Consider the LPP Min $Z=x-y,$ subject to the conditions $x+y ≤3,$ $y-x ≥ 1,$ $x≥0, y≥ 0,$ then minimum value of objective function exists at the point : |
(0, 3) (3, 0) (1, 2) (2, 1) |
(0, 3) |
$Z=x-y$, $x+y ≤3,$ $y-x ≥ 1$ Solving $x+y=3$ $y-x=1$ we get, $x=1,y=2$
min. value at (0, 3) |