Step 1:

Let $x$ calculators be produced by factory I and $y$ calculators be produced by factory II

Models$\qquad$Factory I$\qquad$Factory II$\qquad$Maximum limit

A $\qquad\qquad$50 $\qquad\qquad$40 $\qquad\qquad\;\;$6400

B $\qquad\qquad$50 $\qquad\qquad$20 $\qquad\qquad\;\;$4000

C $\qquad\qquad$30 $\qquad\qquad$40 $\qquad\qquad\;\;$4800

Hence the total cost for $x$ calculators by factory I and $y$ calculators by factory II is $12000x+15000y$

It is given that factory I can produce 50 calculators and factory II can produce 40 calculators of type A and the two factories can produce at 6400 calculators.

$\therefore 50x+40y\leq 6400$

Step 2:

Similarly factory I produces 50 calculators of type B and factory II produces 20 calculators of type B and the two factories can produce at least 4000 calculators.

$\therefore 50x+20y\geq 4000$

Finally factory I produces 30 calculators of type C and factory II produces 40 calculators of type C and the two factories can produce at least 4800 calculators.

$\therefore 30x+40y\geq 4800$

Hence the objective function $Z=12000x+15000y$

Subject to constraints,

$50x+40y\geq 6400$------(1)

$5x+4y\geq 640$

$50x+20y\geq 4000$------(2)

$30x+40y\geq 4800$

$3x+4y\geq 480$------(3)

and $x,y\geq 0$

Step 3:

The graph can be plotted for the three lines to find the feasible region bounded by threes three lines.

Consider the line$AB :5x+4y\geq 640$ when $x=0,y=0\Rightarrow 5\times 0+6\times 0\geq 640$

Which is not true.

Hence $O(0,0)$ does not belong to the region $5x+4y\geq 640$

It is clear that the region above $5x+4y=640$ represents $5x+4y\geq 640$

Similarly consider the line $EF :3x+4y=480$

Put $x=0,y=0\rightarrow 0\geq 480$

Which is not true.

Hence the region $3x+4y \geq 480$ lies above the line $3x+4y\geq 480$

Finally consider the line $CD :5x+2y=400$

Put $x=0,y=0\rightarrow 0\geq 400$

Which is not true.

Hence the region $5x+2y \geq 400$ lies above the line $5x+2y= 400$

Step 4:

On solving the equations $5x+4y=640$ and $3x+4y=480$ we obtain the point of intersection to be $Q(80,60)$

On solving the equations $5x+4y=640$ and $5x+2y=400$ we obtain the point of intersection to be $P(32,120)$

The point C is point of intersection is C(0,200) and the coordinates of F is (160,0) .Let us obtain the values of the objective function as follows :

Let us obtain the value of the objective function as follows :

At the point $(x,y)$ the value of the objective function $Z=12,000x+15000y$

At $C(0,200)$ the value of the objective function $Z=12000\times 0+15000\times 200=30,00,000$

At $C(32,120)$ the value of the objective function $Z=12000\times 32+15000\times 120=2184000$

At $Q(80,60)$ the value of the objective function $Z=12000\times 80+15000\times 60=1860000$

At $F(160,0)$ the value of the objective function $Z=12000\times 160+15000\times 0=1920000$

It is clear that $Z$ is minimum at $Q(80,60)$

Hence the number of days each factory should operate to minimize the operating cost is 80days by factory I and 60 days by factory II