Answer :

let x and y liters of oil be supplied by A to shop D and E respectively. Then 7000 - x - y will be supplied to F. Obviously x ≥ 0, y ≥ 0.


Since requirement at D is 4500L but supplied x quintal hence, 4500 - x is sent form B to D.


Similarly requirement at E is 3000L but supplied y quintal hence, 3000 - y is sent form B to E.


Similarly requirement at E is 3500 quintal but supplied 7000 - x - y quintal hence, 3500 - (7000 - x - y) i.e. x + y - 3500 is sent form B to E.


Diagrammatically it can be explained as:


2nd.JPG


As x ≥ 0, y ≥ 0 and 7000 - x - y ≥ 0 x + y ≤ 7000


4500 - x ≥ 0, 3000 - y ≥ 0 and x + y - 3500 ≥ 0


x ≤ 4500, y ≤ 3000 and x + y ≥ 3500


Cost of delivering 10L of petrol = Re.1


Then Cost of delivering 1L of petrol = Rs.1/10


Total transportation cost Z can be calculated as:


Z =


Z = 0.3x + 0.1y + 3950


Mathematical formulation of given problem is as follows:


Minimize Z = 0.3x + 0.1y + 3950 ….(1)


Subject to the constraints,


x + y ≤ 7000 …..(2)


x ≤ 4500 …..(3)


y ≤ 3000 ….(4)


x + y ≥ 3500 …(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(3500,0),B(4500,0),C(4500,2500), D(4000,3000) and E(500,3000) .



Now, we find the minimum value of Z. According to table the minimum value of Z = 4400 at point E (500,3000).


Hence, the amount of oil delivered from A to D, E and F is 500L, 3000L and 3500L quintals respectively and from B to D,E and F is 4000L, 0 L and 0 L respectively.


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