If the corner points of a feasible region of system of linear inequalities are (15,20) (40,15) and (2,72) and the objective function is Z = 6x + 3y then `Z_(max) - Z_(min)` =

Options

A. 135
B. 78
C. 57
D. 101

Answer

To solve the problem, we need to evaluate the objective function \( Z = 6x + 3y \) at the given corner points of the feasible region, which are (15, 20), (40, 15), and (2, 72). We will then find the maximum and minimum values of \( Z \) and compute \( Z_{max} - Z_{min} \).

### Step-by-Step Solution:

1. **Evaluate \( Z \) at the point (15, 20)**:
\[ Z = 6(15) + 3(20) \]
\[ Z = 90 + 60 = 150 \]

2. **Evaluate \( Z \) at the point (40, 15)**:
\[ Z = 6(40) + 3(15) \]
\[ Z = 240 + 45 = 285 \]

3. **Evaluate \( Z \) at the point (2, 72)**:
\[ Z = 6(2) + 3(72) \]
\[ Z = 12 + 216 = 228 \]

4. **Determine \( Z_{max} \) and \( Z_{min} \)**:
- From the calculations:
- \( Z(15, 20) = 150 \)
- \( Z(40, 15) = 285 \)
- \( Z(2, 72) = 228 \)
- Therefore, \( Z_{max} = 285 \) and \( Z_{min} = 150 \).

5. **Calculate \( Z_{max} - Z_{min} \)**:
\[ Z_{max} - Z_{min} = 285 - 150 = 135 \]

Thus, the final answer is:
\[ Z_{max} - Z_{min} = 135 \]