Minimise $Z=13 x-15 y$ subject to the constraints $x+y \leq 7,2 x-3 y+6 \geq 0, x \geq 0, y \geq 0$.

Shaded region shown as $O A B C$ is bounded and coordinates of its corner points are $(0,0)$, $(7,0),(3,4)$ and $(0,2)$, respectively.

Hence, the minimum value of Z is $(-30)$ at (0, 2).

As clear from the graph, corner points are $O, A, E$ and $D$ with coordinates $(0,0),(52,0)$, $(144,16)$ and $(0,38)$, respectively. Also, given region is bounded.

$$\begin{array}{lrl} \text { Here, } & Z & =3 x+4 y \\ \because & 2 x+y & =104 \text { and } 2 x+4 y=152 \\ \Rightarrow & -3 y & =-48 \\ \Rightarrow & y & =16 \text { and } x=44 \end{array}$$

Hence, $Z$ is at $(44,16)$ is maximum and its maximum value is 196 .

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.