Question

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 8 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 ! 12000 and Z 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.

Answer 1

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.
$\therefore 50x+40y\ge 64005x+4y\ge 640....\left(i\right)$
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.
$\therefore 50x+20y\ge 4000\Rightarrow 5x+2y\ge \mathrm{400....}\left(ii\right)$
Similarly, for model C, $30x+40y\ge 4800$
$\Rightarrow 3x+4y\ge \mathrm{480........}\left(iii\right)$
Also, $x\ge 0,y\ge 0......\left(iv\right)$
[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
$5x+4y\ge 6405x+2y\ge 4003x+4y\ge 480x\ge 0,y\ge 0$
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 kx =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), respectivley.

$\begin{array}{cc}\text{Corner points}& \text{Value of Z =12000x+15000y}\\ (160,0)& 160\times 12000=1920000\\ (80,60)& (80\times 12+60\times 15)\times 1000=1860000\leftarrow Minimum\\ (32,120)& (32\times 12+120\times 15)\times 1000=2184000\\ (0,200)& 0+200\times 15000=3000000\end{array}$

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 and satisfying the given constraints.