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Example 10 - Chapter 12 Class 12 Linear Programming (Term 1)

Last updated at Dec. 12, 2016 by

Example 10 - (Manufacturing problem) A manufacturer has three - Manufacturing problems

Example 10 - Chapter 12 Class 12 Linear Programming - Part 2
Example 10 - Chapter 12 Class 12 Linear Programming - Part 3 Example 10 - Chapter 12 Class 12 Linear Programming - Part 4 Example 10 - Chapter 12 Class 12 Linear Programming - Part 5

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Example 10 (Manufacturing problem) A manufacturer has three machines I, II and III installed in his factory. Machines I and II are capable of being operated for at most 12 hours whereas machine III must be operated for atleast 5 hours a day. She produces only two items M and N each requiring the use of all the three machines. The number of hours required for producing 1 unit of each of M and N on the three machines are given in the following table: Let the number of items of type M be x, & the number of items of type N by y According to Question : Now, Profit on Type M → Rs 600 Profit on Type N → Rs 400 Hence, Z = 600x + 400 y Combining all constraints : Max Z = 600x + 400y Subject to constraints, x + 2y ≤ 12 2x + y ≤ 12 x + 1.5 y ≤ 5 x ≥ 0 , y ≥ 0

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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.