(B) Let factory $I$ operate for $x$ days and 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 every day. The company has orders for at least $6400$ calculators of model $A$.
$\therefore 50x + 40y \geq 6400 \Rightarrow 5x + 4y \geq 640$
At factory $I, 50$ calculators of model $B$ and at factory $II, 20$ calculators of model $B$ are made every day. The company has orders for at least $4000$ calculators of model $B$.
$\therefore 50x + 20y \geq 4000 \Rightarrow 5x + 2y \geq 400$
At factory $I, 30$ calculators of model $C$ and at factory $II, 40$ calculators of model $C$ are made every day. The company has orders for at least $4800$ calculators of model $C$.
$\therefore 30x + 40y \geq 4800 \Rightarrow 3x + 4y \geq 480$
Also,$x \geq 0, y \geq 0.$
We have to minimize the cost $Z = 12000x + 15000y$ subject to the constraints:
$5x + 4y \geq 640$
$5x + 2y \geq 400$
$3x + 4y \geq 480$
$x, y \geq 0$
The feasible region is unbounded with corner points $A(160, 0), B(80, 60), C(32, 120),$ and $D(0, 200).$
| Corner points | Value of $Z = 12000x + 15000y$ |
| $(160, 0)$ | $1920000$ |
| $(80, 60)$ | $1860000$ (Minimum) |
| $(32, 120)$ | $2184000$ |
| $(0, 200)$ | $3000000$ |
To verify the minimum,we plot $12000x + 15000y < 1860000$ or $4x + 5y < 620.$ As shown in the figure,there are no common points with the feasible region,so the minimum value is $1860000.$
Thus,factory $I$ should operate for $80$ days and factory $II$ should operate for $60$ days.