# Ex 12.2, 10 - Chapter 12 Class 12 Linear Programming

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 12.2, 10 There are two types of fertilisers F1 and F2. F1 consists of 10% nitrogen and 6% phosphoric acid and F2 consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, a farmer finds that she needs atleast 14 kg of nitrogen and 14 kg of phosphoric acid for her crop. If F1 costs Rs 6/kg and F2 costs Rs 5/kg, determine how much of each type of fertiliser should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost? Let the fertilizer F1 requirement be x kg fertilizer F2 requirement be y kg According to Question : As we need to Minimize the cost, hence the function used here is Minimize z. Cost of fertilizer F1 = Rs 6 k/g Cost of fertilizer F2 = Rs 5 k/g Minimize z = 6x + 5y Combining all Constraints : Min Z = 6x + 5y Subject to constraints 2x + y ≥ 280 3x + 5y ≥ 700 x, y ≥ 0 As the feasible area is unbounded Hence 1000 may or may not be the minimum value of Z. For this, we need to graph inequality 6x + 5y < 1000 Since, there is no common point between the feasible region and 6x + 5y < 1000 Hence the cost will be minimum, if Fertilizer F1 used = 100 kg Fertilizer F2 used = 80 kg Minimum Cost = Rs 1000

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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.