Corner poins of the feasible region determned by the system of linear constrainsts are (0,3), (1,1), and (3,0). Let Z=px+qy. Where `p, q lt 0` Condition on p and q, so that the minimum of Z occurs at (3,0) and (1,1) is

Options

A. p=2q
B. `p=(q)/(2)`
C. p=3q
D. p=q

Explanation

To solve the problem, we need to find the conditions on \( p \) and \( q \) such that the minimum value of \( Z = px + qy \) occurs at the points \( (3,0) \) and \( (1,1) \). Given that both \( p \) and \( q \) are less than 0, we will derive the necessary conditions step by step.

### Step-by-Step Solution:

1. **Define the function Z at the corner points:**
- For the point \( (3,0) \):
\[ Z(3,0) = p \cdot 3 + q \cdot 0 = 3p \]
- For the point \( (1,1) \):
\[ Z(1,1) = p \cdot 1 + q \cdot 1 = p + q \]

2. **Set the two expressions for Z equal to each other:**
Since we want the minimum value of \( Z \) to occur at both points, we set:
\[ 3p = p + q \]

3. **Rearrange the equation:**
To isolate \( q \), we can rearrange the equation:
\[ 3p - p = q \implies 2p = q \]

4. **Conclusion:**
The condition on \( p \) and \( q \) such that the minimum of \( Z \) occurs at both \( (3,0) \) and \( (1,1) \) is:
\[ q = 2p \]