An oil company has two depots, A and B, with capacities of 7000 L and 4000 L respectively. The company is to supply oil to three pumps, D, E, F, whose requirements are 4500 L, 3000 L, and 3500 L respectively. The distances (in km) between the depots and the petrol pumps are given in the following table: Assuming that the transportation cost per km is re 1 per litre, how should the delivery be scheduled in order that the transportation cost is minimum?

Question

Philoid · Accepted Answer

Let x liters of petrol be transported from A to D and y liters of petrol be transported from A to E.

Therefore, 7000 - (x + y) will be transported to F.

Also, (4500 - x) liters of petrol, (3000 - y) liters of petrol and (3500 – (7000 – (x + y))) liters of petrol will be transported to D, E, F by B.

∴According to the question,

x

Minimize Z = 7x + 3(4500 – x) + 6y + 4(3000 – y) + 3(7000 –(x + y)) + 2 ( (x + y) – 3500)

Z = 3x + y + 39500

The feasible region represented by x is given by

The corner points of feasible region are A(500,3000) , B(4000,3000) , C(4500,2500) , D(4500,0) , E(3500,0)

The minimum value of Z is 44000 at point (500,3000).

Hence, 500,3000,3500 liters of petrol should be transported from A to D, E, F and 4000, 0, 0 liters of petrol should be transported from B to D, E, F.