1. **Stating the problem:** A company manufactures two products, Product A and Product B. Each unit of Product A requires 3 hours of labour and 4 units of raw material, while each unit of Product B requires 2 hours of labour and 5 units of raw material. The company has 90 hours of labour and 70 units of raw material available each day. Each unit of Product A generates a profit of 250, while each unit of Product B generates a profit of 180. The company wants to maximize profit with the resources available. 2. **Writing the objective function:** Let $x$ be the number of units of Product A and $y$ be the number of units of Product B. The profit function to maximize is: $$Z = 250x + 180y$$ 3. **Writing the system of inequalities (constraints):** - Labour constraint: Each unit of A requires 3 hours, each unit of B requires 2 hours, total available is 90 hours: $$3x + 2y \leq 90$$ - Raw material constraint: Each unit of A requires 4 units, each unit of B requires 5 units, total available is 70 units: $$4x + 5y \leq 70$$ - Non-negativity constraints: $$x \geq 0, \quad y \geq 0$$ --- 1. **Stating the problem:** Maximize the objective function $$Z = 5x + 2y$$ subject to the constraints: $$x + y \leq 8$$ $$2x + 3y \leq 18$$ $$x, y \geq 0$$ 2. **Finding the feasible region:** The constraints define a polygon in the first quadrant bounded by the lines: - $x + y = 8$ - $2x + 3y = 18$ - $x = 0$ - $y = 0$ 3. **Finding corner points (vertices) of the feasible region:** - Intersection of $x=0$ and $x + y = 8$ is $(0,8)$ - Intersection of $y=0$ and $x + y = 8$ is $(8,0)$ - Intersection of $x=0$ and $2x + 3y = 18$ is $(0,6)$ - Intersection of $y=0$ and $2x + 3y = 18$ is $(9,0)$ but $x=9$ is outside $x + y \leq 8$, so not feasible - Intersection of $x + y = 8$ and $2x + 3y = 18$: Solve system: $$x + y = 8 \Rightarrow y = 8 - x$$ Substitute into second: $$2x + 3(8 - x) = 18$$ $$2x + 24 - 3x = 18$$ $$-x + 24 = 18$$ $$-x = -6 \Rightarrow x = 6$$ Then $y = 8 - 6 = 2$ So intersection point is $(6,2)$ 4. **Evaluating $Z$ at each vertex:** - At $(0,0)$: $Z = 5(0) + 2(0) = 0$ - At $(0,6)$: $Z = 5(0) + 2(6) = 12$ - At $(6,2)$: $Z = 5(6) + 2(2) = 30 + 4 = 34$ - At $(8,0)$: $Z = 5(8) + 2(0) = 40$ 5. **Conclusion:** The maximum value of $Z$ is $40$ at the point $(8,0)$. --- This completes the problem with new numbers for practice.