A firm manufactures two products A and B. Each product is processed on two machines M1 and M2. Product A requires 4 minutes of processing time on M1 and 8 min. on M2; product B requires 4 minutes on M1 and 4 min. on M2. The machine M1 is available for not more than 8 hrs 20 min. while machine M2 is available for 10 hrs. during any working day. The products A and B are sold at a profit of ₹ 3 and ₹ 4 respectively.Formulate the problem as a linear programming problem and find how many products of each type should be produced by the firm each day in order to get maximum profit.
Let required production of product A and B be x and y respectively.
Since profit on each product A and B are Rs. 3 and Rs. 4 respectively. So, profits on x number of type A and y number of type B are 3x and 4y respectively.
Let Z denotes total output daily, so,
Z = 3x + 4y
Since, each A and B requires 4 minutes each on machine 
. So, x of type A and y of type B require 4x and 4y minutes respectively. But,
Total time available on machine 
is 8 hours 20 minutes = 500 minutes.
So,
4x + 4y 
500
x + y 
125 {First Constraint}
Since, each A and B requires 8 minutes and 4 minutes on machine 
respectively. So, x of type A and y of type B require 8x and 4y minutes respectively. But,
Total time available on machine 
is 10 hours = 600 minutes.
So,
8x + 4y 
600
2x + y 
150 {Second Constraint}
Hence mathematical formulation of the given LPP is,
Max Z = 3x + 4y
Subject to constraints,
x + y 
125
2x + y 
150
x,y 
0 [Since production of A and B can not be less than zero]
Region x + y 
125: line x + y = 125 meets the axes at A(125,0), B(0,125) respectively.
Region containing the origin represents x + y 
125 as origin satisfies x + y 
125.
Region 2x + y 
150: line 2x + y = 150 meets the axes at C(75,0), D(0,150) respectively.
Region containing the origin represents 2x + y 
150 as origin satisfies 2x + y 
150.
Region x,y 
0: it represents the first quadrant.
The corner points are O(0,0), B(0,125), E(25,100), and C(75,0).
The vaues of Z at these corner points are as follows:
The maximum value of Z is 500 which is attained at B(0,125).
Thus, the maximum profit is Rs 500 obtained when no units of product A and 125 units of product B are manufactured.