Find the feasible solution of linear inequations ` 2x+3y le 12,2x+yle 8, x ge 0,yge0 ` by graphically

Answer

To find the feasible solution of the linear inequalities \(2x + 3y \leq 12\), \(2x + y \leq 8\), \(x \geq 0\), and \(y \geq 0\) graphically, follow these steps:

### Step 1: Graph the first inequality \(2x + 3y \leq 12\)

1. **Find the intercepts**:
- Set \(x = 0\):
\[ 2(0) + 3y = 12 \implies 3y = 12 \implies y = 4 \quad \text{(Point: (0, 4))} \]
- Set \(y = 0\):
\[ 2x + 3(0) = 12 \implies 2x = 12 \implies x = 6 \quad \text{(Point: (6, 0))} \]

2. **Draw the line**:
- Plot the points (0, 4) and (6, 0) on the graph.
- Draw a line through these points. Since the inequality is less than or equal to, shade below the line.

### Step 2: Graph the second inequality \(2x + y \leq 8\)

1. **Find the intercepts**:
- Set \(x = 0\):
\[ 2(0) + y = 8 \implies y = 8 \quad \text{(Point: (0, 8))} \]
- Set \(y = 0\):
\[ 2x + 0 = 8 \implies 2x = 8 \implies x = 4 \quad \text{(Point: (4, 0))} \]

2. **Draw the line**:
- Plot the points (0, 8) and (4, 0) on the graph.
- Draw a line through these points. Since the inequality is less than or equal to, shade below the line.

### Step 3: Identify the feasible region

1. **Consider the constraints \(x \geq 0\) and \(y \geq 0\)**:
- This restricts the feasible region to the first quadrant (where both x and y are non-negative).

2. **Find the intersection of the shaded areas**:
- The feasible solution is the area where the shaded regions from both inequalities overlap, bounded by the axes.

### Step 4: Identify the vertices of the feasible region

1. **Vertices**:
- The points of intersection of the lines and the axes will give the vertices of the feasible region.
- The vertices can be found at:
- (0, 0) (origin)
- (0, 4) (from the first inequality)
- (4, 0) (from the second inequality)
- The intersection of the two lines can also be found by solving the equations:
\[ 2x + 3y = 12 \quad \text{and} \quad 2x + y = 8 \]
Subtract the second equation from the first:
\[ 2x + 3y - (2x + y) = 12 - 8 \implies 2y = 4 \implies y = 2 \]
Substitute \(y = 2\) back into \(2x + y = 8\):
\[ 2x + 2 = 8 \implies 2x = 6 \implies x = 3 \]
So, the intersection point is (3, 2).

### Step 5: Conclusion

The feasible solution region is a quadrilateral formed by the points (0, 0), (0, 4), (4, 0), and (3, 2).