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 :
Answer & explanation
Correct answer: option 1
$Z=x-y$, $x+y ≤3,$ $y-x ≥ 1$
Solving $x+y=3$
$y-x=1$
we get, $x=1,y=2$
| points → | (0, 1) | (0, 3 ) | (1, 2) |
| Z value→ | -1 | -3 | -1 |
min. value at (0, 3)