The feasible region of the linear programming problem is represented below: The constraints of this LPP are |
$x + y ≥ 50,3x + y ≥ 90, x, y ≥0$ $x + y < 50,3x + y ≤ 90, x, y ≥0$ $x + y ≥ 50,3x + y ≤ 90, x, y≥0$ $x + y ≥ 50,3x + y < 90, x, y ≥0$ |
$x + y ≥ 50,3x + y < 90, x, y ≥0$ |
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: $x+y\ge 50,\;3x+y<90,\;x,y\ge 0$ |