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Misc 3 - A dietician wishes to mix together two kinds of food

Misc 3 - Chapter 12 Class 12 Linear Programming - Part 2
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Misc 3 - Chapter 12 Class 12 Linear Programming - Part 4
Misc 3 - Chapter 12 Class 12 Linear Programming - Part 5
Misc 3 - Chapter 12 Class 12 Linear Programming - Part 6
Misc 3 - Chapter 12 Class 12 Linear Programming - Part 7

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Misc 3 A dietician wishes to mix together two kinds of food X and Y in such a way that the mixture contains at least 10 units of vitamin A, 12 units of vitamin B and 8 units of vitamin C. The vitamin contents of one kg food is given below: One kg of food X costs Rs 16 and one kg of food Y costs Rs 20. Find the least cost of the mixture which will produce the required diet? Let the mixture Contains x units of Food x The mixture Contains y units of Food y Let the mixture Contains x units of Food x The mixture Contains y units of Food y Vitamin A Food X contains → 1 Food Y Contains → 2 Total Requirement → at least 10 ∴ x + 2y ≥ 10 Vitamin B Food X Contains → 2 Food Y Contains → 2 Total Requirement → at least 12 ∴ 2x + 2y ≥ 12 x + y ≥ 6 Vitamin C Food X Contains → 3 Food Y Contains → 1 Total Requirement → at least 8 ∴ 3x + y ≥ 8 Now, Food X cost → Rs 16 Food Y Cost → Rs 20 ∴ Z = 16x + 20Y Combining all Constraints : Min Z = 16x + 20y Subject to Constraints, x + 2y ≥ 10 x + y ≥ 6 3x + y ≥ 8 x ≥ 0, y ≥ 0 As, the feasible region is unbounded Hence, 112 may or may not be the minimum value of Z. For This, we need to graph inequality : 16x + 20y < 112 4x + 5y < 28 Since, there is no point in Common Between the feasible region & inequality. Hence, minimum value of Z is 112. Thus, Cost of Mixture will be minimum if 2 packets of Food X & 4 Packets of Food Y are used. Minimum cost of mixture is Rs 112.

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.