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Optimization: Maximizing Profit

  1. Apr 22, 2012 #1
    1. The problem statement, all variables and given/known data

    Through market research, a computer manufacturer found that x thousand
    units of its new laptop will sell at a price of 2000 - 5x dollars per unit.
    The cost, C, in dollars of producing this many units is
    C(x) = 15 000 000 +1 800 000x + 75x^2. Determine the level of sales
    that will maximize profit.


    2. Relevant equations
    Profit = Revenue - Cost

    3. The attempt at a solution

    I said Revenue = 2000 - 5x and Cost = 15 000 000 +1 800 000x + 75x^2
    Using the formula Profit = Revenue - Cost I subtracted them from each other and got:

    Profit = -14998000 - 1800005x - 75x^2

    Then I found the derivative which came out to be:

    P'(x) = -1800005 - 150x

    Then I set it equal to zero and solved for x:

    0 = -1800005 - 150x
    1800005 = -150x
    x = -12000.03

    but the answer in the back of the book is 19 704 units.
    Can anyone explain what I did wrong?
     
  2. jcsd
  3. Apr 22, 2012 #2

    I like Serena

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    Welcome to PF, molly16! :smile:

    Your revenue looks to be the price of only 1 unit.
    Shouldn't more units be sold?
     
  4. Apr 22, 2012 #3
    So does that mean I should multiply 2000 - 5x by x to represent the number of units?
     
  5. Apr 22, 2012 #4

    I like Serena

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    I believe your problem statement says "x thousand units" that each sell at a price of "2000 - 5x" per unit.
     
  6. Apr 22, 2012 #5
    So:
    R = (1000x)(2000 - 5x)

    then

    Profit = Revenue - Cost
    P = (1000x)(2000 - 5x) - 15 000 000 +1 800 000x + 75x^2

    and

    P' = 200000 - 10150x
    x= 19.704

    #of units = (1000x) = (1000)(19.704) = 19704 units

    which was the answer in the back of the book

    Thanks for the help!
     
  7. Apr 23, 2012 #6

    I like Serena

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    You're welcome. :smile:
     
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