a manufacturer produce two types of steel trungs he has two machines a and b. the first type of trunk requires 3 hour on machine A and 3 hours on match in B the second type require 3 hour on machine A and 2 hour on machine B machine A and B can work at most for 18 hour and 15 hour per day respectively. he earn a profit of Rs 30 and Rs 25 per trunk on the first type and second type respectively how many trunks of each type must he make each day to make maximum profit.
Answers
Rs. 165
Let the manufacturer manufacture x and y numbers of type 1 and type 2trunks.
∴According to the question,
3X + 3y
Maximize Z = 30x + 25y
The feasible region determined 3X + 3y is given by
(Plot the graph as the points given below i.e., A(0,0) , B(0,6) , C(3,3) , D(5,0). and draw a line from 6 in x axis to 6 in y axis and another line from 8 in x axis to 5 in y axis the mark a point in 3,3 which both the line in one)
The corner points of feasible region are A(0,0) , B(0,6) , C(3,3) , D(5,0).The value of Z at corner point is
Corner point Z = 30x + 25y
1. A(0,0) 0
2. B(0,6) 150
3. C(3,3) 165 (MAXIMUM)
4. D(5,0) 150
The maximum value of Z is 165 and occurs at point (3,3).
The manufacturer should manufacture 3 trunks of each type to earn maximum profit of Rs.165.
I hope this helps u well