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 maximise his profit?

Answer 1

Let number of pedestal lamps to be made is $X$ and number of wooden shades to be made is $Y$

Since, pedestal lamps requires $2$ hours and wooden shades requires $1$ hours of grinding/cutting time. Also, there is maximum $12$ hours of grinding/cutting time.

$\therefore 2X+Y\le 12...\left(1\right)$

Since, pedestal lamps requires $3$ hours and wooden shades requires $2$ hours for spayer. Also, there is maximum $20$ hours for spayer.

$\therefore 3X+2Y\le 20...\left(2\right)$

Since, count of objects can't be negative.

$\therefore X\ge 0,Y\ge 0...\left(3\right)$

We have to maximize profit of the industry.

Here, profit on pedestal lamps is $5Rs$ and on wooden shades is $3Rs$

So, objective function is $Z=5X+3Y$

Plotting all the constraints given by equation (1), (2) and (3), we got the feasible region as shown in the image.

Corner points	Value of $Z=5X+3Y$
A (0,10)	30
B (4,4)	32 (maximum)
C (6,0)	30

Hence, industry should produce 4 pedestal lamps and 4 wooden shades in a day to maximise his profit. Also, maximum profit will be $32Rs$