A factory manufactures two types of screws, A and B. Each type of screw requires the use of two machines, 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 screws B. Each machine is available for at the most 4 hours on any day. The manufacturer can sell a package of screws A at a profit of Rs 7 and screws B at a 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.

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Solution

Consider the manufacturer produce x packages of Screw A and y packages of Screw B. The quantities are always positive, so,

x≥0 y≥0

Tabulate the given data as,

Screw A Screw B Availability
Automatic machine (min) 4 6 4×60=240
Hand Operated machine (min) 6 3 4×60=240

The required constraints are,

4x+6y≤240 6x+3y≤240 x≥0 y≥0

The objective function (profit) which needs to maximize is,

Z=7x+10y

The line 4x+6y≤240 gives the intersection point as,

x 0 60
y 40 0

Also, when x=0,y=0 for the line 4x+6y≤240, then,

0+0≤240 0≤240

This is true, so the graph have the shaded region towards the origin.

The line 6x+3y≤240 gives the intersection point as,

x 0 40
y 80 0

Also, when x=0,y=0 for the line 6x+3y≤240, then,

0+0≤240 0≤240

This is true, so the graph have the shaded region towards the origin.

By the substitution method, the intersection points of the lines 4x+6y≤240 and 6x+3y≤240 is ( 30,20 ).

Plot the points of all the constraint lines,

It can be observed that the corner points are A( 40,0 ),B( 30,20 ),C( 0,40 ).

Substitute these points in the given objective function to find the maximum value of Z.

Corner points Z=7x+10y
A( 40,0 ) 280
B( 30,20 ) 410 (Maximum)
C( 0,40 ) 400

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

Thus, 30 packages of screw A and 20 packages of screw B should be produced each day to get the maximum profit of 410.

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