Graphically, x+y<1 and y>0 represent a feasible region to the left of the line x+y=1 and above the x-axis. Neither the line nor the x-axis themselves are included in this region. It's clear that x<1 because if x=1, then y<0 which counters the constraint y>0.
Now consider the line 2x+z=3. When x=1, z=1. We can build a small table of values using x as the common value for deciding y and z. Since y>0, |y|=y.
| x |
y |
z |
|x|+|y|+|z| |
| -1 |
0<y<2 |
5 |
y+6 |
| -0.5 |
0<y<1.5 |
4 |
y+4.5 |
| 0 |
0<y<1 |
3 |
y+3 |
| 0.5 |
0<y<0.5 |
2 |
y+2.5 |
| 1 |
y<0! |
1 |
|
The last row contains the prohibited value of x. When x=1-δ, then y<δ and |x|+|y|+|z|=1-δ+y+1+2δ=y+2+δ.
Therefore |x|+|y|+|z|→2 or =2+.