The corner points of the feasible region determined by the system of linear inequalities are (0,0),(4,0), (2, 4) and (0, 5). If the maximum value of $Z=ax+by $ where $a, b > 0$ occurs at both (2, 4) and (4, 0), then

a=b a=2b 2a=b a=3b

a=2b

The correct answer is Option (2) → $a=2b$ as max occurs at both (2, 4) and (4, 0) $Z(2,4)=Z(4,0)$ $2a+4b=4a$ so $a=2b$