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Finding the formula

  1. Jul 22, 2008 #1
    I'm not exactly sure, as I just don't know how to do it, but it probably concerns finding a formula to describe cost, then finding dy/dx = 0 to find the minimum. The reason I've shown no working is that I don't know where to start exactly. If I could find a formula I'm sure I could do the rest, but I'm not sure where to go at the moment.
    Thanks very much for your help.
     

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  3. Jul 22, 2008 #2

    HallsofIvy

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    The shortest distance between two points is a straight line but it costs more to build under the water than on land. The minimum cost line will be a straight line, under the water to shore, some place between B and C. Let's call that point "D"

    Suppose that point is a distance x from C. Tnen the distance from B to C is 120-x. You can find the distance from A to D by the Pythagorean formula and then multiply the distances by the costs of building under the water or on ground.
     
  4. Jul 22, 2008 #3
    :O!
    Thank you very much. I'll work on it and get back to you.
     
  5. Jul 22, 2008 #4
    Okay, I've got the distance from A to the beach as 130-x. But it doesn't seem right, I'll need to differentiate at some point to get a minimum.

    I got it by using pythagoras on (120-x)^2+50^2...
    Could you tell me if I'm on the right path or not? I need to head to bed, I wanted to have a crack before then.
    Thanks for your help so far :D
    I'll be back when I'm well rested.
     
  6. Jul 22, 2008 #5

    HallsofIvy

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    The distance from A to the beach is given by c2= (120-x)2+ 502= x2- 240x+ 16900. I have no idea how you could get "130- x" out of that!
     
  7. Jul 22, 2008 #6
    I was tired, that's how.
    Ah, I see what I did. When I have brackets squared, I tend to just square the insides, rather than expanding and simplifying like I should.
    If you've gotten to the stage where c^2= x^2-240x+16900, shouldn't you find the square root of the numbers right of the = sign?
    So that we get c=130-14.49x.

    Well that's what I did. Then I multiplied the two distances by their costs, simplified, and ended up with x=11.74. Which means I can calculate the actual distances of pipeline...and one is a negative.
    So obviously I'm not there yet. I still don't understand where the minimum comes in, if at all.
     
  8. Jul 22, 2008 #7

    Borek

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    So far so good, you should end with pipe_cost = f(x) equation.

    You can't simplify not knowing the pipe_cost.
     
  9. Jul 22, 2008 #8
    Okay then. Well I've got Cost=134799000-132955x. Obtained from (130-14.49x)*9500 + (120-x)*4700.
    And again, I'm not sure where to go next.
     
    Last edited: Jul 22, 2008
  10. Jul 22, 2008 #9

    Borek

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    Check your math, your equation is wrong.

    Calculate separately cost of the land pipe and cost of the sea pipe. Add them.
     
  11. Jul 22, 2008 #10
    Oops. I already did check my math, and edited. Turns out, after a recheck, I was wrong again.
    =(1235000-137655x)+(564000-4700x)
    =1799000-142355x
    Now what. Jeez I hope that's right.
     
  12. Jul 22, 2008 #11

    Borek

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    Doesn't look Ok to me. Where have you lost x squared and square root of both non-hypotenuse edges?
     
  13. Jul 22, 2008 #12
    Well, I had c^2=(120-x)^2+50^2, and, to get c= I square rooted the righthand site. Maybe I should've left it how it was, with the root sign over it, rather than actually square rooting it.
     
  14. Jul 22, 2008 #13
    Oh nevermind, I got a friend from school to help me. I'd gone down the wrong path, what I had up there in my previous post is what I should've been differentiating. That's what I did, I obtained x = 28.46.
    Thanks very much Borek and HallsofIvy for your help :D
     
  15. Jul 23, 2008 #14

    Borek

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    No idea what you did before, all I know is that it is impossible to square root (120-x)2+502 (at least as long as we are talking about real numbers).
     
  16. Jul 25, 2008 #15
    Well I figured the square root of 50^2 was 50...Right?
     
  17. Jul 25, 2008 #16
    Your missing the big picture, I don't know how you are doing calculus without a basic understanding of algebra.

    You have

    c^2=(120-x)^2+50^2

    Lets consider something a bit easier before we tackle this one:

    x^2 = 4^2 + 3^2

    By your logic if I were to solve for x I would just take the squareroot of each term on both sides, so you would do this:

    x = 4 + 3

    so x = 7 right? .... wrong.

    x^2 = 4^2 + 3^2

    Lets work it out shall we.

    x^2 = 16 + 9

    x^2 = 25

    What number squared equals 25? 5 right? Or -5 too right? well why did we get 7 before?

    Because in general it is not true that if

    n^2 = p^2 + q^2

    then

    n = p + q

    What you need to do is take the square root of the ENTIRE side NOT each individual term

    so you have n^2 = p^2 + q^2

    then

    sqrt(n^2) = +/- sqrt(p^2 + q^2)

    And mind you sqrt(p^2 + q^2) DOES NOT EQUAL p + q (well it does sometimes but only in very special cases like p and q are both equal to zero)

    Now back to your problem

    c^2=(120-x)^2+50^2

    If you want to take the square root of both sides then you have

    c = +/- sqrt((120-x)^2 + 50^2)

    You can rule out c = -sqrt((120-x)^2 + 50^2) because from the context of the problem it does not make sense for the cost to be negative.

    Now you can work to reduce what is already inside the square root sign

    c = sqrt((120-x)^2 + 50^2)

    c = sqrt((14400 - 240x + x^2) + 2500)

    c = sqrt(x^2 -240x +16900)

    Hope that helps you. Good luck and review your algebra!
     
  18. Jul 25, 2008 #17
    Ah yes. I grasped that rule a long time ago. But, what I do, I have trouble applying rules to everything. Not trouble as such, but sometimes I just forget that rules apply in certain situations and get messed up, until someone like you shows me the light.
    But yeah you're right, I was doing it wrong and I see how.
    Thanks very much. :D
     
  19. Jul 25, 2008 #18
    In my experience this stems from lack of experience!

    If you are not exposed to all types of situations when learning the original rules, then you will most certainly get confused when said situations actually show their faces! I honestly going back and reading an Algebra book (maybe one different from what you originally used) would do you some good. If you have the spare time of course. If you plan on ending your math career with calculus then maybe not.

    Anyway, good luck!
     
  20. Jul 27, 2008 #19
    I honestly don't have the time, and I probably won't do this kind of maths ever again after this year. I'm hunting for a high score for my last year of school, but that's all really, I'm planning a career in IT.
    And you're right, it probably is lack of experience. But I have a book of notes I can take to the exam, and every time something like this happens I put a big red note to pay attention to these kind of mistakes.
    And it's doing problems in all kinds of situations that helps you learn, which is what I'm doing.
    Thanks for the advice! :D!
     
  21. Jul 27, 2008 #20

    Borek

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    You may be surprised... Important and huge areas of IT rely heavily on math.
     
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