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The corner points of the feasible region...

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`. Condition on p and q so that the maximum of Z occurs at both the points (15,15) and (0,20) is:

A

p=q

B

p=2q

C

q=2p

D

q=3p

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To solve the problem, we need to find the conditions on \( p \) and \( q \) such that the maximum value of \( Z = px + qy \) occurs at both points \( (15, 15) \) and \( (0, 20) \). ### Step-by-Step Solution: 1. **Define the Objective Function**: The objective function is given by: \[ Z = px + qy \] 2. **Calculate \( Z \) at the Point \( (15, 15) \)**: Substitute \( x = 15 \) and \( y = 15 \) into the objective function: \[ Z(15, 15) = p(15) + q(15) = 15p + 15q \] 3. **Calculate \( Z \) at the Point \( (0, 20) \)**: Substitute \( x = 0 \) and \( y = 20 \) into the objective function: \[ Z(0, 20) = p(0) + q(20) = 20q \] 4. **Set the Two Expressions for \( Z \) Equal**: Since we want the maximum value of \( Z \) to occur at both points, we set the two expressions equal to each other: \[ 15p + 15q = 20q \] 5. **Rearrange the Equation**: Rearranging gives: \[ 15p + 15q - 20q = 0 \implies 15p - 5q = 0 \] 6. **Solve for \( q \) in Terms of \( p \)**: From the equation \( 15p - 5q = 0 \), we can express \( q \) in terms of \( p \): \[ 15p = 5q \implies q = 3p \] ### Conclusion: The condition on \( p \) and \( q \) such that the maximum of \( Z \) occurs at both points \( (15, 15) \) and \( (0, 20) \) is: \[ q = 3p \]
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