Question

A medicine company has factories at two places, X and Y. From these places, supply is made to each of its three agencies situated at P, Q and R. the monthly requirement of the agencies are respectively 40 packets, 40 packets and 50 packets of medicine, while the production capacity of the factories at X and Y are 60 packets and 70 packets respectively. 

The transportation costs per packet from the factories to the agencies are given as follows.

How many packets from each factory should be transported to each agency so that the cost of transportation is minimum? Also, find the minimum cost.

Answer 1

Let x packets of medicines be transported from X to P and y packets of medicines be transported from X to Q.

Therefore, 60 - (x + y) will be transported to R.

Also, (40 - x) packets, (40 - y) packets and (50 - (60 - (x + y))) packets will be transported to P, Q, R from Y.

∴According to the question,

x ≥ 0, y ≥ 0, x + y ≤ 60, x ≤ 40, y ≤ 40, x + y ≥ 10

Minimize Z = 5x + 4(40 - x) + 4y + 2(40 - y) + 3(60 - (x + y)) + 5((x + y) – 10)

Z = 3x + 4y + 370

The feasible region represented by x ≥ 0, y ≥ 0, x + y ≤ 60, x ≤ 40, y ≤ 40, x + y ≥ 10 is given by

The corner points of feasible region are A(0,10) , B(0,40) , C(20,40) , D(40,20), E(10,0).

Corner Point	Z = 3x + 4y + 370
A(0, 10)	410
B(0, 40)	530
C(20, 40)	590
D(40, 20)	570
E(10, 0)	400	Minimum

The minimum value of Z is 40 at point (10,0). 

Hence, 10, 0, 50 packets of medicines should be transported from X to P, Q, R and 30, 40, 0 packets of medicines should be transported from Y to P, Q, R.