Minimize Cost: Optimize Cost of Telephone Line Across River

[hallowon]

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()

 

[Dick]

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?

 

[hallowon]

alright ill try.

 

[hallowon]

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

 

[Dick]

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.

 

[hallowon]

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

 

[hallowon]

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

 

[Dick]

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]

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

Yes. That's what you want.

 

[hallowon]

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.

 

[Dick]

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.