Let godown A supply x and y quintals of grain to the shops D and E respectively. Then, (100 − x − y) will be supplied to shop F.
The requirement at shop D is 60 quintals since x quintals are transported from godown
A. Therefore, the remaining (60 −x) quintals will be transported from godown B. Similarly, (50 − y) quintals and 40 − (100 − x − y) = (x + y − 60) quintals will be transported from godown B to shop E and F respectively. The given problem can be represented diagrammatically as follows.
![](https://www.sarthaks.com/?qa=blob&qa_blobid=1847400939656851152)
![](https://www.sarthaks.com/?qa=blob&qa_blobid=555705914151518708)
Total transportation cost z is given by,
![](https://www.sarthaks.com/?qa=blob&qa_blobid=9111496928985708870)
The given problem can be formulated as Minimize z = 2.5x + 1.5y + 410 … (1) subject to the constraints,
![](https://www.sarthaks.com/?qa=blob&qa_blobid=1927598473214652094)
The feasible region determined by the system of constraints is as follows.
![](https://www.sarthaks.com/?qa=blob&qa_blobid=5458060594621956189)
The corner points are A (60, 0), B (60, 40), C (50, 50), and D (10, 50).
The values of z at these corner points are as follows.
![](https://www.sarthaks.com/?qa=blob&qa_blobid=1300741705488916357)
![](https://www.sarthaks.com/?qa=blob&qa_blobid=10144735297232457387)
The minimum value of z is 510 at (10, 50).
Thus, the amount of grain transported from A to D, E, and F is 10 quintals, 50 quintals, and 40 quintals respectively and from B to D, E, and F is 50 quintals, 0 quintals, and 0 quintals respectively.