We are to minimise
Z = - 50x + 20 y subject to the constraints 2x - y ≥ - 5, 3x + y ≥ 3, 2x - 3y ≤ 12, x ≥ 0, y ≥ 0
Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
Now we draw the graph of 2x - y = - 5
For x = 0,   - y = -5   or  y = 5
For y = 0,  2x =  -5,  or  x = -
 line meets OX in 
Now we draw the graph of 3x + y = 3
For x = 0, y = 3
For y = 0, 3x = 3 or x = 1
∴ line meets OX in B(1, 0) and OY in M(0, 3)
Again we draw the graph of 2x - 3 y - 12
For x = 0, - 3 y = 12 or y = - 4
For y = 0, 2 x = 12 or x = 6
∴ line meets OX in C(6, 0) and OY in N(0, - 4)
Since feasible region is the region which satisfies alt the constraints.
∴  feasible region (shaded) is unbounded and has corner points B(1, 0), C(6, 0), L(0, 5), M(0,3)
At B(1, 0), Z = - 50 + 0 = - 50
At C(6, 0), Z = - 300 + 0 = - 300
At L(0, 5), Z = 0 + 100 = 100
At M(0, 3), Z = 0 + 60 = 60
∴ - 300 is the smallest value of Z at the comer point (6, 0).
Since the feasible region is unbounded.
∴ we are to check whether this value is minimum.

For this, we draw the graph of
- 50x + 20 y < - 300 i.e. - 5x + 2y < - 30    ...(1)
Consider - 5x + 2 y = - 30
This line passes through C(6, 0), D(8, 5).
Now (1) has common points with feasible region.
∴  Z = - 50x + 20y has no minimum value.