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Basic algebra: find break-even point

  1. Jan 16, 2013 #1

    939

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    1. The problem statement, all variables and given/known data

    Mike make shirts. He has fixed daily costs of $150. It costs an additional $3 to make each shirt. He would like to make a profit of $750 a day making shirts. If he can make 24 shirts a day, how much must he charge to meet his goal? Find break even point.

    2. Relevant equations

    c(x) = 3x + 150
    p = r(x) - c(x)

    3. The attempt at a solution

    1) p = r(x) - c(x)
    p = r(x) - c(24)
    750 = r(x) - 222
    750 = 972 - 222

    972 = r(24)
    r = 40.5
    r(x) = 40.5x

    He must charge $40.50.

    2)40.5x = 3x + 150
    37.5x = 150
    x = 4

    Break even point is $4
     
    Last edited: Jan 16, 2013
  2. jcsd
  3. Jan 16, 2013 #2

    haruspex

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    It's not clear whether the last part means a break-even price or a break-even output. In your equations you've taken it to be break-even output, so the answer should be a number of shirts, not a number of dollars. And it's exact, not approximate.
     
  4. Jan 16, 2013 #3

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    Thanks.

    Btw, what if for a break even point, the "price" gives something approximate, i.e. 350.55 shirts for one and 350.71 for the other?

    Would you just take the break even point to be the first price that gives more revenue than cost?
     
  5. Jan 16, 2013 #4

    haruspex

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    Yes.
     
  6. Jan 17, 2013 #5

    HallsofIvy

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    The "break even point" is NOT a price. It is the number of shirts he must sell in order to just meet his costs.

    You assumed that yourself when you wrote "40.5x = 3x + 150". $40.50 is the price he is getting for each shirt. so 40.5x is the gross income if x is the number of shirts. Similarly, $3 is the marginal cost of each shirt so 3x is a cost only if x is the number of shirts. Your answer should be "He must make 4 shirts a day to break even", NOT "$4".
     
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