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Math Help - Minimum Cost Word Problem

  1. #1
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    Minimum Cost Word Problem

    Hey guys, I haven't been here in awhile (since taking my last algebra class) and now I'm in calculus. You all were a great help then and I hope to tap into the usefulness of this board again! So here is a word problem that I am having trouble with if someone can work it out for me.

    1) Minimum Cost An offshore oil well is 1 mile off the coast. The oil refinery is 2 miles down the coast. Laying pipe in the ocean is twice as expensive as laying it on land. Find the most economical path for the pipe from the well to the oil refinery.


    I'm drawing a blank for this, I think it requires the use of Pythagorean theorem or use of the hypotenuse to get the function? then take the first derivative to find the critical points and second derivative to check concavity to determine if its a minimum? A complete work through of this problem would be greatly appreciated!
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  2. #2
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    Quote Originally Posted by FactoringAnguish View Post

    1) Minimum Cost An offshore oil well is 1 mile off the coast. The oil refinery is 2 miles down the coast. Laying pipe in the ocean is twice as expensive as laying it on land. Find the most economical path for the pipe from the well to the oil refinery.

    let x = distance "down beach" from the oil well toward the refinery

    water pipe length = \sqrt{1+x^2}

    land pipe length = 2-x

    cost as a function of x ...

    C = 2\sqrt{1+x^2} + (2-x)

    find \frac{dC}{dx} and minimize.
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  3. #3
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    Wow that was quick! you even included a picture thank you skeeter.

    Would you mind if I request a workout of taking the derivatives also?
    -For the first derivative I am getting 2x (x^2+1)^-1/2 -1
    I'm not feeling too confident with that and not sure about finding the critical point, with that -1 dangling off to the side of the fraction.
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  4. #4
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    Quote Originally Posted by FactoringAnguish View Post
    Wow that was quick! you even included a picture thank you skeeter.

    Would you mind if I request a workout of taking the derivatives also?
    -For the first derivative I am getting 2x (x^2+1)^-1/2 -1
    I'm not feeling too confident with that and not sure about finding the critical point, with that -1 dangling off to the side of the fraction.
    correct ...

    \frac{dC}{dx} = \frac{2x}{\sqrt{x^2+1}} - 1

    set \frac{dC}{dx} = 0

    \frac{2x}{\sqrt{x^2+1}} - 1 = 0

    you have (fraction - 1) = 0 ... what does that tell you about the numerator and denominator of the fraction?
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  5. #5
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    I'm not sure, this is trivial algebra and not even calculus but I am extremely rusty! Please fill me in, after you show me I will know from now on what to do. Normally I would incorporate the -1 into the fraction by multiplying it by the denominator but the root being in the denominator is throwing me off.
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  6. #6
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    fraction - 1 = 0

    fraction = 1

    therefore, numerator = denominator

    2x = \sqrt{x^2+1}

    solve for x
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  7. #7
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    Ok I followed the 2x=sqrt(x^2+1)

    I want to say x= 1/2 and x=-1/2 Is this correct or am I way off?
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  8. #8
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    Quote Originally Posted by FactoringAnguish View Post
    Ok I followed the 2x=sqrt(x^2+1)

    I want to say x= 1/2 and x=-1/2 Is this correct or am I way off?
    2x = \sqrt{x^2+1}<br />

    4x^2 = x^2 + 1

    3x^2 = 1

    since x > 0 in the context of the problem ...

    x = \sqrt{\frac{1}{3}}

    you really need to work on your algebra ... it's the hardest part of studying calculus.
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  9. #9
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    Yeah you are 100% correct.

    I took a break from math altogether for a year and I completely forgot how to do problems that I would have not had a problem with prior. I pick it back up quick once I see it done but man do I draw a blank until that happens.

    I wonder if thats something thats exclusive to me or is it the same for others. A year is a relatively short time and I forgot so much during that time period, I didn't use a drop of math so I didn't retain much skills I picked up. I'm halfway through this semester of calculus and so far so good (thankfully).
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