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Misc 6 - Two godowns A and B have grain of 100 quintals - Miscellaneou

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Misc 6 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 are 60, 50 and 40 quintals respectively. The cost of transportation per quintal from the godowns to the shops are given in the following table: How should the supplies be transported in order that the transportation cost is minimum? What is the minimum cost? Let Godown A transport x quintals to shop D Godown A transported y quintals to shop E Since number of quintals transported to each depot must be greater than or equal to zero -: ∴ x ≥ 0, y ≥ 0 100 − (x + y) ≥ 0 x + y ≤ 100 60 – x ≥ 0 x ≤ 60 50 − y ≥ 0 y ≤ 50 x + y − 60 ≥ 0 x + y ≥ 60 As we need to minimize the cost of transportation, Hence the function used is minimize Z. Total transportation cost will be Z = 6x + 3y + 2.50 [100 − (x + y)] + 4 (60 − x) + 2 (50 − y) + 3 [x + y − 60] Z = 2.50x + 1.50y + 410 Combining all Constraints :− Minimize Z = 2.50x + 1.50y + 410 Subject to Constraints : x + y ≤ 100 x ≤ 60, y ≤ 50 x + y ≥ 60 x, y ≥ 0 Hence, transportation cost will be Minimum if : Hence, transportation cost will be Minimum if : Minimum Cost = Rs 510 From A : 10, 50, 40 quintals to shops D, E, F respectively From B : 50, 0, 0 quintals to shops D, E, F respectively

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.