Question

A cottage industry manufactures pedestal lamps and wooden shades, each requiring the use of grinding/cutting machine and a sprayer. It takes 2 hours on the grinding/cutting machine and 3 hours on the sprayer to manufacture a pedestal lamp. It takes one 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? Make an L.P.P. and solve it graphically.

Answer 1

Let the number of padestal lamps and wooden shades manufactured by cottage industry be x and y respectively.
Here profit is the objective function z.
z =5x + 3y … (i)

We have to maximise z subject to the constrains

Since (0, 0) Satisfy 3x + 2y ≤ 20
=> Graph of 3x + 2y≤ 20 is that half plane in which origin lies.
The shaded area OABC is the feasible region whose corner points are O, A, B and C.

For coordinate B.
Equation 2x + y =12 and 3x + 2y = 20 are solved as
3x + 2 (12 -2x) = 20
=> 3x + 24 -4x = 20  => x = 4
=> y =12 -8 = 4
Coordinate of B =(4, 4)
Now we evaluate objective function Z at each corner.

Hence maximum profit is Rs.32 when manufacturer produces 4 lamps and 4 shades.