# Ex 12.2, 10 - Chapter 12 Class 12 Linear Programming (Term 1)

Last updated at Feb. 17, 2020 by Teachoo

Ex 12.2

Ex 12.2, 1
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Chapter 12 Class 12 Linear Programming

Serial order wise

Last updated at Feb. 17, 2020 by Teachoo

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 : Nitrogen Quantity of Nitrogen in F1 = 10% Quantity of Nitrogen in F2 = 5% Minimum Available = 14 kg ∴ 10% of x + 5% of y ≥ 14 10/100 x + 5/100 y ≥ 14 2x + y ≥ 280 Phosphoric Acid Quantity of Acid in F1 = 6% Quantity of Acidin F2 = 10% Minimum Available = 14 kg ∴ 6% of x + 10% of y ≥ 14 6/100 x + 10/100 y ≥ 14 3x + 5y ≥ 700 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 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