.4} \approx e^{3.04} \approx 20.95$$ So maximum production: $$N \approx 10 \times 185.59 \times 20.95 = 38884.81$$ 2. Problem 1(b): Find marginal productivity of money and estimate increase for extra $80000. 2.1 The Lagrange multiplier $\lambda$ from equations corresponds to the marginal increase in production per additional unit of money spent. From above: $$\lambda = \frac{6 x^{-0.4} y^{0.4}}{30} = \frac{6 \times 6000^{-0.4} \times 2000^{0.4}}{30}$$ Calculate powers: $$6000^{-0.4} = \frac{1}{6000^{0.4}} = e^{-0.4 \times 8.6995} = e^{-3.4798} \approx 0.0308$$ $$2000^{0.4} = 20.95$$ Plug values: $$\lambda = \frac{6 \times 0.0308 \times 20.95}{30} = \frac{3.871}{30} = 0.129$$ 2.2 Marginal productivity of money is approx 0.129 units of production per unit money. 2.3 Increase in production for additional 80000: $$\Delta N = \lambda \times 80000 = 0.129 \times 80000 = 10320$$ 3. Problem 2: Maximize output $N(x,y) = 4xy - 8x$ subject to budget 3.1 Budget constraint: $$x + y = 60$$ since $x$ and $y$ are in thousands and total budget is 60000. 3.2 Form Lagrangian: $$\mathcal{L} = 4xy - 8x - \lambda (x + y - 60)$$ 3.3 Partial derivatives and zero: $$\frac{\partial \mathcal{L}}{\partial x} = 4y - 8 - \lambda = 0$$ $$\frac{\partial \mathcal{L}}{\partial y} = 4x - \lambda = 0$$ $$x + y = 60$$ 3.4 From derivatives: $$4y - 8 = \lambda$$ $$4x = \lambda$$ Set equal: $$4y - 8 = 4x$$ $$4y - 4x = 8$$ $$y - x = 2$$ 3.5 Combine with budget: $$x + y = 60$$ $$y - x = 2$$ Add both: $$2y = 62 \Rightarrow y = 31$$ Then, $$x = 60 - 31 = 29$$ 3.6 Calculate maximum output: $$N(29,31) = 4 \times 29 \times 31 - 8 \times 29 = 3596 - 232 = 3364$$ Final answers: - Problem 1 (a) max production approx 38885 units with $x=6000$, $y=2000$ - Problem 1 (b) marginal productivity of money approx 0.129 units/money unit, increase of 10320 units for extra 80000 - Problem 2 max output 3364 units with $x=29$, $y=31$