1. **State the problem:** We are given a Cobb-Douglas production function $$Q = K^{0.4}L^{0.5}$$ subject to the constraint $$120 - 3K - 4L = 0$$. We want to find the values of $K$ and $L$ that maximize production $Q$ under this constraint. 2. **Write the constraint in a usable form:** From $$120 - 3K - 4L = 0$$, we have $$3K + 4L = 120$$. 3. **Set up the Lagrangian:** Define the Lagrangian function $$\mathcal{L} = K^{0.4}L^{0.5} + \lambda(120 - 3K - 4L)$$. 4. **Find partial derivatives and set to zero:** $$\frac{\partial \mathcal{L}}{\partial K} = 0.4K^{-0.6}L^{0.5} - 3\lambda = 0$$ $$\frac{\partial \mathcal{L}}{\partial L} = 0.5K^{0.4}L^{-0.5} - 4\lambda = 0$$ $$\frac{\partial \mathcal{L}}{\partial \lambda} = 120 - 3K - 4L = 0$$ 5. **Express $\lambda$ from the first two equations:** From the first: $$\lambda = \frac{0.4K^{-0.6}L^{0.5}}{3}$$ From the second: $$\lambda = \frac{0.5K^{0.4}L^{-0.5}}{4}$$ 6. **Set the two expressions for $\lambda$ equal:** $$\frac{0.4K^{-0.6}L^{0.5}}{3} = \frac{0.5K^{0.4}L^{-0.5}}{4}$$ 7. **Simplify the equation:** Multiply both sides by 12 to clear denominators: $$12 \times \frac{0.4K^{-0.6}L^{0.5}}{3} = 12 \times \frac{0.5K^{0.4}L^{-0.5}}{4}$$ $$4 \times 0.4K^{-0.6}L^{0.5} = 3 \times 0.5K^{0.4}L^{-0.5}$$ $$1.6K^{-0.6}L^{0.5} = 1.5K^{0.4}L^{-0.5}$$ 8. **Divide both sides by $K^{-0.6}L^{-0.5}$:** $$\cancel{K^{-0.6}}L^{0.5} \div \cancel{K^{-0.6}}L^{-0.5} = L^{0.5 - (-0.5)} = L^{1}$$ $$K^{0.4} \div K^{-0.6} = K^{0.4 - (-0.6)} = K^{1}$$ So the equation becomes: $$1.6L = 1.5K$$ 9. **Solve for $L$ in terms of $K$:** $$L = \frac{1.5}{1.6}K = \frac{15}{16}K$$ 10. **Substitute $L$ into the constraint:** $$3K + 4L = 120$$ $$3K + 4 \times \frac{15}{16}K = 120$$ $$3K + \frac{60}{16}K = 120$$ $$3K + 3.75K = 120$$ $$6.75K = 120$$ 11. **Solve for $K$:** $$K = \frac{120}{6.75} = \frac{120}{6.75} = 17.777...$$ 12. **Find $L$:** $$L = \frac{15}{16} \times 17.777... = 16.666...$$ 13. **Calculate maximum production $Q$:** $$Q = K^{0.4}L^{0.5} = (17.777...)^{0.4} \times (16.666...)^{0.5}$$ Calculate each term: $$17.777...^{0.4} \approx 3.039$$ $$16.666...^{0.5} = \sqrt{16.666...} \approx 4.082$$ So, $$Q \approx 3.039 \times 4.082 = 12.41$$ **Final answer:** $$K \approx 17.78, \quad L \approx 16.67, \quad Q_{max} \approx 12.41$$