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Math Help - Farmer optimization question

  1. #1
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    Farmer optimization question

    Okay, I may be bad at optimization problems, but this question I can't even read it without getting a headache, and I have a test!!

    If a farmer digs up his potatoes on July 1st, the crop is 120 m^3, which sell as new potatoes at $3.00/ m^3. If the crop is allowed to mature further, it increases at 15 m^3/ week. However, the price drops 20 cents/m^3/ week. When should the farmer dig the crop for the maximum cash return??

    So am I supposed to find the profit or something or the day hes supposed to dig it up aghhhh
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  2. #2
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    Quote Originally Posted by Mathforum View Post
    Okay, I may be bad at optimization problems, but this question I can't even read it without getting a headache, and I have a test!!

    If a farmer digs up his potatoes on July 1st, the crop is 120 m^3, which sell as new potatoes at $3.00/ m^3. If the crop is allowed to mature further, it increases at 15 m^3/ week. However, the price drops 20 cents/m^3/ week. When should the farmer dig the crop for the maximum cash return??

    So am I supposed to find the profit or something or the day hes supposed to dig it up aghhhh
    cash = (potato volume)(price/volume)

    let t = number of weeks after July 1

    set up the equation as a function of t ... let's see what you can do.
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  3. #3
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    Okay, so to start you'd have the original amount, 120 x 3. Plus 15t minus 20 cents for each week.
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  4. #4
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    So if the crop increases at 15m^3/ week...then thats $3.00 more for each m^3...minus 20 cents for each m^3
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  5. #5
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    C(t) = (360 + 15t) (3 - 20t) ?? am i on the right track
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  6. #6
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    C(t) = (360 + 15t)(3- .20t)
    = 1080 - 72t +45t - 3t^2
    = -3t^2 - 27t- 1080
    C'(t) = -6t - 27
    Set C'(t) = 0
    therforee... -6t -27 = 0

    t = - 9/ 2 :S
    That seems so wrong
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  7. #7
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    Waiiit...I did the volume wrong hmm... C(t) = (120 + 15t) (3- .20t)

    C(t) = -3t^2 + 21t +360
    C'(t) = -6t +21
    t = 3.5 weeks That seems better!!
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