# Optimization Help

## 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

## The Attempt at a Solution

Cost =2000()+1000()

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Dick
Homework Helper
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?

alright ill try.

i can't seem to put it in terms of x do i have to use pythagorean theorum using 12 and 5 km?

Dick
Homework Helper
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.

So, Cost =1000x + 2000(12^2 + 5^2)^(1/2)
Cost' = 1000?
x= 0?

Or, Cost =1000x + 2000(12^2 + (5-x)^2)^1(/2)

Dick
Homework Helper
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?

Dick
Homework Helper
Or, Cost =1000x + 2000(12^2 + (5-x)^2)^1(/2)
Yes. That's what you want.

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:
Dick