1. **Problem 1:** A company produces two products X and Y with a labor constraint of 240 hours. Product X requires 4 hours per unit and Y requires 2 hours per unit. Profit per unit of X is 10, profit per unit of Y is 8. Maximize profit under the labor constraint. 2. **Step 1:** Define variables. Let $x$ = units of product X, $y$ = units of product Y. 3. **Step 2:** Set up constraints. Labor constraint: $$4x + 2y \leq 240$$ 4. **Step 3:** Define profit function. $$P = 10x + 8y$$ 5. **Step 4:** Use the labor constraint as equality to maximize profit (since more production means more profit): $$4x + 2y = 240$$ 6. **Step 5:** Express $y$ in terms of $x$: $$2y = 240 - 4x \Rightarrow y = 120 - 2x$$ 7. **Step 6:** Substitute $y$ into profit function: $$P = 10x + 8(120 - 2x) = 10x + 960 - 16x = 960 - 6x$$ 8. **Step 7:** Maximize profit by choosing $x$. Since profit decreases as $x$ increases (coefficient is -6), profit is maximized at the smallest $x$ that satisfies production non-negativity. 9. **Step 8:** Check if $x=0$ is valid solution: $$y = 120 - 2(0) = 120$$ Labor use: $$4(0) + 2(120) = 240$$ 10. **Step 9:** Profit at $x=0, y=120$: $$P=10(0) + 8(120)=960$$ Since profit decreases as $x$ increases, producing $0$ units of X and 120 units of Y maximizes profit. 11. **Problem 2:** A firm produces X and Y with labor and capital constraints. Profit per unit X = 15, for Y = 20. Labor constraint: $$2x + 3y \leq 240$$ Capital constraint: $$3x + 2y \leq 300$$ 12. **Step 1:** Define the objective function: $$P = 15x + 20y$$ 13. **Step 2:** Use Lagrangian method. Set up Lagrangian: $$\mathcal{L} = 15x + 20y + \lambda(240 - 2x - 3y) + \mu(300 - 3x - 2y)$$ 14. **Step 3:** Compute partial derivatives and set to zero: $$\frac{\partial \mathcal{L}}{\partial x} = 15 - 2\lambda - 3\mu = 0$$ $$\frac{\partial \mathcal{L}}{\partial y} = 20 - 3\lambda - 2\mu = 0$$ $$\frac{\partial \mathcal{L}}{\partial \lambda} = 240 - 2x - 3y = 0$$ $$\frac{\partial \mathcal{L}}{\partial \mu} = 300 - 3x - 2y = 0$$ 15. **Step 4:** Solve the system: From first two equations: $$15 = 2\lambda + 3\mu$$ $$20 = 3\lambda + 2\mu$$ 16. **Step 5:** Multiply first equation by 3: $$45 = 6\lambda + 9\mu$$ Multiply second by 2: $$40 = 6\lambda + 4\mu$$ Subtract second from first: $$5 = 5\mu \Rightarrow \mu = 1$$ 17. **Step 6:** Substitute $\mu$ back to first equation: $$15 = 2\lambda + 3(1) \Rightarrow 15 = 2\lambda + 3 \Rightarrow 2\lambda = 12 \Rightarrow \lambda = 6$$ 18. **Step 7:** Use constraints equations: $$240 = 2x + 3y$$ $$300 = 3x + 2y$$ 19. **Step 8:** Solve for $x,y$. Multiply first by 3: $$720 = 6x + 9y$$ Multiply second by 2: $$600 = 6x + 4y$$ Subtract second from first: $$120 = 5y \Rightarrow y = 24$$ 20. **Step 9:** Substitute $y=24$ into first constraint: $$240 = 2x + 3(24) = 2x + 72 \Rightarrow 2x=168 \Rightarrow x=84$$ 21. **Step 10:** Calculate profit: $$P=15(84) + 20(24) = 1260 + 480 = 1740$$ **Final answers:** 1. Produce $x=0$, $y=120$ units to maximize profit $960$. 2. Produce $x=84$, $y=24$ units to maximize profit $1740$ using Lagrangian method.