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Optimization Help

  1. Mar 17, 2010 #1
    1. The problem statement, all variables and given/known data

    A telephone company has to run a line from point A on one side of a river to another point B that is on the other side, 5km down from the point opposite A. The river is uniformly 12 km wide. The company can run the line along the shoreline to a point C and then under the river to B. The cost of the line along the shore is $1000 per km and the cost under the river is twice as much. Where should point C be to minimize the cost?

    Heres the recreated diagram that came qith question : http://smg.photobucket.com/albums/v28/pokemon123/?action=view&current=opimzation.gif

    2. Relevant equations



    3. The attempt at a solution
    Cost =2000()+1000()
     
  2. jcsd
  3. Mar 17, 2010 #2

    Dick

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    Ok, let x be the distance between A and C. Now I think you want to put expressions for distances in terms of x into () and (). Can you do that?
     
  4. Mar 17, 2010 #3
    alright ill try.
     
  5. Mar 17, 2010 #4
    i can't seem to put it in terms of x do i have to use pythagorean theorum using 12 and 5 km?
     
  6. Mar 17, 2010 #5

    Dick

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    Yes, you do. Start with filling in the () in 1000(). You don't need the pythagorean theorem for that one. Then try the () in 2000(). You do for that one.
     
  7. Mar 17, 2010 #6
    So, Cost =1000x + 2000(12^2 + 5^2)^(1/2)
    Cost' = 1000?
    x= 0?
     
  8. Mar 17, 2010 #7
    Or, Cost =1000x + 2000(12^2 + (5-x)^2)^1(/2)
     
  9. Mar 17, 2010 #8

    Dick

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    1000(x) is good. 2000(12^2 + 5^2)^(1/2) is less than good. The distance across the river is 12. The distance along the bank isn't 5. The distance along the bank between A and B is 5. What's the distance along the bank between C and B?
     
  10. Mar 17, 2010 #9

    Dick

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    Yes. That's what you want.
     
  11. Mar 17, 2010 #10
    So far ,Cost =1000x + 2000(12^2 + (5-x)^2)^1(/2)
    Cost'= ((2000x-10 000/sqrt(x^2-10x+169)) +1000)
    0 =
    -1000sqrt(x^2-10x+169) = 2000x-10 000
    sqrt(x^2-10x+169) = -2x+10
    x^2-10x+169 =(-2x+10)^2
    x^2-10x+169 = 4x^2-40+100
    0 = x^2-10x-23

    The answer on the back of my book says it is x=0 but this one yield no nice numbers.
     
    Last edited: Mar 17, 2010
  12. Mar 17, 2010 #11

    Dick

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    If you don't get any minima between x=0 and x=5 then the cost minimum must be at either x=0 or x=5. Test them both and see which is less. If it's any encouragment, I got the same quadratic as you did for the minimum.
     
    Last edited: Mar 17, 2010
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