       1. Chapter 12 Class 12 Linear Programming
2. Serial order wise
3. Miscellaneous

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. 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

Miscellaneous 