The feasible region of a LPP under the constraints ` x- y le 1 ,x + y ge 3, x ge 0, y ge 0`

Options

A. is bounded and lies in first quadrant
B. Is unbounded and lies first quadrant
C. does not exist
D. is not in the first quadrant

Answer

To find the feasible region of the Linear Programming Problem (LPP) under the given constraints, we will follow these steps:

### Step 1: Identify the constraints
The constraints provided are:
1. \( x - y \leq 1 \)
2. \( x + y \geq 3 \)
3. \( x \geq 0 \)
4. \( y \geq 0 \)

### Step 2: Convert inequalities to equations
To graph the constraints, we first convert the inequalities into equations:
1. \( x - y = 1 \)
2. \( x + y = 3 \)

### Step 3: Graph the first constraint
For the equation \( x - y = 1 \):
- When \( x = 0 \): \( 0 - y = 1 \) → \( y = -1 \) (point: \( (0, -1) \))
- When \( y = 0 \): \( x - 0 = 1 \) → \( x = 1 \) (point: \( (1, 0) \))

Plot these points on the graph and draw the line. Since the inequality is \( x - y \leq 1 \), shade the region below this line.

### Step 4: Graph the second constraint
For the equation \( x + y = 3 \):
- When \( x = 0 \): \( 0 + y = 3 \) → \( y = 3 \) (point: \( (0, 3) \))
- When \( y = 0 \): \( x + 0 = 3 \) → \( x = 3 \) (point: \( (3, 0) \))

Plot these points on the graph and draw the line. Since the inequality is \( x + y \geq 3 \), shade the region above this line.

### Step 5: Identify the feasible region
The feasible region is where the shaded areas of the two constraints overlap, while also considering the non-negativity constraints \( x \geq 0 \) and \( y \geq 0 \).

### Step 6: Analyze the feasible region
- The region defined by \( x - y \leq 1 \) is unbounded as it extends infinitely in the positive x-direction and downward.
- The region defined by \( x + y \geq 3 \) is also unbounded as it extends infinitely in the positive y-direction and rightward.

### Conclusion
The feasible region is unbounded and lies in the first quadrant. Thus, the answer is that the feasible region is unbounded and lies in the first quadrant.

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