The corner points of the bounded feasible region of the LPP: Maximize $Z = x + y$ subject to constraints $2x+5y ≤ 100, 8x+5y ≤200, x ≥ 0, y ≥ 0$ are

$(0, 0), (25, 0), (0, 20), (\frac{50}{3},\frac{40}{3})$

The correct answer is Option (1) → $(0, 0), (25, 0), (0, 20), (\frac{50}{3},\frac{40}{3})$

Constraints:

1) $2x + 5y \le 100$

2) $8x + 5y \le 200$

3) $x \ge 0, \ y \ge 0$

Find intersections:

Intersection of $2x+5y=100$ and $8x+5y=200$:

Subtract: $(8x+5y)-(2x+5y)=200-100 \Rightarrow 6x=100 \Rightarrow x = \frac{100}{6} = \frac{50}{3}$

Substitute $x=50/3$ into $2x+5y=100$: $2*(50/3)+5y=100 \Rightarrow 100/3 + 5y =100 \Rightarrow 5y=200/3 \Rightarrow y = 40/3$

Intersection with axes:

$2x+5y=100$: x=0 → y=20, y=0 → x=50

$8x+5y=200$: x=0 → y=40, y=0 → x=25

Check feasible region (satisfies all constraints):

Points inside first quadrant and satisfy both inequalities: $(0,0), (0,20), (25,0), (\frac{50}{3},\frac{40}{3})$

Answer: Corner points: $(0,0), (0,20), (25,0), (\frac{50}{3},\frac{40}{3})$