A company makes 3 model of calculators; A, B and C at factory I and factory II. The company has orders for atleast 6400 calculators of model A, 4000 calculators of model B and 4800 calculators of model C. At factory I, 50 calculators of model A, 50 of model B and 30 of model C are made everyday; at factory II, 40 calculators of model A, 20 of model B and 40 of model C are made everyday. It costs Rs.12000 and Rs.15000 each day to operate factory I and II, respectively. Find the number of days each factory should operate to minimise the operating costs and still meet the demand.

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.

$\therefore$ $50x + 40y \ge 6400 \Rightarrow 5x + 4y \ge 640$ ……….(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.

$\therefore$ $50x + 20y \ge 4000 \Rightarrow 5x + 2y \ge 400$ ….(ii)

Similarly, for model $C,$ $30x + 40y \ge 4800$ …..(iii)

$\Rightarrow$ $3x + 4y \ge 480$ ……(iv)

Also, $x \ge 0,y \ge 0$

[since, $x$ and $y$ are non-negative]

It costs Rs.12000 and Rs.15000 each day to operate factories I and II, respectively.

$\therefore$ Corresponding LPP is,

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

$5x + 4y \ge 640$

$5x + 2y \ge 400$

$3x + 4y \ge 480$

$x \ge 0,y \ge 0$

On solving $3x + 4y = 480$ and $5x + 4y = 640$,

we get $x = 80,y = 60$

On solving $5x + 4y = 640$ and $5x + 2y = 400$,

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

$12000x + 15000y < 1860000$

$\Rightarrow$ $4x + 5y < 620$

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,

$Z = 12000x + 15000y$ 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