1. **Problem Statement:** A company produces widgets with production function $$Q = \alpha L^{0.5} K^{0.5}$$ where $L$ is labor and $K$ is capital. The market price per widget is 100, labor cost is 5 per unit, and capital cost is 8 per unit. We want to find the profit-maximizing quantity $Q^*$ and associated profit using the Lagrangian multiplier method. 2. **Formulas and Setup:** Profit $$\pi = pQ - wL - rK$$ where $p=100$, $w=5$, $r=8$, and $$Q = \alpha L^{0.5} K^{0.5}$$. We maximize profit subject to the production function constraint by setting up the Lagrangian: $$\mathcal{L} = 100Q - 5L - 8K + \lambda (\alpha L^{0.5} K^{0.5} - Q)$$ 3. **First Order Conditions:** Take partial derivatives and set to zero: $$\frac{\partial \mathcal{L}}{\partial L} = -5 + \lambda \alpha \cdot 0.5 L^{-0.5} K^{0.5} = 0$$ $$\frac{\partial \mathcal{L}}{\partial K} = -8 + \lambda \alpha \cdot 0.5 L^{0.5} K^{-0.5} = 0$$ $$\frac{\partial \mathcal{L}}{\partial Q} = 100 - \lambda = 0 \implies \lambda = 100$$ $$\frac{\partial \mathcal{L}}{\partial \lambda} = \alpha L^{0.5} K^{0.5} - Q = 0$$ 4. **Solve for $L$ and $K$:** From the first two equations: $$-5 + 100 \alpha \cdot 0.5 L^{-0.5} K^{0.5} = 0 \implies 5 = 50 \alpha K^{0.5} L^{-0.5}$$ $$-8 + 100 \alpha \cdot 0.5 L^{0.5} K^{-0.5} = 0 \implies 8 = 50 \alpha L^{0.5} K^{-0.5}$$ Divide the two equations: $$\frac{5}{8} = \frac{50 \alpha K^{0.5} L^{-0.5}}{50 \alpha L^{0.5} K^{-0.5}} = \frac{K^{0.5} L^{-0.5}}{L^{0.5} K^{-0.5}} = \frac{K}{L}$$ So, $$\frac{K}{L} = \frac{5}{8}$$ 5. **Express $K$ in terms of $L$:** $$K = \frac{5}{8} L$$ 6. **Use production constraint:** $$Q = \alpha L^{0.5} K^{0.5} = \alpha (L K)^{0.5} = \alpha \sqrt{L \cdot \frac{5}{8} L} = \alpha L \sqrt{\frac{5}{8}}$$ 7. **Profit function:** $$\pi = 100 Q - 5 L - 8 K = 100 \alpha L \sqrt{\frac{5}{8}} - 5 L - 8 \cdot \frac{5}{8} L = 100 \alpha L \sqrt{\frac{5}{8}} - 5 L - 5 L = 100 \alpha L \sqrt{\frac{5}{8}} - 10 L$$ 8. **Maximize profit w.r.t $L$:** Set derivative to zero: $$\frac{d\pi}{dL} = 100 \alpha \sqrt{\frac{5}{8}} - 10 = 0$$ Solve for $\alpha$: $$100 \alpha \sqrt{\frac{5}{8}} = 10 \implies \alpha = \frac{10}{100 \sqrt{5/8}} = \frac{0.1}{\sqrt{5/8}}$$ Calculate numeric value: $$\sqrt{\frac{5}{8}} = \sqrt{0.625} \approx 0.7906$$ So, $$\alpha \approx \frac{0.1}{0.7906} \approx 0.1265$$ 9. **Calculate $L^*$ and $K^*$:** From step 7, profit is linear in $L$ with slope zero at optimum, so any $L$ is optimal if $\alpha$ is fixed. Usually, $\alpha$ is given, so assume $\alpha=1$ for calculation. Then, $$\frac{d\pi}{dL} = 100 \cdot 1 \cdot 0.7906 - 10 = 79.06 - 10 = 69.06 > 0$$ Profit increases with $L$, so no finite maximum unless constrained. 10. **Final answers:** - Profit maximizing ratio: $$\frac{K}{L} = \frac{5}{8}$$ - Profit maximizing quantity: $$Q^* = \alpha L^{0.5} K^{0.5} = \alpha L \sqrt{\frac{5}{8}}$$ - Profit: $$\pi = 100 Q^* - 5 L - 8 K$$ **(b) and (c) require qualitative and design answers based on above results, but only first question solved here.**