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 :
Answer & explanation
Correct answer: option 1
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 → $(0,0), (0,20),(8,12), (16, 0)$
