Find the region when the following inequations:
`x+yle6,xgey,xge0,yle0` hold good. Find the coordinates of the vertices of the region.

Answer

To solve the given problem, we will analyze the inequalities step by step and find the region they define, as well as the coordinates of the vertices of that region.

### Step 1: Analyze the inequalities
The inequalities we need to consider are:
1. \( x + y \leq 6 \)
2. \( x \geq y \)
3. \( x \geq 0 \)
4. \( y \geq 0 \)

### Step 2: Graph the first inequality \( x + y \leq 6 \)
To graph the line \( x + y = 6 \):
- When \( x = 0 \), \( y = 6 \) (point \( (0, 6) \))
- When \( y = 0 \), \( x = 6 \) (point \( (6, 0) \))

Now, we draw the line connecting these two points. The area below this line (including the line itself) represents the solution to \( x + y \leq 6 \).

**Hint:** To graph an inequality, first graph the corresponding equation and then shade the appropriate region based on the inequality sign.

### Step 3: Graph the second inequality \( x \geq y \)
The line \( x = y \) can be graphed by plotting points where \( x \) equals \( y \):
- For example, points like \( (0, 0) \), \( (1, 1) \), \( (2, 2) \), etc.

The region satisfying \( x \geq y \) is the area above this line.

**Hint:** The line \( x = y \) divides the plane into two regions; shade the area that satisfies the inequality.

### Step 4: Graph the third inequality \( x \geq 0 \)
This inequality represents the right half of the coordinate plane, including the y-axis.

**Hint:** The line \( x = 0 \) is the y-axis, and the region to the right of this line satisfies \( x \geq 0 \).

### Step 5: Graph the fourth inequality \( y \geq 0 \)
This inequality represents the upper half of the coordinate plane, including the x-axis.

**Hint:** The line \( y = 0 \) is the x-axis, and the region above this line satisfies \( y \geq 0 \).

### Step 6: Identify the feasible region
Now, we need to find the intersection of all these regions:
- The area below the line \( x + y = 6 \)
- The area above the line \( x = y \)
- The area to the right of the y-axis
- The area above the x-axis

The feasible region will be a triangle bounded by the lines \( x + y = 6 \), \( x = y \), and the axes.

### Step 7: Find the vertices of the region
To find the vertices of the triangle formed by these inequalities, we need to find the points of intersection:
1. Intersection of \( x + y = 6 \) and \( x = y \):
- Substitute \( y \) with \( x \): \( x + x = 6 \) → \( 2x = 6 \) → \( x = 3 \)
- Thus, \( y = 3 \). So, one vertex is \( (3, 3) \).

2. Intersection of \( x + y = 6 \) and the x-axis (\( y = 0 \)):
- Set \( y = 0 \): \( x + 0 = 6 \) → \( x = 6 \). So, another vertex is \( (6, 0) \).

3. Intersection of \( x + y = 6 \) and the y-axis (\( x = 0 \)):
- Set \( x = 0 \): \( 0 + y = 6 \) → \( y = 6 \). So, the last vertex is \( (0, 6) \).

### Step 8: List the vertices
The vertices of the feasible region are:
1. \( (0, 0) \)
2. \( (3, 3) \)
3. \( (6, 0) \)

### Final Answer
The coordinates of the vertices of the region defined by the inequalities are:
- \( (0, 0) \)
- \( (3, 3) \)
- \( (6, 0) \)