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Math Help - Finding the maximum

  1. #1
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    Finding the maximum

    Hello, I am a freshman in college at Western MI, I know how to use derivatives to find the minimum and maximum critical values of a function, but I am having a very hard time solving the following problem:

    Borrowed from Calculus by Hughes-Hallett, et. al.
    "The cost of fuel to propel a boat through the water, in dollars per hour, is proportional to the cube of the speed. A certain ferry uses $200 worth of fuel per hour when cruising at 10 miles per hour. Apart from fuel, the cost of running this ferry is $1050 per hour. At what speed should the ferry travel to minimize the cost per mile traveled?"

    So far I have figured the following:

    Cost (relative to speed) = 1/5s^3

    and therefore:

    Cost (per hour) = 1/5s^3 + 1050

    I am having a hard time manipulating this equation to give me cost per mile traveled. Please help me!
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  2. #2
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    Quote Originally Posted by w7west View Post
    Hello, I am a freshman in college at Western MI, I know how to use derivatives to find the minimum and maximum critical values of a function, but I am having a very hard time solving the following problem:

    Borrowed from Calculus by Hughes-Hallett, et. al.
    "The cost of fuel to propel a boat through the water, in dollars per hour, is proportional to the cube of the speed. A certain ferry uses $200 worth of fuel per hour when cruising at 10 miles per hour. Apart from fuel, the cost of running this ferry is $1050 per hour. At what speed should the ferry travel to minimize the cost per mile traveled?"

    So far I have figured the following:

    Cost (relative to speed) = 1/5s^3

    and therefore:

    Cost (per hour) = 1/5s^3 + 1050

    I am having a hard time manipulating this equation to give me cost per mile traveled. Please help me!


    Ok I found it!

    in case you are interested what I did was:

    let f(x) = (1/5)x^3 + 1050

    let g(x) = f(x) / x

    g'(x) = (2(x^3 - 2625)) / (5x^2)


    and I just solved for g'(x) = 0 to find the minimum which turned out to be:

    5*21^(1/3)
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