First, we draw the graph of line 3x + 2y = 150
x                0            50
y                75           0

Now, we consider a point O(0, 0), i.e,  origin.
Substitute it in inequality 3x + 2y ≤ 150, i.e., 3(0) + 2(0) ≤ 150 or 0 ≤ 150 which is true.
∴ we mark that region which contains origin.
We draw the graph of line x + 4y = 80

x            0            80
y            20           0

Now, we consider a point O(0, 0), i.e, origin.
Substituting it in inequality x + 4y ≤ 80, we get
0 + 4(0) ≤ 80  or 0 ≤ 80 which is true.
∴ we mark that region which contains origin.
For x ≥ 15, we mark the region on right side of the line x = 15
For y ≥ 0, we mark the region above the line y = 0
Now, we shade the common region.

The required solution is given by shaded region ABCD as shown in the graph.