# Ex 12.2, 9 - 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, 9 A diet is to contain at least 80 units of vitamin A and 100 units of minerals. Two foods F1 and F2 are available. Food F1 costs Rs 4 per unit food and F2 costs Rs 6 per unit. One unit of food F1 contains 3 units of vitamin A and 4 units of minerals. One unit of food F2 contains 6 units of vitamin A and 3 units of minerals. Formulate this as a linear programming problem. Find the minimum cost for diet that consists of mixture of these two foods and also meets the minimal nutritional requirements. Let the quantity of food F1 = x units and quantity of Food F2 = y units According to Question: Since we need to minimize the cost Hence, function used here is minimize Z Cost of Food F1 Per unit → Rs 4 Cost of Food F2 Per unit → Rs 6 ∴ Minimize Z = 4x + 6y Combining all constraints Minimize Z = 4x + 6y Subject to constraints 3x + 6y ≥ 80 4x + 3y ≥ 100 x ≥ 0, y ≥ 0 Since, the region that is feasible is unbounded So, 104 can be or cannot be the minimum value. For this, we need to graph inequality Since, there is no point common between feasible region & 4x + 6y < 104 Hence, minimum cost of two foods in Rs 104

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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.