Question: User: Question 3 [14 marks] An electronics manufacturer produces two types of components: Circuit Boards and Microcontrollers. Each Circuit Board requires 3 hours of soldering time and 2 hours of quality inspection. Each Microcontroller requires 4 hours of soldering time and 1 hour of quality inspection. The factory has a maximum of 120 hours available for soldering and 50 hours for inspection per week. Warehouse space limits the total combined output to no more than 35 units per week. The profit per Circuit Board is N20, and the profit per Microcontroller is N30. 3.1 Formulate a linear programming model. [5] 3.2 Determine how many units of each component should be produced to maximize total weekly profit. [6] 3.3 Calculate the maximum total weekly profit. [3]

1. **State the problem:** We want to maximize the total weekly profit from producing Circuit Boards and Microcontrollers given constraints on soldering time, inspection time, and warehouse space. 2. **Define variables:** Let $x$ = number of Circuit Boards produced per week. Let $y$ = number of Microcontrollers produced per week. 3. **Formulate the linear programming model:** - Objective function (maximize profit): $$\text{Maximize } Z = 20x + 30y$$ - Subject to constraints: - Soldering time: $$3x + 4y \leq 120$$ - Inspection time: $$2x + y \leq 50$$ - Warehouse space: $$x + y \leq 35$$ - Non-negativity: $$x \geq 0, y \geq 0$$ 4. **Solve the system to find optimal $x$ and $y$:** We check the corner points of the feasible region defined by the constraints. - Intersection of soldering and inspection constraints: Solve: $$3x + 4y = 120$$ $$2x + y = 50$$ From second: $$y = 50 - 2x$$ Substitute into first: $$3x + 4(50 - 2x) = 120$$ $$3x + 200 - 8x = 120$$ $$-5x = -80$$ $$x = 16$$ Then: $$y = 50 - 2(16) = 50 - 32 = 18$$ Check warehouse constraint: $$16 + 18 = 34 \leq 35$$ valid. - Intersection of soldering and warehouse constraints: $$3x + 4y = 120$$ $$x + y = 35$$ From second: $$y = 35 - x$$ Substitute: $$3x + 4(35 - x) = 120$$ $$3x + 140 - 4x = 120$$ $$-x = -20$$ $$x = 20$$ Then: $$y = 35 - 20 = 15$$ Check inspection constraint: $$2(20) + 15 = 40 + 15 = 55 > 50$$ invalid. - Intersection of inspection and warehouse constraints: $$2x + y = 50$$ $$x + y = 35$$ Subtract second from first: $$(2x + y) - (x + y) = 50 - 35$$ $$x = 15$$ Then: $$y = 35 - 15 = 20$$ Check soldering constraint: $$3(15) + 4(20) = 45 + 80 = 125 > 120$$ invalid. - Check vertices at axes: At $x=0$: $$y \leq \min\left(\frac{120}{4}, 50, 35\right) = \min(30, 50, 35) = 30$$ At $y=0$: $$x \leq \min\left(\frac{120}{3}, \frac{50}{2}, 35\right) = \min(40, 25, 35) = 25$$ 5. **Evaluate objective function at feasible vertices:** - At $(0,0)$: $$Z=0$$ - At $(0,30)$: $$Z=20(0)+30(30)=900$$ - At $(15,20)$ invalid - At $(16,18)$ valid: $$Z=20(16)+30(18)=320+540=860$$ - At $(25,0)$: $$Z=20(25)+30(0)=500$$ 6. **Conclusion:** The maximum profit is at $(0,30)$ with profit $$900$$. **Final answers:** - 3.1 Model: $$\max Z=20x+30y$$ subject to $$3x+4y \leq 120$$ $$2x+y \leq 50$$ $$x+y \leq 35$$ $$x,y \geq 0$$ - 3.2 Optimal production: $$x=0, y=30$$ - 3.3 Maximum profit: $$Z=900$$