Let's assume that godown A supplies
X quintals grain to ration shop D and
Y quintals grain to ration shop E.
So, godown A will supply remaining grains 100−X−Y quintals to ration shop F.
Also, godown B will supply 60−X quintals grain to ration shop D, 50−Y quintals grain to ration shop E and 40−(100−X−Y)=X+Y−60 quintals grain to ration shop F.
Now total transportation cost is Z=6×X+3×Y+2.5×(100−X−Y)+4×(60−X)+2×(50−Y)+3×(X+Y−60)
⇒Z=6X+3Y+250−2.5X−2.5Y+240−4X+100−2Y+3X+3Y−180
⇒Z=2.5X+1.5Y+410 ...(1)
Now, Since godown A can supply maximum 60 quintals to ration shop D and 50 quintals to ration shop E and have maximum 100 quintals capacity to supply.
∴X+Y≤100,X≤60 and Y≤50 ...(2)
Also, if godown A supplies all 40 quintals to ration shop F, then remaining 60 quintals will be supplied to ration shop D and E.
So, X+Y≥60 ...(3)
Since, X and Y is amount of grains. It can never be negative.
So, X≥0,Y≥0 ...(4)
We have to minimise transportation cost given by equation (1).
After plotting all the constraints given by equation (2), (3) and (4), we got the feasible region as shown in the image.
Corner points | Value of Z=2.5X+1.5Y+410 |
A (60, 0) | 560 |
B (60, 40) | 620 |
C (50, 50) | 610 |
D (10, 50) | 510 (Minimum) |
Hence minimum transportation cost is 510 Rs. Transporter needs to supply as follows:
Godown A to Ration shop D =X=10 quintals
Godown A to Ration shop E =Y=50 quintals
Godown A to Ration shop F =100−X−Y=40 quintals
Godown B to Ration shop D =60−X=50 quintals
Godown B to Ration shop E =50−Y=0 quintals
Godown B to Ration shop F =X+Y−60=0 quintals