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 ≥ 0, y ≥ 0, x + y ≤ 100, x ≤ 60, y ≤ 50, x + y ≥ 60
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 ≥ 0, y ≥ 0, x + y ≤ 100, x ≤ 60, y ≤ 50, x + y ≥ 60 is given by

The corner points of feasible region are A(10,50) , B(50,50) , C(60,40) , D(60,0)
Corner Point |
Z = 2.5x + 1.5y + 210 |
|
A(10, 50) |
310 |
Minimum |
B(50, 50) |
410 |
|
C(60, 40) |
420 |
|
D(60, 0) |
360 |
|
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.