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 Oct. 1, 2019 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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.