The corner points of the bounded feasible region determined by the system of linear inequalities are (0, 0), (4, 0), (2, 4) and (0, 5). If 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 (1) → $a= 2b$

At (2,4) and (4,0):

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

$\Rightarrow 2a+4b=4a\Rightarrow 4b=2a\Rightarrow a=2b$

Hence $a:b=2:1$ (i.e. $a=2b$)