Maximize $z=2x-y$ Subject to constraints $x+y ≤4, x+3y ≤ 6, x ≥ 0, y ≥ 0$. The corner points of the feasible region are :

(0, 4), (4,0), (0, 0), (2, 3) (0, 2), (6,0), (0, 0), (1, 2) (0, 2), (4,0), (0, 0), (3, 1) (0, 2), (4,0), (0, 0), (1, 3)

(0, 2), (4,0), (0, 0), (3, 1)

The correct answer is Option (3) → (0, 2), (4,0), (0, 0), (3, 1) intersection point for  $x+y=4$ ...(1) $x+3y=6$ ...(2) Eq. (2) - Eq. (1) $2y=2⇒y=1$ from (1) = $x=3$ corner points → $(0, 2), (4,0), (0, 0), (3, 1)$