The constraints of an LPP are `x+yle6,3x+2yge6,xge0 and yge0` Determine the vertices of the feasible region formed by them

Options

A. `(6,6),(0,0),(2,3),(3,2)`
B. `(0,0),(5,6),(6,5),(0,5)`
C. `(0,0),(5,0),(3,2),(6,6)`
D. `(0,6),(0,3),(2,0),(6,0)`

Answer

To determine the vertices of the feasible region formed by the given constraints of the Linear Programming Problem (LPP), we will follow these steps:

### Step 1: Write down the constraints
The constraints given are:
1. \( x + y \leq 6 \)
2. \( 3x + 2y \geq 6 \)
3. \( x \geq 0 \)
4. \( y \geq 0 \)

### Step 2: Convert inequalities to equations
To find the boundary lines, we convert the inequalities into equations:
1. \( x + y = 6 \)
2. \( 3x + 2y = 6 \)

### Step 3: Find the intercepts for the first equation
For the equation \( x + y = 6 \):
- When \( x = 0 \), \( y = 6 \) (y-intercept)
- When \( y = 0 \), \( x = 6 \) (x-intercept)

So, the intercepts are \( (0, 6) \) and \( (6, 0) \).

### Step 4: Find the intercepts for the second equation
For the equation \( 3x + 2y = 6 \):
- When \( x = 0 \), \( 2y = 6 \) → \( y = 3 \) (y-intercept)
- When \( y = 0 \), \( 3x = 6 \) → \( x = 2 \) (x-intercept)

So, the intercepts are \( (0, 3) \) and \( (2, 0) \).

### Step 5: Plot the lines on a graph
Now, we will plot the lines:
1. The line \( x + y = 6 \) connects the points \( (0, 6) \) and \( (6, 0) \).
2. The line \( 3x + 2y = 6 \) connects the points \( (0, 3) \) and \( (2, 0) \).

### Step 6: Determine the feasible region
- For \( x + y \leq 6 \), the feasible region is below the line \( x + y = 6 \).
- For \( 3x + 2y \geq 6 \), the feasible region is above the line \( 3x + 2y = 6 \).
- Since \( x \geq 0 \) and \( y \geq 0 \), we are restricted to the first quadrant.

### Step 7: Identify the vertices of the feasible region
The vertices of the feasible region can be found at the intersections of the lines and the axes:
1. Intersection of \( x + y = 6 \) and \( 3x + 2y = 6 \):
- Solve the equations simultaneously:
- From \( x + y = 6 \), we can express \( y = 6 - x \).
- Substitute into \( 3x + 2(6 - x) = 6 \):
\[ 3x + 12 - 2x = 6 \Rightarrow x + 12 = 6 \Rightarrow x = -6 \text{ (not feasible)} \]
- Since there's no intersection in the first quadrant, we check the vertices at the intercepts:
2. The vertices are:
- \( (0, 6) \)
- \( (0, 3) \)
- \( (2, 0) \)
- \( (6, 0) \)

### Step 8: Verify the feasible region
The feasible region is bounded by the points:
- \( (0, 3) \)
- \( (2, 0) \)
- \( (0, 6) \)
- \( (6, 0) \)

### Final Vertices
The vertices of the feasible region formed by the constraints are:
1. \( (0, 3) \)
2. \( (2, 0) \)
3. \( (0, 6) \)
4. \( (6, 0) \)