Two godowns A and B have grain capacity of 100 quintals and 50 quintals respectively. They supply to 3 ration shops, D, E and F whose requirements arc 60, 50 and 40 quintals respectively. The cost of transportation per quintal from the godowns to the shops arc given in the following table:

Transportation cost per quintal (in Rs.)
From/To	A	B
D	6	4
E	3	2
F	2.50	3

How should the supplies be transported in order that the transportation cost is minimum ? What is the minimum cost?

Let x quintals of grain be transported from godown A to shop D and y quintals of grain to shop E, then 100 - (x + y) quintals will be transported to shop E.
This means that (60 - x) quintals of grain will be transported from godown B to shop D, (50 - y) quintals of grain to shop E and 40 -{100 - (x + y)} = x + y - 60 quintals will be transported to shop F. For transportation cost, we construct the table
Table

We are to maximise
                               
subject to constraints
                     

Consider a set of rectangular cartesian axes OXY in the plane.
It is clear that any point which satisfies x ≥ 0, y ≥ 0 lies in the first quadrant.
x = 60 is a straight line AL parallel to y-axis at a distance of 60.
y = 50 is a straight line BM parallel to .r-axis at a distance of 50.
Now we draw the graph of x + y = 100.
For x = 0, y = 100
For y = 0, x = 100
∴ line meets OX in C(100, 0) and OY in N(0, 100).
Again we draw the graph of x + y = 60.
For x = 0, y = 60
For y = 0, x = 60
∴ line meets OX in A(60, 0) and OY in P(0, 60).

Since feasible region satisfies all the constraints.
∴ AQRS is the feasible region.
The corner points are A(60, 0), Q(60, 40), R(50, 50), S( 10, 50).
 
∴ the minimum cost of transportation is Rs. 510, when from godown A. 10 quintals of grain are sent to shop D, 50 quintals of grain are sent to shop E and 40 quintals ot grain are sent to shop F and from godown B, whole of 50 quintals are sent to shop D.