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CBSE Class 12-science Answered

A company manufactures two articles A and B. There are two departments I and II through which these articles are processed. The maximum capacity of department I is 60 hours a week and that of department II is 48 hours a week. The production of each article A requires 4 hours by department I and 2 hours by department II and that of each unit of B requires 2 hours in department I and 4 hours in department II. If the profit of Rs 60 for each unit of A Rs 80 for each unit of B find the number of units of A and B to be produced per week to have maximum profit.
Asked by Topperlearning User | 01 Sep, 2014, 01:19: PM
answered-by-expert Expert Answer

 

If x units of article A and y units of articles B are produced every week, then we have following LPP maximum z = 60x + 80y
          st 4x + 2y  60
          2x + 4y  48
          x  0, y  0
          Problem can be rewritten as
          Maximum z = 60x + 80y
          st 2x + y  30
          x + 2y  24
          If x  0, y  0
         

Points

Z = 60x + 80y

O(0, 0)
A(15, 0)
P(12, 6)
D(0, 12)

0
900
1200
960

 Since 1200 is highest in the second column, Maximum profit occurs at P(12,6)

Hence the number of units of article A produced is 12 and number of units of article B produced is 6 to get a maximum profit of Rs.1200.

Answered by | 01 Sep, 2014, 03:19: PM

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CBSE 12-science - Maths
A company manufactures two articles A and B. There are two departments I and II through which these articles are processed. The maximum capacity of department I is 60 hours a week and that of department II is 48 hours a week. The production of each article A requires 4 hours by department I and 2 hours by department II and that of each unit of B requires 2 hours in department I and 4 hours in department II. If the profit of Rs 60 for each unit of A Rs 80 for each unit of B find the number of units of A and B to be produced per week to have maximum profit.
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