Consider the following L.P.P minimize z = x - 7y + 190 subject to x + y ≤ 8, x + y ≥ 4, x ≤ 5, y ≤ 5 and x, y ≥ 0. Then which of the following is/are…

We observe that the constraints $x \ge 0, y \ge 0, x \le 5, y \le 5$x≥0,y≥0,x≤5,y≤5 already confine the region to a small square in the first quadrant, so the feasible region is bounded.

This makes (B) true and (A) false.

---

To find the corner points, we look at the intersections of the boundary lines:

• Intersection of $x+y=4$x+y=4 with axes: $(4,0)$(4,0) and $(0,4)$(0,4).
• Intersection of $x+y=8$x+y=8 with $x=5$x=5: $(5,3)$(5,3).
• Intersection of $x+y=8$x+y=8 with $y=5$y=5: $(3,5)$(3,5).
• Intersection of $x=5$x=5 and $y=5$y=5 is $(5,5)$(5,5), but this is outside $x+y \le 8$x+y≤8 ($5+5=10 > 8$5+5=10>8).
• The actual corner points are: $(4,0), (5,0), (5,3), (3,5), (0,5), (0,4)$(4,0),(5,0),(5,3),(3,5),(0,5),(0,4).

Counting these points, there are 6 corner points. Therefore, (D) is true.

The correct choice is Option 4: (B) and (D) only.