# A Factory Manufactures Two Types of Screws a and B, Each Type Requiring the Use of Two Machines, an Automatic and a Hand-operated. Assuming that He Can Sell All the Screws He Manufactures, How Many Packets of Each Type Should the Factory Owner Produce in a Day in Order to Maximize His Profit? Formulate the Above Lpp and Solve It Graphically and Find the Maximum Profit. - Mathematics

A factory manufactures two types of screws A and B, each type requiring the use of two machines, an automatic and a hand-operated. It takes 4 minutes on the automatic and 6 minutes on the hand-operated machines to manufacture a packet of screws 'A' while it takes 6 minutes on the automatic and 3 minutes on the hand-operated machine to manufacture a packet of screws 'B'. Each machine is available for at most 4 hours on any day. The manufacturer can sell a packet of screws 'A' at a profit of 70 paise and screws 'B' at a profit of Rs 1. Assuming that he can sell all the screws he manufactures, how many packets of each type should the factory owner produce in a day in order to maximize his profit? Formulate the above LPP and solve it graphically and find the maximum profit.

#### Solution

Let the factory manufacture x screws of type A and y screws of type B on each day. Therefore,

x ≥ 0 and y ≥ 0

The given information can be compiled in a table as follows.

 Screw A Screw B Availability Automatic Machine (min) 4 6 4x60 = 240 Hand Operated Machine (min) 6 3 4x60 = 240

The profit on a package of screws A is 70 paise and on the package of screws, B is Rs 1. Therefore, the constraints are

4x + 6y <= 240

6x + 3y  < =- 240

Total profit, Z = 0.7x + y

The mathematical formulation of the given problem is

Maximize Z = 0.7x + y … (1)

subject to the constraints,

4x + 6y <= 240  ...2

6x  + 3y <= 240 ...3

x,y >= 0    ....4

The feasible region determined by the system of constraints is

The corner points are A (40, 0), B (30, 20), and C (0, 40).

The values of Z at these corner points are as follows.

 Corner point Z = 0.7x + y A(40, 0) 28 B(30, 20) 41 → Maximum C(0, 40) 40

The maximum value of Z is 41 at (30, 20).

Thus, the factory should produce 30 packages of screws A and 20 packages of screws B to get the maximum profit of Rs 41.

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