The corner points of the bounded feasible region for an LPP are (0,5), (6,6), (9,9) and (0,12). If the objective function is \( Z=px+qy;\ p,q\gt 0 \), then the condition on \( p \) and \( q \) so that the maximum of \( Z \) occurs at (9,9) and (0,12) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( p=\displaystyle\frac{1}{3}q \)
If the maximum occurs at both (9, 9) and (0, 12), Z must have the same value at these two corner points:
\( \displaystyle Z(9,9)=9p+9q \)
\( \displaystyle Z(0,12)=12q \)
\( \displaystyle 9p+9q=12q \)
\( \displaystyle 9p=3q \)
\( \displaystyle p=\frac{1}{3}q \)
This common value is indeed the maximum. With \( p=\frac{q}{3} \):
\( \displaystyle Z(0,5)=5q \)
\( \displaystyle Z(6,6)=2q+6q=8q \)
\( \displaystyle Z(9,9)=3q+9q=12q \)
\( \displaystyle Z(0,12)=12q \)
Since q > 0, the largest value 12q occurs at (9, 9) and (0, 12).
NCERT Exemplar:
Linear Programming
