Question

A manufacturer produces nuts ad bolts. It takes 1 hour of work on machine A and 3 hours on machine B to produce a package of nuts. It takes 3 hours on machine A and 1 hour on machine B to produce a package of bolts. He earns a profit, of Rs 17.50 per package on nuts and Rs. 7.00 per package on bolts. How many packages of each should be produced each day so as to maximize his profit, if he operates his machines for at the most 12 hours a day?

Answer 1

Consider that the manufacturer produce x packages of nuts and y packages of bolts. The quantities are always positive, so,

x≥0 y≥0

Tabulate the given data as,

	Nuts	Bolts	Availability
Machine A (hr)	1	3	12
Machine B (hr)	3	1	12

The required constraints are,

x+3y≤12 3x+y≤12 x≥0 y≥0

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

Z=17.5x+7y

The line x+3y≤12 gives the intersection point as,

x	0	3
y	4	0

Also, when x=0,y=0 for the line x+3y≤12, then,

0+0≤12 0≤12

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

The line 3x+y≤12 gives the intersection point as,

x	0	4
y	12	0

Also, when x=0,y=0 for the line 3x+y≤12, then,

0+0≤12 0≤12

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

By the substitution method, the intersection points of the lines x+3y≤12 and 3x+y≤12 is ( 3,3 ).

Plot the points of all the constraint lines,

It can be observed that the corner points are O( 0,0 ),A( 4,0 ),B( 3,3 ),C( 0,4 ).

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

Corner Points	Z=17.5x+7y
O( 0,0 )	0
A( 4,0 )	70
B( 3,3 )	73.5 (maximum)
C( 0,4 )	28

The maximum value of Z is 73.5 at the point ( 3,3 ).

Thus, to maximize the profit of 73.5, 3 packages of nuts and 3 packages of bolts should be produced each day.