Given a linear programming problem max z= 22x + 18y subject to x + y ≤ 20, 360x + 240 y ≤ 5760 and x ≥ 0, y ≥0. Its corner points are :

$(0,0), (16,0),(8,12), (0, 20)$

The correct answer is Option (1) → $(0,0), (16,0),(8,12), (0, 20)$

$x + y ≤ 20, 360x + 240 y ≤ 5760$

$\frac{x}{16}+\frac{y}{24}≤1$

finding intersection of $x+y=20$  ...(1)

$\frac{x}{16}+\frac{y}{24}=1$

$3x+2y=48$  ...(2)

eq. (2) - 2 × eq. (1)

$x=8$ from (1)

$y=12$

corner points → $(0,0), (0,20),(8,12), (16, 0)$