The corner points of the feasible region are `(0,3), (3,0), (8,0), (12/5,38/5)` and `(0,10)`, then the point of maximum `z = 6x + 4y= 48` is at

Options

A. `(0,3)`
B. `(8,0)`
C. `(12//5, 38//5)`
D. `(3,0)`

Answer

To find the point at which the maximum value of \( z = 6x + 4y \) occurs given the corner points of the feasible region, we will evaluate \( z \) at each of the provided corner points: \( (0,3) \), \( (3,0) \), \( (8,0) \), \( \left(\frac{12}{5}, \frac{38}{5}\right) \), and \( (0,10) \).

### Step-by-Step Solution:

1. **Evaluate \( z \) at the point \( (0, 3) \)**:
\[ z = 6(0) + 4(3) = 0 + 12 = 12 \]

2. **Evaluate \( z \) at the point \( (3, 0) \)**:
\[ z = 6(3) + 4(0) = 18 + 0 = 18 \]

3. **Evaluate \( z \) at the point \( (8, 0) \)**:
\[ z = 6(8) + 4(0) = 48 + 0 = 48 \]

4. **Evaluate \( z \) at the point \( \left(\frac{12}{5}, \frac{38}{5}\right) \)**:
\[ z = 6\left(\frac{12}{5}\right) + 4\left(\frac{38}{5}\right) = \frac{72}{5} + \frac{152}{5} = \frac{224}{5} = 44.8 \]

5. **Evaluate \( z \) at the point \( (0, 10) \)**:
\[ z = 6(0) + 4(10) = 0 + 40 = 40 \]

### Summary of \( z \) values at each point:
- At \( (0, 3) \): \( z = 12 \)
- At \( (3, 0) \): \( z = 18 \)
- At \( (8, 0) \): \( z = 48 \)
- At \( \left(\frac{12}{5}, \frac{38}{5}\right) \): \( z = 44.8 \)
- At \( (0, 10) \): \( z = 40 \)

### Conclusion:
The maximum value of \( z \) is \( 48 \), which occurs at the point \( (8, 0) \).