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Thread: need your assistance!

  1. #1

    Thread Starter
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    need your assistance!

    Can someone help me with this?




    Cost Control...

    An appliance repair shop has 5 vacuum cleaners, 12 TV sets, and 18 VCRs to be repaired. The store employs two part-time repairmen. One repairman can repair one vacuum cleaner, three TV sets, and three VCRs in 1 week, while the second repairman can repair one cacuum cleaner, two TV sets and six VCRs in 1 week. The first employee is paid $250 a week and the second employee is paid $220 a week. To minimize the cost, how many weeks should each of the two repairmen be employed?

  2. #2
    Fanatic Member alkatran's Avatar
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    Apr 2002
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    Step 1: Change the problem into a formula
    Need to repair: (5,12,18)
    A -> Number of weeks with repairman one (1,3,3)
    B -> Number of weeks with repairman two (1,2,6)

    They both have (1,#,#) so we'll eliminate vacuums and say the minimum number of weeks is 5:
    A + B >= 5
    Also, for TVs and VCRs:
    3A + 2B >= 12
    3A + 6B >= 18 == A + 2b >= 9

    Since we have two equations with 2b, might as well make the first as well:
    A + 2B >= 9
    2A + 2B >= 5
    3A + 2B >= 12
    and also:
    A >= 0
    B >= 0

    Now, the easiest way to solve this is: Plot a graph of all these limits and shade the spots that aren't allowed. Then check all the points where limits intersect. One of them will be the best answer.
    Don't pay attention to this signature, it's contradictory.

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