Let the factory I operate for $x$ days and the factory II operate for $y$ days.

At factory I, 50 calculators of model A and at factory II, 40 calculators of model $A$ are made everyday. Also, company has ordered for atleast 6400 calculators of model $A$.

$$\therefore \quad 50 x+40 y \geq 6400 \Rightarrow 5 x+4 y \geq 640\quad\text{.... (i)}$$

Also, at factory I, 50 calculators of model $B$ and at factory II, 20 calculators of modal $B$ are made everyday.

Since, the company has ordered atleast 4000 calculators of model $B$.

$\therefore \quad 50 x+20 y \geq 4000 \Rightarrow 5 x+2 y \geq 400\quad\text{.... (ii)}$

$$ \begin{array}{lc} \text { Similarly, for model C, } & 30 x+40 y \geq 4800 \\ \Rightarrow & 3 x+4 y \geq 480 \quad\text{.... (iii)}\\ \text { Also, } & x \geq 0, y \geq 0\quad\text{.... (iv)} \end{array}$$

[since, $x$ and $y$ are non-negative] It costs ₹ 12000 and ₹ $15000$ each day to operate factories I and II, respectively.

$\therefore$ Corresponding LPP is,

Minimise $Z=12000 x+15000 y$, subject to

$$\begin{gathered} 5 x+4 y \geq 640 \\ 5 x+2 y \geq 400 \\ 3 x+4 y \geq 480 \\ x \geq 0, y \geq 0 \end{gathered}$$

On solving $3 x+4 y=480$ and $5 x+4 y=640$, we get $x=80, y=60$.

On solving $5 x+4 y=640$ and $5 x+2 y=400$, we get $x=32, y=120$

Thus, from the graph, it is clear that feasible region is unbounded and the coordinates of corner points $A, B, C$ and $D$ are $(160,0),(80,60),(32,120)$ and $(0,200)$, respectively.

From the above table, it is clear that for given unbounded region the minimum value of $Z$ may or may not be 1860000.

Now, for deciding this, we graph the inequality

$$\begin{gathered} 12000 x+15000 y<1860000 \\ 4 x+5 y<620 \end{gathered}$$

and check whether the resulting open half plane has points in common with feasible region or not.

Thus, as shown in the figure, it has no common points so, $Z=12000 x+15000 y$ has minimum value 1860000 .

So, number of days factory I should be operated is 80 and number of days factory II should be operated is 60 for the minimum cost and satisfying the given constraints.

Given LPP is,

maximise and minimise $Z=3 x-4 y$ subject to $x-2 y \leq 0,-3 x+y \leq 4, x-y \leq 6, x, y \geq 0$.

[on solving $x-y=6$ and $x-2 y=0$, we get $x=12, y=6$ ]

From the shown graph, for the feasible region, we see that it is unbounded and coordinates of corner points are $(0,0),(12,6)$ and $(0,4)$.

For given unbounded region the minimum value of $Z$ may or may not be $-16$ . So, for deciding this, we graph the inequality.

$$3 x-4 y<-16$$

and check whether the resulting open half plane has common points with feasible region or not.

Thus, from the figure it shows it has common points with feasible region, so it does not have any minimum value.

Also, similarly for maximum value, we graph the inequality $3 x-4 y>12$ and see that resulting open half plane has no common points with the feasible region and hence maximum value 12 exist for $Z=3 x-4 y$.