# Misc 6 - Chapter 12 Class 12 Linear Programming

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

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) x + y 100 100 (x + y) 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 + y 60 x + y 60 x, y 0 Hence, transportation cost will be Minimum if : Minimum Cost = Rs 510

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.