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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 =)

Respuesta :

erum123

Answer:

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

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