Example 3.1 Use the graphical method to solve the following LP problem.
Maximize Z = 15x1 + 10x2
subject to the constraints
(i) 4x1 + 6x2 ≤ 360, (ii) 3x1 + 0x2 ≤ 180, (iii) 0x1 + 5x2 ≤ 200
and x1, x2 ≥ 0.
Solution 1. The given LP problem is already in mathematical form.

### Problem

You are asked to solve the following Linear Programming (LP) problem graphically:

**Objective:**
Maximize \( Z = 15x_1 + 10x_2 \)

**Subject to the constraints:**
1. \( 4x_1 + 6x_2 \leq 360 \)
2. \( 3x_1 \leq 180 \) (or equivalently, \( x_1 \leq 60 \))
3. \( 5x_2 \leq 200 \) (or equivalently, \( x_2 \leq 40 \))
4. \( x_1, x_2 \geq 0 \)

### Step-by-Step Solution Using the Graphical Method

1. **Plotting the constraints:**

   - **Constraint 1: \( 4x_1 + 6x_2 \leq 360 \)**  
     First, convert this inequality into an equation:  
     \( 4x_1 + 6x_2 = 360 \).  
     To find the intercepts:
     - When \( x_1 = 0 \):  
       \( 6x_2 = 360 \implies x_2 = 60 \)
     - When \( x_2 = 0 \):  
       \( 4x_1 = 360 \implies x_1 = 90 \)

     So, the line passes through \( (90, 0) \) and \( (0, 60) \).

   - **Constraint 2: \( 3x_1 \leq 180 \) (or \( x_1 \leq 60 \))**  
     This is a vertical line at \( x_1 = 60 \). The region of interest is to the left of this line.

   - **Constraint 3: \( 5x_2 \leq 200 \) (or \( x_2 \leq 40 \))**  
     This is a horizontal line at \( x_2 = 40 \). The region of interest is below this line.

2. **Feasible Region:**
   - The feasible region is determined by the intersection of the constraints. The region must satisfy all three inequalities and lies within the first quadrant (since \( x_1 \geq 0 \) and \( x_2 \geq 0 \)).
   - The corner points of the feasible region are where the constraint lines intersect.

3. **Finding the Corner Points:**
   - **Intersection of \( 4x_1 + 6x_2 = 360 \) and \( x_1 = 60 \):**  
     Substituting \( x_1 = 60 \) into \( 4x_1 + 6x_2 = 360 \):  
     \( 4(60) + 6x_2 = 360 \implies 240 + 6x_2 = 360 \implies x_2 = 20 \).  
     So, one corner point is \( (60, 20) \).

   - **Intersection of \( 4x_1 + 6x_2 = 360 \) and \( x_2 = 40 \):**  
     Substituting \( x_2 = 40 \) into \( 4x_1 + 6(40) = 360 \):  
     \( 4x_1 + 240 = 360 \implies 4x_1 = 120 \implies x_1 = 30 \).  
     So, another corner point is \( (30, 40) \).

   - **Intersection of \( x_1 = 60 \) and \( x_2 = 0 \):**  
     This is the point \( (60, 0) \).

   - **Intersection of \( x_1 = 0 \) and \( x_2 = 40 \):**  
     This is the point \( (0, 40) \).

4. **Evaluating the Objective Function:**
   Now, evaluate \( Z = 15x_1 + 10x_2 \) at each corner point:

   - At \( (0, 40) \):  
     \( Z = 15(0) + 10(40) = 400 \)
   - At \( (60, 0) \):  
     \( Z = 15(60) + 10(0) = 900 \)
   - At \( (60, 20) \):  
     \( Z = 15(60) + 10(20) = 900 + 200 = 1100 \)
   - At \( (30, 40) \):  
     \( Z = 15(30) + 10(40) = 450 + 400 = 850 \)

5. **Optimal Solution:**
   The maximum value of \( Z = 1100 \) occurs at the point \( (60, 20) \).

### Conclusion:
The optimal solution is \( x_1 = 60 \) and \( x_2 = 20 \), and the maximum value of \( Z \) is 1100.

Would you like more details or have any further questions?

### Expanding Questions:
1. How do you interpret the constraints graphically in linear programming?
2. What happens if the feasible region is unbounded in a graphical LP solution?
3. Can you have more than one optimal solution in a linear programming problem?
4. How do you handle non-linear constraints in an LP problem?
5. Why is the feasible region always a polygon in a two-variable LP problem?

**Tip:** Always verify that your solution satisfies all constraints before concluding the optimal value in a graphical LP method.

Example 3.1 Use the graphical method to solve the following LP problem. Maximize Z = 15x1 + 10x2 subject to the constraints (i) 4x1 + 6x2 ≤ 360, (ii) 3x1 ≤ 180, (iii) 0x1 + 5x2 ≤ 200 and x1, x2 ≥ 0.

This page provides a detailed step-by-step solution to the linear programming problem: Maximize Z = 15x1 + 10x2, subject to constraints 4x1 + 6x2 ≤ 360, x1 ≤ 60, x2 ≤ 40, and x1, x2 ≥ 0. Learn how to solve it using the graphical method, with a focus on finding corner points and optimizing the objective function. Suitable for high school students in Grades 10-12.

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