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Optimization Help

  • Thread starter hallowon
  • Start date
  • #1
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Homework Statement



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

Homework Equations





The Attempt at a Solution


Cost =2000()+1000()
 

Answers and Replies

  • #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?
 
  • #3
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alright ill try.
 
  • #4
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i can't seem to put it in terms of x do i have to use pythagorean theorum using 12 and 5 km?
 
  • #5
Dick
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i can't seem to put it in terms of x do i have to use pythagorean theorum using 12 and 5 km?
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.
 
  • #6
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So, Cost =1000x + 2000(12^2 + 5^2)^(1/2)
Cost' = 1000?
x= 0?
 
  • #7
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Or, Cost =1000x + 2000(12^2 + (5-x)^2)^1(/2)
 
  • #8
Dick
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So, Cost =1000x + 2000(12^2 + 5^2)^(1/2)
Cost' = 1000?
x= 0?
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?
 
  • #9
Dick
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Or, Cost =1000x + 2000(12^2 + (5-x)^2)^1(/2)
Yes. That's what you want.
 
  • #10
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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:
  • #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:

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