Math Problem Statement

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.

Solution

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

Problem Statement:

  • Objective Function: Maximize Z=2x1+3x2Z = 2x_1 + 3x_2
  • Constraints:
    1. x1x22x_1 - x_2 \leq 2
    2. x1+x24x_1 + x_2 \geq 4
    3. x10x_1 \geq 0
    4. x20x_2 \geq 0

Step 1: Convert Inequalities to Equations

  1. For x1x22x_1 - x_2 \leq 2:
    • Equation: x1x2=2x_1 - x_2 = 2
  2. For x1+x24x_1 + x_2 \geq 4:
    • Equation: x1+x2=4x_1 + x_2 = 4

Step 2: Plot the Constraints on the Graph

  • Equation 1: x1x2=2x_1 - x_2 = 2

    • Find intercepts:
      • If x1=0x_1 = 0: 0x2=20 - x_2 = 2x2=2x_2 = -2 (out of feasible region since x20x_2 \geq 0)
      • If x2=0x_2 = 0: x10=2x_1 - 0 = 2x1=2x_1 = 2
    • Points: (2,0)(2, 0) and another point, for example (4,2)(4, 2) if x1=4x_1 = 4: 4x2=24 - x_2 = 2x2=2x_2 = 2
  • Equation 2: x1+x2=4x_1 + x_2 = 4

    • Find intercepts:
      • If x1=0x_1 = 0: 0+x2=40 + x_2 = 4x2=4x_2 = 4
      • If x2=0x_2 = 0: x1+0=4x_1 + 0 = 4x1=4x_1 = 4
    • Points: (0,4)(0, 4) and (4,0)(4, 0)

Step 3: Identify the Feasible Region

  • The feasible region is the area where all constraints intersect while considering x10x_1 \geq 0 and x20x_2 \geq 0.
  • x1x22x_1 - x_2 \leq 2: This is the area below the line x1x2=2x_1 - x_2 = 2.
  • x1+x24x_1 + x_2 \geq 4: This is the area above the line x1+x2=4x_1 + x_2 = 4.

Step 4: Find Intersection Points

  1. Intersection of x1x2=2x_1 - x_2 = 2 and x1+x2=4x_1 + x_2 = 4:
    • Solve the system of equations: x1x2=2x_1 - x_2 = 2 x1+x2=4x_1 + x_2 = 4
    • Add the equations: 2x1=6x1=32x_1 = 6 \quad \Rightarrow \quad x_1 = 3
    • Substitute x1=3x_1 = 3 into one of the equations: 3+x2=4x2=13 + x_2 = 4 \quad \Rightarrow \quad x_2 = 1
    • Intersection Point: (3,1)(3, 1)

Step 5: Evaluate the Objective Function at the Vertices

  • Vertices of the Feasible Region: (2,0)(2, 0), (3,1)(3, 1), and (0,4)(0, 4)
  • Calculate Z=2x1+3x2Z = 2x_1 + 3x_2 at each vertex:
    1. At (2,0)(2, 0): Z=2(2)+3(0)=4Z = 2(2) + 3(0) = 4
    2. At (3,1)(3, 1): Z=2(3)+3(1)=6+3=9Z = 2(3) + 3(1) = 6 + 3 = 9
    3. At (0,4)(0, 4): Z=2(0)+3(4)=12Z = 2(0) + 3(4) = 12

Step 6: Identify the Optimal Solution

  • Maximum value of ZZ is 12 at the point (0,4)(0, 4).

Conclusion

The optimal solution to the LPP is:

  • Optimal Point: (0,4)(0, 4)
  • Maximum Value: Z=12Z = 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 x10x_1 \geq 0 and x20x_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.

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Math Problem Analysis

Mathematical Concepts

Linear Programming
Graphical Method
Optimization

Formulas

Objective function: Z = 2x1 + 3x2
Constraint equations: x1 - x2 = 2, x1 + x2 = 4

Theorems

Fundamental theorem of linear programming (optimal solution lies at a vertex of the feasible region)

Suitable Grade Level

Grades 11-12 or early university (Introductory Linear Programming)