To solve the linear programming (LP) problem using the graphical method, we will follow these steps:

1. Identify the constraints and plot them on a graph.
2. Determine the feasible region that satisfies all constraints.
3. Find the corner points (vertices) of the feasible region.
4. Calculate the value of the objective function at each corner point.
5. Identify the maximum value of the objective function from these points.

Step-by-Step Solution

Given LP Problem:

Maximize:
$Z = x_1 - 2x_2$

Subject to constraints:

1. $-x_1 + x_2 \leq 1$
2. $6x_1 + 4x_2 \geq 24$
3. $0 \leq x_1 \leq 5$
4. $2 \leq x_2 \leq 4$
5. $x_1, x_2 \geq 0$

Step 1: Plot the Constraints

Let's rewrite the inequalities to plot them on the graph:

1. $-x_1 + x_2 \leq 1$

   • Rewrite as $x_2 \leq x_1 + 1$

2. $6x_1 + 4x_2 \geq 24$

   • Rewrite as $x_2 \geq \frac{24 - 6x_1}{4}$
   • Simplify to $x_2 \geq 6 - 1.5x_1$

3. $0 \leq x_1 \leq 5$

   • These are vertical lines at $x_1 = 0$ and $x_1 = 5$.

4. $2 \leq x_2 \leq 4$

   • These are horizontal lines at $x_2 = 2$ and $x_2 = 4$.

Step 2: Determine the Feasible Region

To find the feasible region, we need to plot each constraint on the graph and identify the overlapping area that satisfies all constraints.

Step 3: Find the Corner Points of the Feasible Region

Let's find the intersection points of the lines representing the constraints:

• Intersection of $x_2 = x_1 + 1$ and $x_2 = 6 - 1.5x_1$:

  Solve $x_1 + 1 = 6 - 1.5x_1$: $x_1 + 1 = 6 - 1.5x_1$ $x_1 + 1.5x_1 = 6 - 1$ $2.5x_1 = 5$ $x_1 = 2$

  Then, $x_2 = x_1 + 1 = 2 + 1 = 3$.
  Point: $(2, 3)$

• Intersection of $x_1 = 0$ and $x_2 = 2$:

  $(0, 2)$

• Intersection of $x_1 = 0$ and $x_2 = 4$:

  $(0, 4)$

• Intersection of $x_1 = 5$ and $x_2 = 2$:

  $(5, 2)$

• Intersection of $x_1 = 5$ and $x_2 = 4$:

  $(5, 4)$

• Intersection of $x_1 = 0$ and $x_2 = 6 - 1.5x_1$:

  Set $x_1 = 0$: $x_2 = 6 - 1.5(0) = 6$. However, since $x_2 = 6$ is outside the $2 \leq x_2 \leq 4$ range, this point is not valid in our feasible region.

Step 4: Calculate the Objective Function at Each Corner Point

We now evaluate $Z = x_1 - 2x_2$ at each valid corner point:

1. At $(0, 2)$:
   $Z = 0 - 2(2) = -4$

2. At $(0, 4)$:
   $Z = 0 - 2(4) = -8$

3. At $(5, 2)$:
   $Z = 5 - 2(2) = 1$

4. At $(5, 4)$:
   $Z = 5 - 2(4) = -3$

5. At $(2, 3)$:
   $Z = 2 - 2(3) = -4$

Step 5: Identify the Maximum Value of the Objective Function

The maximum value of $Z$ occurs at point $(5, 2)$ with $Z = 1$.

Conclusion:

The solution to the LP problem is:

• Optimal Point: $(x_1, x_2) = (5, 2)$
• Maximum Value of $Z$: $1$

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

Follow-up Questions

1. How would the solution change if the objective function was to minimize instead of maximize?
2. What would happen if one of the constraints was modified or removed?
3. How does the graphical method handle problems with more than two variables?
4. Can you provide an example of an unbounded solution in a graphical method?
5. How do you verify if a given point is feasible for the problem?

Tip: When using the graphical method, always make sure to accurately plot all constraints and label intersection points clearly for easier analysis of the feasible region.