Question

1) 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 1 and that from a shade is 2. 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? Formulate an LP. and solve it graphically.

• A. Z is maximum at (3,9)
• B. Z is maximum at (0,10)
• C. Z is maximum at (3,5)
• D. Z is maximum at (4,4)

Answer 1

B. Z is maximum at (0,10)

Let us assume that the number of pedestal lamps = X

Number of Wooden shades = Y

The objective function is profit maximization: Max Z = X + 2Y

Constraint 1: Hours available on Sprayer : 3X + 2Y <= 20

Constraint 2: Hours available on Grinding/ Cutting Machine : 2X + Y <= 12

Also, X, Y >=0

So, the Linear Programming model is as follows:

Max Z = X + 2Y

subject to:

3X + 2Y <= 20

2X + Y <= 12

X, Y >=0

The constraints have to be plotted on the graph as straight lines:

The feasible area has been highlighted in the graph.

The value of the objective function has to be calculated on the extreme points of the feasible area - (0,0), (6,0), (4,4), (0,10)

At (0,0): Z = ( 1 x 0) + ( 2 x 0) = 0 + 0 = 0

At (6,0): Z = ( 1 x 6) + ( 2 x 0) = 6 + 0 = 6

At (4,4): Z = ( 1 x 4) + ( 2 x 4) = 4 + 8 = 12

At (0,10): Z = ( 1 x 0) + ( 2 x 10) = 0 + 20 = 20

Therefore, the objective function is maximum (ie 20) at (0,10)

So, X = 0, and Y =10

Number of Pedestal lamps = X = 0

Number of Wooden shades = Y = 10

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