Question

A cottage industry manufactures pedestal lamps and wooden shades, each requiring the use of a grinding/cutting machine and a sprayer. It takes 2 hours on grinding/cutting machine and 3 hours on the sprayer to manufacture a pedestal lamp. It takes 1 hour on the grinding/cutting machine and 2 hours on the sprayer to manufacture a shade. On any day, the sprayer is available for at the most 20 hours and the grinding/cutting machine for at the most 12 hours. The profit from the sale of a lamp is Rs 5 and that from a shade is Rs 3. Assuming that the manufacturer can sell all the lamps and shades that he produces, how should he schedule his daily production in order to maximize his profit?

Answer 1

Assume that the cottage industry manufacture x pedestal lamps and y wooden shades. The quantities are always positive, so,

x≥0 y≥0

Tabulate the given data as,

	Lamps	Shades	Availability
Cutting machine (hr)	2	1	12
Sprayer (hr)	3	2	20

The required constraints are,

2x+y≤12 3x+2y≤20 x≥0 y≥0

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

Z=5x+3y

The line 2x+y≤12 gives the intersection point as,

x	0	6
y	12	0

Also, when x=0,y=0 for the line 2x+y≤12, then,

0+0≤12 0≤12

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

The line 3x+2y≤20 gives the intersection point as,

x	0	20 3
y	10	0

Also, when x=0,y=0 for the line 3x+2y≤20, then,

0+0≤20 0≤20

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

By the substitution method, the intersection points of the lines 2x+y≤12 and 3x+2y≤20 is ( 4,4 ).

Plot the points of all the constraint lines,

It can be observed that the corner points are A( 6,0 ),B( 4,4 ),C( 0,10 ).

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

Corner points	Z=5x+3y
A( 6,0 )	30
B( 4,4 )	32 (Maximum)
C( 0,10 )	30

The maximum value of Z is 32 at the point ( 4,4 ).

Thus, 4 numbers of pedestal lamps and 4 numbers of wooden shades should be produced each day to get the maximum profit of 32.