Two godowns, A and B, have a grain storage capacity of 100 quintals and 50 quintals respectively. Their supply goes to three ration shops, D, E and F, whose requirements are 60, 50 and 40 quintals respectively. The costs of transportation per quintal from the godowns to the shops are given in the following table.


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



Let x quintals of supplies be transported from A to D and y quintals be transported from A to E.


Therefore, 100 - (x + y) will be transported to F.


Also, (60 - x) quintals, (50 - y) quintals and (40 – (100 – (x + y))) quintals will be transported to D, E, F by godown B.


According to the question,


x


Minimize Z = 6x + 4(60 - x) + 3y + 2(50 – y) + 2.50(100 - (x + y)) + 3((x + y ) – 60 )


Z = 6x + 240 – 4x + 3y + 100 – 2y + 250 - 2.5x - 2.5y + 3x + 3y – 180


Z = 2.5x + 1.5y + 210


The feasible region represented by x is given by



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



The minimum value of Z is 310 at point (10,50).


Hence, 10, 50, 40 quintals of supplies should be transported from A to D, E, F and 50, 0, 0 quintals of supplies should be transported from B to D, E, F.


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