Given equation, z=x+4y and constraints, x+2y≤2→ (1) ⇒
x
2
+
y
2
=1 x+2y≥8 → (2) ⇒
x
8
+
y
4
=1 x,y≥0 → (3)
Putting (0, 0) in Eq. (1), we get 0 ≤ 2 which is True. Putting (0, 0) in Eq. (2), we get 0 ≥ 8 which is False. Equation (3) implies that the solution is in first quadrant. From the graph, it has no feasible solution.