Use the same region as in the last problem: x + 2y ≥ 8, 3x + y ≥ 9, x ≥ 0, y ≥ 0, with corners (0…

Step 1. The candidate. The corner values are $27$, $13$ and $16$. The largest is $M = 27$.

Step 2. A feasible point beyond it. The point $(0, 20)$ obeys $0 + 40 \ge 8$ and $0 + 20 \ge 9$, so it is feasible. There $Z = 3 \times 20 = 60$, which is more than $27$.

Step 3. Conclude. The half-plane $Z > 27$ has a point in common with the region. In fact $(0, y)$ gives $Z = 3y$, which grows without limit. So there is no maximum.