For L.P.P. maximize `z = 4x_(1) + 2X_(2)` subject to `3x_(1) + 2x_(2) ge 9, x_(1) - x_(2) le 3,x_(1) ge 0, x_(2) ge 0` has ……

Options

A. infinite number of optimal solutions
B. unbounded solution
C. no solution
D. one optimal solution

Explanation

To solve the linear programming problem (L.P.P.) of maximizing the objective function \( z = 4x_1 + 2x_2 \) subject to the given constraints, we will follow these steps:

### Step 1: Write down the constraints
The constraints given are:
1. \( 3x_1 + 2x_2 \geq 9 \)
2. \( x_1 - x_2 \leq 3 \)
3. \( x_1 \geq 0 \)
4. \( x_2 \geq 0 \)

### Step 2: Convert inequalities to equalities for graphing
To graph the constraints, we convert the inequalities into equalities:

1. \( 3x_1 + 2x_2 = 9 \)
2. \( x_1 - x_2 = 3 \)

### Step 3: Find the intercepts for each line
**For the first equation \( 3x_1 + 2x_2 = 9 \):**
- When \( x_1 = 0 \):
\[ 2x_2 = 9 \implies x_2 = 4.5 \]
- When \( x_2 = 0 \):
\[ 3x_1 = 9 \implies x_1 = 3 \]
Thus, the intercepts are \( (0, 4.5) \) and \( (3, 0) \).

**For the second equation \( x_1 - x_2 = 3 \):**
- When \( x_1 = 0 \):
\[ -x_2 = 3 \implies x_2 = -3 \quad (\text{not feasible since } x_2 \geq 0) \]
- When \( x_2 = 0 \):
\[ x_1 = 3 \]
Thus, the intercept is \( (3, 0) \).

### Step 4: Graph the constraints
1. The line from \( (0, 4.5) \) to \( (3, 0) \) represents the first constraint.
2. The line \( x_1 - x_2 = 3 \) can be rewritten as \( x_2 = x_1 - 3 \). It intersects the x-axis at \( (3, 0) \) and has a slope of 1.

### Step 5: Determine the feasible region
- The region satisfying \( 3x_1 + 2x_2 \geq 9 \) is above the line.
- The region satisfying \( x_1 - x_2 \leq 3 \) is below the line.
- Both \( x_1 \geq 0 \) and \( x_2 \geq 0 \) restrict the feasible region to the first quadrant.

### Step 6: Identify the vertices of the feasible region
The vertices of the feasible region are:
1. \( (3, 0) \) (intersection of both lines)
2. The point where \( 3x_1 + 2x_2 = 9 \) intersects the axes, which we already calculated as \( (0, 4.5) \).

### Step 7: Evaluate the objective function at the vertices
Now we evaluate \( z = 4x_1 + 2x_2 \) at the vertices:
1. At \( (3, 0) \):
\[ z = 4(3) + 2(0) = 12 \]
2. At \( (0, 4.5) \):
\[ z = 4(0) + 2(4.5) = 9 \]

### Step 8: Determine the maximum value
The maximum value of \( z \) occurs at \( (3, 0) \) with \( z = 12 \).

### Step 9: Check for boundedness
The feasible region is unbounded because the constraints allow for infinitely large values of \( z \) as \( x_1 \) and \( x_2 \) can increase indefinitely while still satisfying the constraints.

### Conclusion
The feasible region is unbounded, and the maximum value of \( z \) cannot be determined as it can increase indefinitely.