24 June, 06:45

# A company manufactures two products X and Y. Each product has to be processed in three departments: welding, assembly and painting. Each unit of X spends 2 hours in the welding department, 3 hours in assembly and 1 hour in painting. The corresponding times for a unit of Y are 3, 2 and 1 hours respectively. The employee hours available in a month are 1,500 for the welding department, 1,500 in assembly and 550 in painting. The contribution to profits are 100 USD for product X and 120 USD for product Y. What is the objective function (Z) to be maximised in this linear programming problem (where Z is total profit in USD) ? (note : means = )

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1. 24 June, 08:37
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100x+120y = z

z = \$ 63000

Step-by-step explanation:

Product Welding Assembly Painting Cont. to profit

X 2x hours 3x hours 1xhour = \$100x

Y 3y hours 2y hours 1y hour = \$120y

Total hours 1500 hours 1500 hours 550 hours

available

Let X represent product X

Let Y represent Product Y

2x + 3y = 1500

x + y = 550

y = 550-x

2x + 3 (550-x) = 1500

2x + 1650 - 3x = 1500

150 = x

y = 550-150

y = 400

Objective Function Z = 100x + 120y

Z = 100 (150) + 120 (400)

Z = 15000+48000

Z = \$63000