Question

A factory manufactures two types of screws,A and B each type of screw requires the use of two machine,an automatic and a hand operated.It takes 4 minutes on the automatic and 6 minutes on hand operated machines to manufacture a package of screws A,while it takes 6 minutes on automatic and 3 minutes on the hand operated machines to manufacture a package of screw B.Each machine is available for at the most 4 hours on any day.The manufacturer can sell a package of screw A at a profit of Rs7 and screw B at profit of Rs10.Assuming that he can sell all the screws he manufactures,how many packages of each type should the factory owner produce in a day in order to maximize his profit?Determine the maximum profit.

Answer 1

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

Therefore, x≥0,y≥0 The given information can be compiled in a table as follows.

| Screw A| Screw B| Availability
---|---|---|---
Automatic machine(min)| 4| 6| 4×60=120
Hand Operated Machine(min)| 6| 3| 4×60=120

The profit on a package of screws A is Rs7 on the package of screws B is Rs10.

Therefore,the constraints are 4x+6y≤240 6x+3y≤240 Total profit,Z=7x+10y

The mathematical formulation of the given problem is
Maximize Z=7x+10y....(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.

The maximum value of Z is 410 at (30,20).

Thus, the factory should produce 30 packages of screw A and 20 packages of screw B to get a maximum profit of Rs410.