Solve the following LPP by graphical method Max Z = 2x1 + 3x2
Subject to constraints x1 - x2 ≤ 2 and x1 + x2 ≥ 4
and x1
, x2 ≥ 0.

To solve the given Linear Programming Problem (LPP) using the graphical method, follow these steps:

### Problem Statement:
- **Objective Function**: Maximize \( Z = 2x_1 + 3x_2 \)
- **Constraints**:
  1. \( x_1 - x_2 \leq 2 \)
  2. \( x_1 + x_2 \geq 4 \)
  3. \( x_1 \geq 0 \)
  4. \( x_2 \geq 0 \)

### Step 1: Convert Inequalities to Equations
1. **For** \( x_1 - x_2 \leq 2 \):
   - Equation: \( x_1 - x_2 = 2 \)
2. **For** \( x_1 + x_2 \geq 4 \):
   - Equation: \( x_1 + x_2 = 4 \)

### Step 2: Plot the Constraints on the Graph
- **Equation 1**: \( x_1 - x_2 = 2 \)
  - Find intercepts:
    - If \( x_1 = 0 \): \( 0 - x_2 = 2 \) ⟹ \( x_2 = -2 \) (out of feasible region since \( x_2 \geq 0 \))
    - If \( x_2 = 0 \): \( x_1 - 0 = 2 \) ⟹ \( x_1 = 2 \)
  - Points: \( (2, 0) \) and another point, for example \( (4, 2) \) if \( x_1 = 4 \): \( 4 - x_2 = 2 \) ⟹ \( x_2 = 2 \)

- **Equation 2**: \( x_1 + x_2 = 4 \)
  - Find intercepts:
    - If \( x_1 = 0 \): \( 0 + x_2 = 4 \) ⟹ \( x_2 = 4 \)
    - If \( x_2 = 0 \): \( x_1 + 0 = 4 \) ⟹ \( x_1 = 4 \)
  - Points: \( (0, 4) \) and \( (4, 0) \)

### Step 3: Identify the Feasible Region
- The feasible region is the area where all constraints intersect while considering \( x_1 \geq 0 \) and \( x_2 \geq 0 \).
- \( x_1 - x_2 \leq 2 \): This is the area below the line \( x_1 - x_2 = 2 \).
- \( x_1 + x_2 \geq 4 \): This is the area above the line \( x_1 + x_2 = 4 \).

### Step 4: Find Intersection Points
1. **Intersection of \( x_1 - x_2 = 2 \) and \( x_1 + x_2 = 4 \)**:
   - Solve the system of equations:
     \[
     x_1 - x_2 = 2
     \]
     \[
     x_1 + x_2 = 4
     \]
   - Add the equations:
     \[
     2x_1 = 6 \quad \Rightarrow \quad x_1 = 3
     \]
   - Substitute \( x_1 = 3 \) into one of the equations:
     \[
     3 + x_2 = 4 \quad \Rightarrow \quad x_2 = 1
     \]
   - Intersection Point: \( (3, 1) \)

### Step 5: Evaluate the Objective Function at the Vertices
- **Vertices of the Feasible Region**: \( (2, 0) \), \( (3, 1) \), and \( (0, 4) \)
- Calculate \( Z = 2x_1 + 3x_2 \) at each vertex:
  1. At \( (2, 0) \): \( Z = 2(2) + 3(0) = 4 \)
  2. At \( (3, 1) \): \( Z = 2(3) + 3(1) = 6 + 3 = 9 \)
  3. At \( (0, 4) \): \( Z = 2(0) + 3(4) = 12 \)

### Step 6: Identify the Optimal Solution
- Maximum value of \( Z \) is 12 at the point \( (0, 4) \).

### Conclusion
The optimal solution to the LPP is:
- **Optimal Point**: \( (0, 4) \)
- **Maximum Value**: \( Z = 12 \)

Would you like more details on any step, or do you have any questions?

### Relative Questions to Expand Understanding
1. How do we interpret the feasible region in a linear programming problem?
2. Why do we only evaluate the objective function at the vertices of the feasible region?
3. What happens if the feasible region is unbounded?
4. Can we use this graphical method for LPPs with more than two variables?
5. What is the significance of having constraints \( x_1 \geq 0 \) and \( x_2 \geq 0 \)?

### Tip
When solving an LPP graphically, always ensure that the feasible region is accurately plotted and identify all vertices correctly, as the optimal solution lies at one of these vertices.

Solve the following LPP by graphical method Max Z = 2x1 + 3x2 Subject to constraints x1 - x2 ≤ 2 and x1 + x2 ≥ 4 and x1, x2 ≥ 0.

Learn how to solve a linear programming problem (LPP) graphically with the objective function Max Z = 2x1 + 3x2, subject to constraints x1 - x2 ≤ 2, x1 + x2 ≥ 4, and x1, x2 ≥ 0. Detailed step-by-step solution with graphical representation. Suitable for Grades 11-12 and early university students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation