The feasible region of the linear programming problem is represented…

The correct answer is Option (4) → $x + y ≥ 50,3x + y < 90, x, y ≥0$

The two boundary lines shown in the graph are:

$x+y=50$ passing through $(50,0)$ and $(0,50)$

$3x+y=90$ passing through $(30,0)$ and $(0,90)$

The shaded feasible region lies:

above the line $x+y=50$

below the line $3x+y=90$

in the first quadrant

Therefore the constraints are:

$x+y\ge 50$

3x + y < 90

$x\ge 0,\; y\ge 0$

Correct answer: $x+y\ge 50,\;3x+y<90,\;x,y\ge 0$