Answer :

let x and y quintal of grain supplied by A to shop d and E respectively. Then 100 - x - y will be supplied to shop F. Obviously x ≥ 0, y ≥ 0.


Since requirement at shop D is 60 quintal but supplied x quintal hence, 60 - x is sent form B to D.


Similarly requirement at shop E is 50 quintal but supplied y quintal hence, 50 - y is sent form B to E.


Similarly requirement at shop E is 40 quintal but supplied 100 - x - y quintal hence, 40 - (100 - x - y) i.e. x + y - 60 is sent form B to E.


Diagrammatically it can be explained as:


godown.png


As x ≥ 0, y ≥ 0 and 100 - x - y ≥ 0 x + y ≤ 100


60 - x ≥ 0, 50 - y ≥ 0 and x + y - 60 ≥ 0


x ≤ 60, y ≤ 50 and x + y ≥ 60


Total transportation cost Z can be calculated as:


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


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


Z = 2.5x + 1.5y + 410


Mathematical formulation of given problem is as follows:


Minimize Z = 2.5x + 1.5y + 410 ….(1)


Subject to the constraints,


x + y ≤ 100 …..(2)


x ≤ 60 …..(3)


y ≤ 50 ….(4)


x + y ≥ 60 …(5)


x ≥ 0, y ≥ 0 ….(6)


Now let us graph the feasible region of the system of inequalities (2) to (6). The feasible region (shaded) is shown in the fig. Here, we can observe that the feasible region is bounded.



The coordinates of the corner points A(60,0),B(60,40),C(50,50) and D(10,50).



Now, we find the minimum value of Z. According to table the minimum value of Z = 510 at point D (10,50).


Hence, the amount of grain delivered from A to D, E and F is 10 quintal, 50 quintals and 40 quintals respectively and from B to D,E and F is 50 quintal, 0 quintal and 0 quintal respectively.


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