KCET 2018 Math Solved Paper

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.