# A company makes 3 model of calculators:A,B and c at factory I and factory II.The company has orders for at least 6400 calculators of model A;4000 calculator of model B and 4800 calculator of model C.At factory I ,50 calculators of model A,50 of model B and 30 of model C are made every day;at factory II,40 calculators of model A,20 of model B and 40 of model C are made every day.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 minimize the operating costs and still meet the demand.

Toolbox:
• Let $R$ be the feasible region for a linear programming problem and let $z=ax+by$ be the objective function.When $z$ has an optimum value (maximum or minimum),where variables $x$ and $y$ are subject to constraints described by linear inequalities,this optimum value must occur at a corner point of the feasible region.
• If R is bounded then the objective function Z has both a maximum and minimum value on R and each of these occur at corner points of R
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