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

$a=2b$

The correct answer is Option (3) → $a=2b$

Given corner points (2,4) and (4,0) give same maximum of $Z=ax+by$

$2a+4b=4a+0\cdot b$

$2a+4b=4a$

$4b=2a$

$a=2b$

Relation: $a=2b$