Let the supply of wheat is `x` quintal from A to D and `y` quital from A to E. Then wheat supply will be `(100-x-y)` quintal from A to F. Similarly, `(60-x),(50-y),(x+y-60)` quintals of wheat will be supplied from B to D,E,F respectively.
Now minimum transportation cost
`Z=6x+3y+2.50(100-x-y)+4(60-x)`
`+2(50-y)+3(x+y-60)`
`=2.50x+1.50y+410`
and constraints `xge0, yge0`
`100-x-yge0impliesx+yle100`
`60-xge0impliesxle60`
`50-yge0impliesyle50`
`x+y-60ge0impliesx+yge60`
First we draw the graph of the lines `x+y=100, x=60,y=50,x+y=60`
Now, we find the feasible region by constraints `x+yle100,xle60,yle50,x+yge60,xge0,yge0` and shade it. Its vertices are `A(10,50),B(60,0),C(60,40),D(50,50)`, at which we find the value of `Z`
Therefore, minimum transportation cost `Rs. 510`
For this 10,50,40 quintals will supply from A to D E,F respectively and 50,0,0 quintals will supply from B to D,E,F respectively.