The feasible region of the linear programming problem is represented…

The feasible region of the linear programming problem is represented below:

$Feasible region of the LPP$

The constraints of this LPP are

Answer & explanation

Correct answer: option 4

The correct answer is Option (4) → $x + y \ge 50, 3x + y \lt 90, x, y \ge 0$

Note: The given answer is as per the official NTA Answer Key. However, Option (3) could also be a possible answer, as explained below.

The boundary lines pass through the marked points:

Through (0, 50) and (50, 0): $x + y = 50$ (drawn solid)
Through (0, 90) and (30, 0): $3x + y = 90$ (drawn dashed)

The shaded region lies in the first quadrant, between (0, 50) and (0, 90) on the y-axis. Testing the origin, which is not in the region:

$x + y = 0$, which is less than 50. The region is on the other side, so $x + y \ge 50$.
$3x + y = 0$, which is less than 90. The region is on the same side, so $3x + y$ is below 90.

The only question is whether the points of the line $3x + y = 90$ belong to the region. In a graph, a dashed boundary usually marks a line that is not part of the region (a strict inequality). In linear programming, however, constraints are normally written with $\le$ or $\ge$, so the boundary lines are part of the feasible region.

Option (1) $x + y \ge 50, 3x + y \ge 90, x, y \ge 0$ — Incorrect. $3x + y \ge 90$ is the side of the dashed line away from the origin, but the shaded region is on the origin’s side.

Option (2) $x + y \lt 50, 3x + y \le 90, x, y \ge 0$ — Incorrect. $x + y \lt 50$ is the side of the solid line containing the origin, but the shaded region is on the other side.

Option (3) $x + y \ge 50, 3x + y \le 90, x, y \ge 0$ — Should also be correct, though not the answer given by NTA. It describes the shaded region with the usual non-strict LPP constraints, reading the dashed line only as a boundary of the region. It differs from Option (4) only in whether the points of $3x + y = 90$ are included.

Option (4) $x + y \ge 50, 3x + y \lt 90, x, y \ge 0$ — Correct. The solid line $x + y = 50$ belongs to the region, so the first constraint has $\ge$; the dashed line $3x + y = 90$ does not, so the second constraint is strict.