Which of the following is a corner point of the feasible region of system of linear inequations 2x + 3y `lt=6 , x + 4y lt= 4 ` and x, y `gt= 0` ?

Options

A. `(1,0)`
B. `(1,1) `
C. `(0,3) `
D. `((12)/(5) , (2)/(5))`

Answer

To find the corner points of the feasible region defined by the system of linear inequalities \(2x + 3y \leq 6\), \(x + 4y \leq 4\), and \(x, y \geq 0\), we will follow these steps:

### Step 1: Graph the inequalities
1. **Graph the line \(2x + 3y = 6\)**:
- Set \(x = 0\):
\[ 2(0) + 3y = 6 \implies y = 2 \quad \text{(Point: (0, 2))} \]
- Set \(y = 0\):
\[ 2x + 3(0) = 6 \implies x = 3 \quad \text{(Point: (3, 0))} \]
- Draw the line connecting (0, 2) and (3, 0).

2. **Graph the line \(x + 4y = 4\)**:
- Set \(x = 0\):
\[ 0 + 4y = 4 \implies y = 1 \quad \text{(Point: (0, 1))} \]
- Set \(y = 0\):
\[ x + 4(0) = 4 \implies x = 4 \quad \text{(Point: (4, 0))} \]
- Draw the line connecting (0, 1) and (4, 0).

3. **Identify the feasible region**:
- Since both inequalities are less than or equal to, shade the area below each line in the first quadrant (where \(x, y \geq 0\)).

### Step 2: Find the intersection points
1. **Find the intersection of the two lines**:
- Set the equations equal to each other:
\[ 2x + 3y = 6 \quad \text{(1)} \]
\[ x + 4y = 4 \quad \text{(2)} \]

- From equation (2), express \(x\) in terms of \(y\):
\[ x = 4 - 4y \]

- Substitute \(x\) in equation (1):
\[ 2(4 - 4y) + 3y = 6 \]
\[ 8 - 8y + 3y = 6 \]
\[ 8 - 5y = 6 \]
\[ -5y = -2 \implies y = \frac{2}{5} \]

- Substitute \(y\) back to find \(x\):
\[ x = 4 - 4\left(\frac{2}{5}\right) = 4 - \frac{8}{5} = \frac{20}{5} - \frac{8}{5} = \frac{12}{5} \]

- Thus, the intersection point is \(\left(\frac{12}{5}, \frac{2}{5}\right)\).

### Step 3: Identify the corner points
- The corner points of the feasible region are:
1. \((0, 0)\) (origin)
2. \((0, 2)\) from the line \(2x + 3y = 6\)
3. \((3, 0)\) from the line \(2x + 3y = 6\)
4. \((0, 1)\) from the line \(x + 4y = 4\)
5. \((4, 0)\) from the line \(x + 4y = 4\)
6. \(\left(\frac{12}{5}, \frac{2}{5}\right)\) from the intersection of the two lines.

### Conclusion
The corner points of the feasible region are \((0, 0)\), \((0, 2)\), \((3, 0)\), \((0, 1)\), \((4, 0)\), and \(\left(\frac{12}{5}, \frac{2}{5}\right)\).