CBSE Class 12 Sample Paper for 2019 Boards

Class 12
Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Question 28

A manufacturer makes two types of toys A and B. Three machine are needed for this purpose and the time (in minutes) required for each toy on the machines is given below:

 Types of Toys Machines I II III A 20 10 10 B 10 20 30

The machines I, II and III are available for a maximum of 3 hours, 2 hours and 2 hours 30 minutes respectively. The profit on each toy of type A is Rs 50 and that of type B is Rs 60. Formulate the above problem as a L.P.P and solve it graphically to maximize profit.

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### Transcript

Question 28 A manufacturer makes two types of toys A and B. Three machine are needed for this purpose and the time (in minutes) required for each toy on the machines is given below: The machines I, II and III are available for a maximum of 3 hours, 2 hours and 2 hours 30 minutes respectively. The profit on each toy of type A is Rs 50 and that of type B is Rs 60. Formulate the above problem as a L.P.P and solve it graphically to maximize profit. Let Number of toys of Type A be x, Number of toys of Type B be y Given that The profit on each toy of type A is Rs 50 and that of type B is Rs 60 We need to maximize the Profit ∴ Z = 50x + 60y Now, according to question Max time available on Machine I = 3 hours = 180 minutes Max time available on Machine II = 2 hours = 120 minutes Max time available on Machine III = 2.5 hours = 150 minutes Machine I Max Time = 180 mins ∴ 20x + 10 y ≤ 180 2x + y ≤ 18 Machine II Max Time = 120 mins ∴ 10x + 20y ≤ 120 x + 2y ≤ 12 Machine III Max Time = 150 mins ∴ 10x + 30 y ≤ 150 x + 3y ≤ 15 Machine III Max Time = 150 mins ∴ 10x + 30 y ≤ 150 x + 3y ≤ 15 Combining all constraints : Max Z = 50x + 60y Subject to constraints, 2x + y ≤ 18, x + 2y ≤ 12 x + 3y ≤ 15 & x ≥ 0 , y ≥ 0 2x + y ≤ 18 X 9 0 Y 0 18 x + 2y ≤ 12 X 12 0 Y 0 6 x + 3y ≤ 15 X 15 0 Y 0 5 Corner Point Z = 50x + 60y (0, 0) 0 A (0, 5) 300 B (6, 3) 480 C (8, 2) 520 -> Maximum D (9, 0) 450 Hence, profit will be maximum if Number of toys of Type A = 8 Number of toys of Type B = 2 Maximum Profit = Rs. 520

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.