For an objective function Z = ax + by, where `a, bgt0`, the corner points of the feasible region determined by a set of constraints (linear inequalities) are (0, 20). (10, 10), (30, 30) and (0, 40). The condition on a and b such that the maximum z occurs at both the points (30, 30) and (0, 40) is:

Options

A. `b - 3a =0`
B. `a=3b`
C. `a+2b=0`
D. `2a - b =0 `

Answer

To solve the problem, we need to find the conditions on \( a \) and \( b \) such that the maximum value of the objective function \( Z = ax + by \) occurs at both points \( (30, 30) \) and \( (0, 40) \).

### Step-by-step Solution:

1. **Substituting the first point (30, 30) into the objective function:**
\[ Z_1 = a(30) + b(30) = 30a + 30b \]

2. **Substituting the second point (0, 40) into the objective function:**
\[ Z_2 = a(0) + b(40) = 0 + 40b = 40b \]

3. **Setting the two expressions for \( Z \) equal to each other since both points yield the maximum value:**
\[ 30a + 30b = 40b \]

4. **Rearranging the equation:**
\[ 30a + 30b - 40b = 0 \]
\[ 30a - 10b = 0 \]

5. **Factoring out the common terms:**
\[ 10(3a - b) = 0 \]

6. **Setting the factor equal to zero:**
\[ 3a - b = 0 \]

7. **Rearranging to find the relationship between \( a \) and \( b \):**
\[ b = 3a \]

### Conclusion:
The condition on \( a \) and \( b \) such that the maximum \( Z \) occurs at both points \( (30, 30) \) and \( (0, 40) \) is:
\[ b - 3a = 0 \]