Question 28 - CBSE Class 12 Sample Paper for 2019 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at March 30, 2023 by Teachoo

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.

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 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.

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