The feasible region of an LPP Max $Z=3x+2y$ subject to $x ≥0, y≥0, x-2y≤3$ is:
Answer & explanation
Correct answer: option 3
$Z=3x+2y$
$x ≥0, y≥0$ → solution in first quadrant plotting $x-2y=3$ first
| x | 3 | 0 |
| y | 0 | -3/2 |
now checking for (0, 0)
$x - 2y ≤3$
$⇒0≤3$
So solution has two side of $x-2y=3$ containing (0, 0)