Find the feasible solutions of in equations ` 3x+2yle 18 , 2x +yle 10 , X ge 0 Y ge0`

Answer

To find the feasible solutions of the inequalities \(3x + 2y \leq 18\), \(2x + y \leq 10\), \(x \geq 0\), and \(y \geq 0\), we will follow these steps:

### Step 1: Rewrite the inequalities
We have the following inequalities:
1. \(3x + 2y \leq 18\)
2. \(2x + y \leq 10\)
3. \(x \geq 0\)
4. \(y \geq 0\)

### Step 2: Find the intercepts of the first inequality
For the inequality \(3x + 2y \leq 18\):
- To find the x-intercept, set \(y = 0\):
\[ 3x + 2(0) = 18 \implies 3x = 18 \implies x = 6 \]
- To find the y-intercept, set \(x = 0\):
\[ 3(0) + 2y = 18 \implies 2y = 18 \implies y = 9 \]

### Step 3: Find the intercepts of the second inequality
For the inequality \(2x + y \leq 10\):
- To find the x-intercept, set \(y = 0\):
\[ 2x + 0 = 10 \implies 2x = 10 \implies x = 5 \]
- To find the y-intercept, set \(x = 0\):
\[ 2(0) + y = 10 \implies y = 10 \]

### Step 4: Graph the inequalities
1. **Graph the line \(3x + 2y = 18\)**:
- Plot the points (6, 0) and (0, 9).
- Shade the area below this line.

2. **Graph the line \(2x + y = 10\)**:
- Plot the points (5, 0) and (0, 10).
- Shade the area below this line.

3. **Include the axes**:
- Since \(x \geq 0\) and \(y \geq 0\), only the first quadrant is considered.

### Step 5: Identify the feasible region
The feasible region is where the shaded areas of both inequalities overlap in the first quadrant. This region will be bounded by the lines and the axes.

### Step 6: Determine the vertices of the feasible region
To find the vertices of the feasible region, we need to find the points of intersection of the lines:
1. Solve the equations \(3x + 2y = 18\) and \(2x + y = 10\) simultaneously.

From \(2x + y = 10\), we can express \(y\) in terms of \(x\):
\[ y = 10 - 2x \]

Substituting this into the first equation:
\[ 3x + 2(10 - 2x) = 18 \]
\[ 3x + 20 - 4x = 18 \]
\[ -x + 20 = 18 \]
\[ -x = -2 \implies x = 2 \]

Now substitute \(x = 2\) back into \(y = 10 - 2x\):
\[ y = 10 - 2(2) = 10 - 4 = 6 \]

So one vertex is \((2, 6)\).

### Step 7: Check other vertices
- The vertices from the intercepts are:
- \((6, 0)\)
- \((0, 9)\)
- \((5, 0)\)
- \((0, 10)\)
- \((2, 6)\)

### Conclusion
The feasible region is bounded by the points \((0, 0)\), \((2, 6)\), \((5, 0)\), and \((0, 9)\).