Diet problems

Chapter 12 Class 12 Linear Programming
Concept wise

### Transcript

Misc 2 A farmer mixes two brands P and Q of cattle feed. Brand P, costing Rs 250 per bag, contains 3 units of nutritional element A, 2.5 units of element B and 2 units of element C. Brand Q costing Rs 200 per bag contains 1.5 units of nutritional element A, 11.25 units of element B, and 3 units of element C. The minimum requirements of nutrients A, B and C are 18 units, 45 units and 24 units respectively. Determine the number of bags of each brand which should be mixed in order to produce a mixture having a minimum cost per bag? What is the minimum cost of the mixture per bag? Let the mixture contain x bags of brand P & y bags of brand Q According to Question : As we need to Minimize the cost, Hence the function used here is Minimum Z. Element A Brand P → 3 units Brand Q → 15 units Least Requirement → 18 units ∴ 3x + 1.5y ≥ 18 2x + y ≥ 12 Element B Brand P → 2.5 units Brand Q → 11.25 units Least Requirement → 45 units ∴ 2.5x + 11.25y ≥ 45 2x + 9y ≥ 36 Element C Brand P → 2 units Brand Q → 3 units Least Requirement → 24 units ∴ 2x + 3y ≥ 24 Now, Cost of brand P → Rs 250 Cost of brand Q → Rs 200 ∴ Minimize Z = 250 x + 200 y Combining all constraints : Min z = 250 x + 200 y Subject to constraints 2x + y ≥ 12 2x + 9y ≥ 36 2x + 3y ≥ 24 x ≥ 0, y ≥ 0 As the feasible region is unbounded, hence, 1950 may or may not be the minimum value of Z For this we need to graph inequality 250 x + 200y < 1950 Since there is no common point between the feasible region f the inequality. Hence, the cost will be minimum if The mixture contains = 3 bags of brand P mixture contains = 6 bags of brand Q Minimum cost is Rs 1950

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.