9.Ans: $(0, 0)$, $(4, 0)$, and $(0, 4)$
10.Ans: $(0, 0)$, $(3, 0)$, and $(0, 6)$
11.Ans: Adding both equations: $2x = 6 \Rightarrow x = 3$. Then $3 + y = 5 \Rightarrow y = 2$. The point is $(3, 2)$.
12.Ans: Multiply first by 3: $3x + 6y = 24$. Subtract second: $5y = 15 \Rightarrow y = 3$. Substitute $y$: $x + 6 = 8 \Rightarrow x = 2$. The point is $(2, 3)$.
13.Ans: The corner points are $(0, 0)$, $(3, 0)$, $(2, 3)$, and $(0, 4)$.
14.Ans: Values are $0, 12, 18, 20$. Maximum value is $20$ at $(0, 5)$.
15.Ans: Values are $50, 110, 120$. Minimum value is $50$ at $(0, 5)$.
16.Ans: Corner points are $(0, 0)$, $(10, 0)$, $(0, 10)$. At $(10, 0)$, $Z = 50$. At $(0, 10)$, $Z = 30$. Max value is $50$ at $(10, 0)$.
17.Ans: At $(0, 3)$, $Z = 0 + 6 = 6$. At $(6, 0)$, $Z = 6 + 0 = 6$. Minimum value is $6$ at all points on the segment joining $(0, 3)$ and $(6, 0)$.
18.Ans: At $(0, 10)$, $Z = 1900$. At $(5, 5)$, $Z = 1500 + 950 = 2450$. The point $(5, 5)$ yields the higher value.
19.Ans: The inequalities are $y \ge x + 1$ and $y \le x$. There is no region where $y$ is simultaneously greater than $x+1$ and less than $x$. Therefore, there is NO feasible region.
20.Ans: The corner points are $(0,0), (2,0), (2,3), (0,3)$. Max $Z$ occurs at $(2,3)$ which is $3(2) + 2(3) = 12$.
21.Ans: Yes, $m=12$ is indeed the true minimum.
22.Ans: The objective function has no maximum value (the LPP is unbounded above).
23.Ans: At $(2,0), Z = 6$. At $(0,2), Z = 10$. The region is unbounded. Check $3x+5y < 6$. This half-plane shares no points with $x+y \ge 2, x,y \ge 0$. So minimum is $6$ at $(2,0)$.
25.Ans: Infinitely many optimal solutions.
26.Ans: Infeasible Solution (No feasible region).
27.Ans: True (If the line segment forms part of the optimal boundary).
29.Ans: $Z = -3 + 2(2) = -3 + 4 = 1$.
30.Ans: It does not have a maximum or minimum value.
