The carner points of the feasible region formed by the system of linear inequations `x-yge-1,-x+yge0,x+yle2andx,yge0`, are

Options

A. `(0,0),(-1,0),(0,1)`
B. `(0,0),(2,0),(1,1)`
C. `(0,1),(1,1),(0,0)`
D. `(0,0),(1,1),((1)/(2),(3)/(2)),(0,1)`

Answer

To find the corner points of the feasible region formed by the given system of linear inequalities, we will follow these steps:

### Step 1: Write the inequalities
The system of inequalities is:
1. \( x - y \geq -1 \)
2. \( -x + y \geq 0 \) (or \( y \geq x \))
3. \( x + y \leq 2 \)
4. \( x \geq 0 \)
5. \( y \geq 0 \)

### Step 2: Convert inequalities to equations
To find the boundary lines, we convert the inequalities into equations:
1. \( x - y = -1 \)
2. \( -x + y = 0 \) (or \( y = x \))
3. \( x + y = 2 \)

### Step 3: Find the intersection points of the lines
We will find the intersection points of these lines:

**Intersection of Line 1 and Line 2:**
1. From \( x - y = -1 \) (Line 1) and \( y = x \) (Line 2):
\[ x - x = -1 \implies 0 = -1 \text{ (no solution)} \]

**Intersection of Line 1 and Line 3:**
2. From \( x - y = -1 \) (Line 1) and \( x + y = 2 \) (Line 3):
\[ x - y = -1 \quad (1) \\ x + y = 2 \quad (2) \]
Adding (1) and (2):
\[ 2x = 1 \implies x = \frac{1}{2} \]
Substituting \( x = \frac{1}{2} \) into (2):
\[ \frac{1}{2} + y = 2 \implies y = 2 - \frac{1}{2} = \frac{3}{2} \]
So, the intersection point is \( \left( \frac{1}{2}, \frac{3}{2} \right) \).

**Intersection of Line 2 and Line 3:**
3. From \( y = x \) (Line 2) and \( x + y = 2 \) (Line 3):
\[ x + x = 2 \implies 2x = 2 \implies x = 1 \]
Substituting \( x = 1 \) into \( y = x \):
\[ y = 1 \]
So, the intersection point is \( (1, 1) \).

### Step 4: Identify the corner points
The corner points of the feasible region are:
1. The origin \( (0, 0) \) (from \( x \geq 0 \) and \( y \geq 0 \)).
2. The point \( (0, 1) \) (from the line \( x - y = -1 \) when \( x = 0 \)).
3. The point \( \left( \frac{1}{2}, \frac{3}{2} \right) \) (intersection of Line 1 and Line 3).
4. The point \( (1, 1) \) (intersection of Line 2 and Line 3).
5. The point \( (2, 0) \) (from the line \( x + y = 2 \) when \( y = 0 \)).

### Step 5: List the corner points
The corner points of the feasible region are:
1. \( (0, 0) \)
2. \( (0, 1) \)
3. \( \left( \frac{1}{2}, \frac{3}{2} \right) \)
4. \( (1, 1) \)
5. \( (2, 0) \)

### Conclusion
The corner points of the feasible region formed by the given system of linear inequalities are:
- \( (0, 0) \)
- \( (0, 1) \)
- \( \left( \frac{1}{2}, \frac{3}{2} \right) \)
- \( (1, 1) \)
- \( (2, 0) \)