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# Two godowns A and B have grain capacity of 100 quintals and 50 quintals respectively. They supply to 3 ration shops D, E and F whose requirements are 60, 50 and 40 quintals respectively.

6. Two godowns A and B have grain capacity of 100 quintals and 50 quintals respectively. They supply to 3 ration shops, D, E and F whose requirements are 60, 50 and 40 quintals respectively. The cost of  transportation per quintal from the godowns to the shops are given in the following table:

 Transportation cost per quintal (in Rs) From/To A B D 6 4 E 3 2 F 2.50 3

How should the supplies be transported in order that the transportation cost is minimum? What is the                   minimum cost?

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Let godown A supply  x and y quintals of grain to shops D and E respectively. Then , (100-x-y) will be  supplied to shop F. Requirements at shop D is  60 since godown A supply  x .Therefore remaining (60-x) quintals of grain will be transported from godown B.

Similarly, (50-y) quintals and 40-(100-x-y)=(x+y-60) will be transported from godown B to shop E and F respectively. The problem can be represented diagrammatically as follows:

$x,y\geq 0$  and  $100-x-y\geq 0$

$x,y\geq 0$  and  $x+y\leq 100$

$60-x\geq 0,50-y\geq 0\, \, \, and\, \, x+y-60\geq 0$

$\Rightarrow \, \, \, \, x\leq 60,y\leq 50,x+y\geq 60$

Total transportation cost z is given by ,

$z=6x+3y+2.5(100-x-y)+4(60-x)+2(50-y)+3(x+y-60)$

$z=2.5x+1.5y+410$

Mathematical formulation of given problem is as  follows:

Minimize : $z=2.5x+1.5y+410$

Subject to constraint ,

$x+y\leq 100$

$x\leq 60$

$y\leq 50$

$x+y\geq 60$

$x,y\geq 0$

The feasible  region determined by constraints is as follows:

The corner points of feasible region are $A(60,0),B(60,40),C(50,50),D(10,50)$

The value of Z at corner points is as shown :

 corner points $z=2.5x+1.5y+410$ $A(60,0)$ 560 $B(60,40)$ 620 $C(50,50)$ 610 $D(10,50)$ 510 minimum

therefore  510  may or may not be minimum value of Z .

Hence , Z has miniimum value  510  at  point  $D(10,50)$

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