the corner points of the feasible region of a system of linear inequalities `3x + 5y le 15, 5x + 2y ge 10, x ge 0, y ge 0`, are:

Options

A. `(2,0), ((20)/(19), (45)/(19)), (0,3)`
B. `(0,3), ((20)/(19), (45)/(19)), (0,5)`
C. `(2,0), ((20)/(19), (45)/(19)), (5,0)`
D. `(0,0), (2,0),((20)/(19), (45)/(19)), (0,3)`

Answer

To find the corner points of the feasible region defined by the system of linear inequalities:

1. **Identify the inequalities**:
- \(3x + 5y \leq 15\)
- \(5x + 2y \geq 10\)
- \(x \geq 0\)
- \(y \geq 0\)

2. **Convert inequalities to equations**:
- For \(3x + 5y = 15\)
- For \(5x + 2y = 10\)

3. **Find the intercepts for the first equation \(3x + 5y = 15\)**:
- Set \(x = 0\):
\[ 3(0) + 5y = 15 \implies y = 3 \quad \text{(Point: (0, 3))} \]
- Set \(y = 0\):
\[ 3x + 5(0) = 15 \implies x = 5 \quad \text{(Point: (5, 0))} \]

4. **Find the intercepts for the second equation \(5x + 2y = 10\)**:
- Set \(x = 0\):
\[ 5(0) + 2y = 10 \implies y = 5 \quad \text{(Point: (0, 5))} \]
- Set \(y = 0\):
\[ 5x + 2(0) = 10 \implies x = 2 \quad \text{(Point: (2, 0))} \]

5. **Plot the lines**:
- Plot the points (0, 3), (5, 0) for the line \(3x + 5y = 15\).
- Plot the points (0, 5), (2, 0) for the line \(5x + 2y = 10\).

6. **Find the intersection of the two lines**:
- Solve the equations simultaneously:
\[ 3x + 5y = 15 \quad \text{(1)} \]
\[ 5x + 2y = 10 \quad \text{(2)} \]
- From (1), express \(y\):
\[ 5y = 15 - 3x \implies y = \frac{15 - 3x}{5} \]
- Substitute \(y\) in (2):
\[ 5x + 2\left(\frac{15 - 3x}{5}\right) = 10 \]
\[ 5x + \frac{30 - 6x}{5} = 10 \]
\[ 25x + 30 - 6x = 50 \quad \text{(Multiply through by 5)} \]
\[ 19x = 20 \implies x = \frac{20}{19} \]
- Substitute \(x\) back to find \(y\):
\[ y = \frac{15 - 3\left(\frac{20}{19}\right)}{5} = \frac{15 - \frac{60}{19}}{5} = \frac{\frac{285 - 60}{19}}{5} = \frac{225}{19 \cdot 5} = \frac{45}{19} \]
- Thus, the intersection point is \(\left(\frac{20}{19}, \frac{45}{19}\right)\).

7. **Identify the corner points**:
- The corner points of the feasible region are:
- (0, 3)
- (5, 0)
- (2, 0)
- (0, 5)
- \(\left(\frac{20}{19}, \frac{45}{19}\right)\)

8. **Conclusion**:
- The corner points of the feasible region are:
- (2, 0)
- (5, 0)
- \(\left(\frac{20}{19}, \frac{45}{19}\right)\)
- (0, 3)

**Final Answer**: The corner points are \((2, 0)\), \((5, 0)\), and \(\left(\frac{20}{19}, \frac{45}{19}\right)\).