The feasible region represented by the constraints \(x + y \le 50, 3x + y \le 90, x \ge 0, y \ge 0\) of an LPP is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Region A
The origin O(0, 0) decides the side of each line:
\(x + y \le 50\): O gives \(0 \le 50\), true, so the region is on the side of O.
\(3x + y \le 90\): O gives \(0 \le 90\), true, so the region is on the side of O.
With \(x \ge 0,\ y \ge 0\), the region lies in the first quadrant, below both lines. The two lines meet where
\(3x + y - (x + y) = 90 - 50\)
\(2x = 40,\quad x = 20,\quad y = 30\)
So the feasible region is the quadrilateral with corners (0, 0), (30, 0), (20, 30) and (0, 50). This is Region A.
Each of the other regions breaks at least one constraint. For example:
Region B, point (40, 5): \(3(40) + 5 = 125 \gt 90\)
Region C, point (5, 48): \(5 + 48 = 53 \gt 50\)
Region D, point (40, 50): \(40 + 50 = 90 \gt 50\) and \(3(40) + 50 = 170 \gt 90\)