Q. 125.0( 2 Votes )

# A firm manufactures two products A and B. Each product is processed on two machines M_{1} and M_{2}. Product A requires 4 minutes of processing time on M_{1} and 8 min. on M_{2}; product B requires 4 minutes on M_{1} and 4 min. on M_{2}. The machine M_{1} is available for not more than 8 hrs 20 min. while machine M_{2} 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.

Answer :

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.

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