The corner points of the bounded feasible region for a linear programming problem (LPP) are (0, 3/2), (1, 2) and (4, 0). If the objective function is \(Z = ax + by\), where 'a' and 'b' are positive, then the condition on 'a' and 'b' so that the maximum of Z occurs at (1, 2) and (4, 0) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(3a = 2b\)
Values of Z at the corners:
\( \displaystyle Z\left(0, \tfrac{3}{2}\right) = \frac{3b}{2} \)
\( Z(1, 2) = a + 2b \)
\( Z(4, 0) = 4a \)
The maximum occurs at both (1, 2) and (4, 0) only if Z has the same value at both points:
\( a + 2b = 4a \)
\( 3a = 2b \)
This common value must also be the largest. With \(b = \frac{3a}{2}\), Z at both points is 4a, while at \(\left(0, \frac{3}{2}\right)\) it is \(\frac{9a}{4} \lt 4a\). So the condition is \(3a = 2b\).
NCERT Exemplar:
Linear Programming
