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solve the following linear programming problem using graphical method?

maximize z=110x1+110x2
subject to
x1+x2</=9
x1>/=2
20x1+50x2</=360
x2>/=3
x1,x2>/=0

1 answer

To solve the linear programming problem using the graphical method, follow these steps:

Problem Formulation

We need to maximize the objective function: \[ z = 110x_1 + 110x_2 \]

Subject to the constraints:

  1. \( x_1 + x_2 \leq 9 \)
  2. \( x_1 \geq 2 \)
  3. \( 20x_1 + 50x_2 \leq 360 \)
  4. \( x_2 \geq 3 \)
  5. \( x_1, x_2 \geq 0 \)

Step 1: Convert Inequalities to Equalities

To graph the constraints, we convert the inequalities to equalities:

  1. \( x_1 + x_2 = 9 \)
  2. \( x_1 = 2 \)
  3. \( 20x_1 + 50x_2 = 360 \)
  4. \( x_2 = 3 \)

Step 2: Find the Intersection Points

Let's plot each of the lines defined by these equations:

  1. For \( x_1 + x_2 = 9 \):

    • When \( x_1 = 0 \), \( x_2 = 9 \) (point A: (0, 9))
    • When \( x_2 = 0 \), \( x_1 = 9 \) (point B: (9, 0))

  2. For \( x_1 = 2 \):

    • This is a vertical line at \( x_1 = 2 \).

  3. For \( 20x_1 + 50x_2 = 360 \):

    • When \( x_1 = 0 \), \( 50x_2 = 360 \) leads to \( x_2 = 7.2 \) (point C: (0, 7.2))
    • When \( x_2 = 0 \), \( 20x_1 = 360 \) leads to \( x_1 = 18 \) (point D: (18, 0))

  4. For \( x_2 = 3 \):

    • This is a horizontal line at \( x_2 = 3 \).

Step 3: Plot the Constraints on the Graph

Next, plot the points and lines on a graph:

  • The line from (0, 9) to (9, 0) represents the constraint \( x_1 + x_2 \leq 9 \).
  • The vertical line at \( x_1 = 2 \).
  • The line from (0, 7.2) to (18, 0) represents the constraint \( 20x_1 + 50x_2 \leq 360 \).
  • The horizontal line at \( x_2 = 3 \).

Step 4: Identify the Feasible Region

The feasible region is the area that satisfies all constraints.

Starting from the inequalities:

  • The area below or on the line \( x_1 + x_2 = 9 \).
  • The area to the right of the line \( x_1 = 2 \).
  • The area below or on the line \( 20x_1 + 50x_2 = 360 \).
  • The area above the line \( x_2 = 3 \).

Finding Vertices of the Feasible Region

Evaluate the intersections of the constraint lines:

  1. Intersection of \( x_1 + x_2 = 9 \) and \( x_2 = 3 \): \[ x_1 + 3 = 9 \implies x_1 = 6 \quad \Rightarrow (6, 3) \]

  2. Intersection of \( x_1 + x_2 = 9 \) and \( x_1 = 2 \): \[ 2 + x_2 = 9 \implies x_2 = 7 \quad \Rightarrow (2, 7) \]

  3. Intersection of \( 20x_1 + 50x_2 = 360 \) and \( x_2 = 3 \): \[ 20x_1 + 50 \cdot 3 = 360 \implies 20x_1 + 150 = 360 \implies 20x_1 = 210 \implies x_1 = 10.5 \quad \Rightarrow (10.5, 3) \] But since \( x_1 \leq 9 \) here, this point is not in the feasible region.

  4. Intersection of \( 20x_1 + 50x_2 = 360 \) and \( x_1 = 2 \): \[ 20 \cdot 2 + 50x_2 = 360 \implies 40 + 50x_2 = 360 \implies 50x_2 = 320 \implies x_2 = 6.4 \quad \Rightarrow (2, 6.4) \]

Step 5: Evaluate the Objective Function at the Vertices

Now evaluate \( z = 110x_1 + 110x_2 \) at the feasible vertex points:

  1. \( (6, 3) \): \[ z = 110(6) + 110(3) = 660 + 330 = 990 \]
  2. \( (2, 7) \): \[ z = 110(2) + 110(7) = 220 + 770 = 990 \]
  3. \( (2, 6.4) \): \[ z = 110(2) + 110(6.4) = 220 + 704 = 924 \]

Step 6: Conclusion

The optimal solution occurs at both points \( (6, 3) \) and \( (2, 7) \), yielding the maximum value of: \[ \text{Max } z = 990 \] Thus, the maximum value of the objective function, within the feasible region defined by the constraints, is 990 at two points: \( (6, 3) \) and \( (2, 7) \).