Question

A company manufactures two types of novelty souvenirs made of plywood. Souvenirs of type A require 5 minutes each for cutting and 10 minutes each for assembling. Souvenirs of type B require 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours 20 minutes available for cutting and 4 hours of assembling. The profit is Rs 5 each for type A and Rs 6 each for type B souvenirs. How many souvenirs of each type should the company manufacture in order to maximize the profit?

Answer 1

Assume that the company manufactures x souvenirs of type A and y souvenirs of type B. The quantities are always positive, so,

x≥0 y≥0

Tabulate the given data as,

	Type A	Type B	Availability
Cutting (min)	5	8	3×60+20=200
Assembling (min)	10	8	4×60=240

The required constraints are,

5x+8y≤200

And,

10x+8y≤240 5x+4y≤120

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

Z=5x+6y

The line 5x+8y≤200 gives the intersection point as,

x	0	40
y	25	0

Also, when x=0,y=0 for the line 5x+8y≤200, then,

0+0≤200 0≤200

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

The line 5x+4y≤120 gives the intersection point as,

x	0	24
y	30	0

Also, when x=0,y=0 for the line 5x+4y≤120, then,

0+0≤120 0≤120

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

By the substitution method, the intersection points of the lines 5x+8y≤200 and 5x+4y≤120 is ( 8,20 ).

Plot the points of all the constraint lines,

It can be observed that the corner points are A( 24,0 ),B( 8,20 ),C( 0,25 ).

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

Corner points	Z=5x+6y
A( 24,0 )	120
B( 8,20 )	160 (Maximum)
C( 0,25 )	150

The maximum value of Z is 160 at the point ( 8,20 ).

Thus, to get the maximum profit of 160, 8 souvenirs of type A and 20 souvenirs of type B should be produced each day.