The corner points of the feasible region determined by the system of linear constraints are (0, 10), (5, 5) (15, 15), (0, 20). Let Z = px + qy , where `p,q gt 0`. Then, the condition on p and q so that the maximum of Z occurs at both the points (15, 15) and (0, 20), is

Options

A. p=q
B. p=2q
C. q=2p
D. q=3p

Explanation

To solve the problem, we need to find the condition on \( p \) and \( q \) such that the maximum of \( Z = px + qy \) occurs at both points \( (15, 15) \) and \( (0, 20) \).

### Step-by-step Solution:

1. **Identify the Objective Function**:
We have the objective function \( Z = px + qy \).

2. **Evaluate Z at the Points**:
- At the point \( (15, 15) \):
\[ Z(15, 15) = p(15) + q(15) = 15p + 15q \]
- At the point \( (0, 20) \):
\[ Z(0, 20) = p(0) + q(20) = 20q \]

3. **Set the Two Expressions for Z Equal**:
Since we want the maximum of \( Z \) to be the same at both points, we set the two expressions equal to each other:
\[ 15p + 15q = 20q \]

4. **Rearrange the Equation**:
Rearranging gives:
\[ 15p + 15q - 20q = 0 \]
Simplifying this, we get:
\[ 15p - 5q = 0 \]

5. **Solve for the Relationship Between p and q**:
Dividing the entire equation by 5:
\[ 3p = q \]

6. **Conclusion**:
The condition on \( p \) and \( q \) such that the maximum of \( Z \) occurs at both points \( (15, 15) \) and \( (0, 20) \) is:
\[ q = 3p \]