Let’s systematically break down and solve this Linear Programming (LP) problem:

---

a) Define Variables (2 Marks)

• $X_1$: The number of air conditioners produced.
• $X_2$: The number of heaters produced.

---

b) Derive the Objective Function (3 Marks)

The objective is to maximize profit.
Each air conditioner yields $25, and each heater yields $15.

The objective function is:
$\text{Maximize } Z = 25X_1 + 15X_2$

---

c) Formulate Constraints (7 Marks)

1. Wiring Time Constraint:
   Each air conditioner takes 3 hours, and each heater takes 2 hours. The total available wiring time is 240 hours.
   $3X_1 + 2X_2 \leq 240$

2. Drilling Time Constraint:
   Each air conditioner takes 2 hours, and each heater takes 1 hour. The total available drilling time is 140 hours.
   $2X_1 + X_2 \leq 140$

3. Non-Negativity Constraints:
   $X_1 \geq 0, \, X_2 \geq 0$

---

d) Graphically Illustrate the Feasible Region (4 Marks)

Let’s solve for the boundary lines of each constraint.

1. From $3X_1 + 2X_2 \leq 240$:

   • If $X_1 = 0$: $X_2 = 120$ (point: $(0, 120)$).
   • If $X_2 = 0$: $X_1 = 80$ (point: $(80, 0)$).

2. From $2X_1 + X_2 \leq 140$:

   • If $X_1 = 0$: $X_2 = 140$ (point: $(0, 140)$).
   • If $X_2 = 0$: $X_1 = 70$ (point: $(70, 0)$).

Feasible Region

The feasible region is the intersection of the constraints, bounded by these lines and the non-negativity constraints. I will create a graph to show this region.

---

e) Corner Points (8 Marks)

Using the intersection points of the lines:

1. $(0, 0)$ – Origin.

2. Intersection of $3X_1 + 2X_2 = 240$ and $2X_1 + X_2 = 140$. Solve simultaneously:

   • From $2X_1 + X_2 = 140$: $X_2 = 140 - 2X_1$.
   • Substitute into $3X_1 + 2(140 - 2X_1) = 240$: $3X_1 + 280 - 4X_1 = 240 \implies -X_1 = -40 \implies X_1 = 40.$ $X_2 = 140 - 2(40) = 60$. Point: $(40, 60)$.

3. $(80, 0)$ – From $3X_1 + 2X_2 = 240$.

4. $(0, 120)$ – From $3X_1 + 2X_2 = 240$.

Corner points: $(0, 0), (0, 120), (40, 60), (80, 0)$.

---

f) Determine the Optimal Solution (4 Marks)

Evaluate the objective function $Z = 25X_1 + 15X_2$ at each corner point:

1. At $(0, 0)$: $Z = 25(0) + 15(0) = 0$.
2. At $(0, 120)$: $Z = 25(0) + 15(120) = 1800$.
3. At $(40, 60)$: $Z = 25(40) + 15(60) = 1000 + 900 = 1900$.
4. At $(80, 0)$: $Z = 25(80) + 15(0) = 2000$.

Optimal solution: $X_1 = 80, X_2 = 0$.

---

g) Determine the Profit at the Optimal Solution (2 Marks)

The maximum profit is: $Z = 25(80) + 15(0) = 2000$

---

Would you like me to graph the feasible region or clarify any part?

5 Questions to Expand Knowledge

1. How does the graphical method help in visualizing LP problems?
2. Can the simplex method handle more than two variables?
3. What if there were additional constraints—how would that change the solution?
4. How can sensitivity analysis be applied to this LP problem?
5. What real-world factors could affect the accuracy of the LP model?

Tip

When solving LP problems, always check if the constraints form a bounded feasible region to ensure the existence of an optimal solution.