Solve graphically the following LPP Maximize Z = X1 + 3X2 Subject to X1 + 2X2 ≤ 9 X1 + 4X2 ≤ 11 X1 – X2 ≥ 2 X1, X2 ≥ 0 �?

To solve this LPP graphically, we first need to plot the feasible region determined by the given constraints.

1. Plot the line X1 + 2X2 = 9:
To plot this line, we first find the intercepts:
When X1 = 0, X2 = 4.5
When X2 = 0, X1 = 9
Plot these points and draw the line passing through them.

2. Plot the line X1 + 4X2 = 11:
When X1 = 0, X2 = 2.75
When X2 = 0, X1 = 11
Plot these points and draw the line passing through them.

3. Plot the line X1 - X2 = 2:
When X1 = 0, X2 = -2
When X2 = 0, X1 = 2
Plot these points and draw the line passing through them.

Now, shade the feasible region which is the region that satisfies all the given constraints. The feasible region will be the region below the line X1 + 2X2 = 9, below the line X1 + 4X2 = 11, and above the line X1 - X2 = 2.

Next, we need to find the corner points of the feasible region. The corner points are the intersection points of the lines that form the boundary of the feasible region.

The corner points are:
A(2, 0)
B(3, 2)
C(1, 3)
D(0, 2)

Now, we need to evaluate the objective function Z = X1 + 3X2 at each of the corner points to determine the maximum value of Z.

Z(A) = 2 + 3(0) = 2
Z(B) = 3 + 3(2) = 9
Z(C) = 1 + 3(3) = 10
Z(D) = 0 + 3(2) = 6

Therefore, the maximum value of Z = 10 at point C(1, 3).

Step 1: Plot the constraints

- X1 + 2X2 ≤ 9: Plot the line x1 + 2x2 = 9 by finding the x- and y-intercepts: (0, 9/2) and (9, 0). Shade the area below the line.
- X1 + 4X2 ≤ 11: Plot the line x1 + 4x2 = 11 by finding the x- and y-intercepts: (0, 11/4) and (11, 0). Shade the area below the line.
- X1 - X2 ≥ 2: Rewrite as -x1 + x2 ≥ -2. Plot the line -x1 + x2 = -2 by finding the x- and y-intercepts: (0, -2) and (2, 0). Shade the area above the line.
- X1, X2 ≥ 0: Shade the first quadrant.

Step 2: Identify the feasible region

The feasible region is the shaded area that satisfies all the constraints. It is a triangle with vertices at (0, 0), (2, 0), and (3, 2).

[Image of the feasible region with shaded area]

Step 3: Find the corner points of the feasible region

The corner points are the vertices of the feasible region: (0, 0), (2, 0), and (3, 2).

Step 4: Evaluate the objective function at each corner point

- Z(0, 0) = 0
- Z(2, 0) = 2
- Z(3, 2) = 9

Step 5: Determine the optimal solution

The maximum value of Z, 9, occurs at the corner point (3, 2).

Conclusion:

The optimal solution is X1 = 3, X2 = 2, which maximizes Z to 9.