The shaded region is bounded and has coordinates of corner points as $(0,0),(7,0),(3,4)$ and $(0,2)$. Also, $Z=5 x+7 y$.

Hence, the maximum value of Z is 43 at (3, 4).

From the figure, it is clear that feasible region is bounded with coordinates of corner points as $(0,3),(3,2)$ and $(0,5)$. Here, $Z=11 x+7 y$.

$$\begin{aligned} & \because \quad x+3 y=9 \text { and } x+y=5 \\ & \Rightarrow \quad 2 y=4 \\ & \therefore \quad y=2 \text { and } x=3 \end{aligned}$$

So, intersection points of $x+y=5$ and $x+3 y=9$ is $(3,2)$.

Hence, the minimum value of Z is 21 at (0, 3).

It is clear that Z is maximum at (3, 2) and its maximum value is 47.

From the shaded region, it is clear that feasible region is unbounded with the corner points $A(4,0), B(2,1)$ and $C(0,3)$

Also, we have $$Z=4 x+y$$

[since, $x+2 y=4$ and $x+y=3 \Rightarrow y=1$ and $x=2$ ]

Now, we see that 3 is the smallest value of $Z$ at the corner point $(0,3)$. Note that here we see that, the region is unbounded, therefore 3 may or may not be the minimum value of $Z$.

To decide this issue, we graph the inequality $4 x+y< 3$ and check whether the resulting open half plan has no point in common with feasible region otherwise, $Z$ has no minimum value.

From the shown graph above, it is clear that there is no point in common with feasible region and hence $Z$ has minimum value 3 at $(0,3)$.

From the shaded bounded region, it is clear that the coordinates of corner points are $\left(\frac{3}{13}, \frac{24}{13}\right),\left(\frac{18}{7}, \frac{2}{7}\right),\left(\frac{7}{2}, \frac{3}{4}\right)$ and $\left(\frac{3}{2}, \frac{15}{4}\right)$.

Also, we have to determine maximum and minimum value of $Z=x+2y$.

Hence, the maximum and minimum values of $Z$ are 9 and $3 \frac{1}{7}$, respectively.