1. **Problem Statement:** A firm produces two products, X and Y. Each kilogram of X requires 8 hours of machine time (T) and 10 MJ of energy (K). Each kilogram of Y requires 12 hours of T and 6 MJ of K. The profit per unit for X is 20.50 and for Y is 31. The firm has a maximum of 480 hours of T and 300 units of K available. We need to determine the production quantities of X and Y to maximize profit. 2. **Define variables:** Let $x$ = kilograms of product X produced Let $y$ = kilograms of product Y produced 3. **Formulate the objective function:** Maximize profit $P = 20.50x + 31y$ 4. **Formulate the constraints:** Machine time constraint: $8x + 12y \leq 480$ Energy constraint: $10x + 6y \leq 300$ Non-negativity constraints: $x \geq 0$, $y \geq 0$ 5. **Solve constraints for boundary lines:** From $8x + 12y = 480$, divide by 4: $2x + 3y = 120$ or $y = \frac{120 - 2x}{3}$ From $10x + 6y = 300$, divide by 2: $5x + 3y = 150$ or $y = \frac{150 - 5x}{3}$ 6. **Find corner points of feasible region:** - Intersection with axes: - For $8x + 12y \leq 480$: - If $x=0$, $y=40$ - If $y=0$, $x=60$ - For $10x + 6y \leq 300$: - If $x=0$, $y=50$ - If $y=0$, $x=30$ - Intersection of two constraints: Solve system: $$\begin{cases} 8x + 12y = 480 \\ 10x + 6y = 300 \end{cases}$$ Multiply second equation by 2: $$20x + 12y = 600$$ Subtract first equation: $$(20x + 12y) - (8x + 12y) = 600 - 480$$ $$12x = 120 \Rightarrow x = 10$$ Substitute $x=10$ into $8x + 12y = 480$: $$8(10) + 12y = 480 \Rightarrow 80 + 12y = 480 \Rightarrow 12y = 400 \Rightarrow y = \frac{400}{12} = 33.33$$ 7. **Evaluate profit at corner points:** - At $(0,0)$: $P=0$ - At $(0,40)$: $P=20.50(0) + 31(40) = 1240$ - At $(30,0)$: $P=20.50(30) + 31(0) = 615$ - At $(10,33.33)$: $P=20.50(10) + 31(33.33) = 205 + 1033.33 = 1238.33$ 8. **Conclusion:** Maximum profit is at $(0,40)$ producing 0 kg of X and 40 kg of Y with profit 1240. **Final answer:** Produce 0 kg of X and 40 kg of Y to maximize profit at 1240.