# Modeling Problem

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• May 2nd 2009, 11:42 AM
sk8erboyla2004
Modeling Problem
Ive encounter this problem which is very confusing for me

Tim needs to run a pipeline to his camp that is 100 feet west along the shore and 30 feet north across a lake from his starting position. It cost $10 per foot to run pipe over land and$20 to run pipe over water What path should Tim take to minimize his cost? What his minimum cost ?

I so if I draw a mental picture you a height of 30 feet on the y axis and 100 feet on the x axis

I know I can use the distance formula somehow but i am still lost
• May 2nd 2009, 02:15 PM
Jester
Quote:

Originally Posted by sk8erboyla2004
Ive encounter this problem which is very confusing for me

Tim needs to run a pipeline to his camp that is 100 feet west along the shore and 30 feet north across a lake from his starting position. It cost $10 per foot to run pipe over land and$20 to run pipe over water What path should Tim take to minimize his cost? What his minimum cost ?

I so if I draw a mental picture you a height of 30 feet on the y axis and 100 feet on the x axis

I know I can use the distance formula somehow but i am still lost

Suppose you move $l$ft in the $x$ direction. The cost for this would be $C = 10 l$. Now the distance between you (now) and the point across the river is $\sqrt{(100-l)^2 + 30^2}$ and the cost for this portion is $C = 20\sqrt{(100-l)^2 + 30^2}$ and thus the total cost is

$C = 10 l + 20\sqrt{(100-l)^2 + 30^2}$

Now minimize C (i.e. $\frac{dC}{dl} = 0$ and find l). Don't forget to check the endpoints $l = 0$ (all river) and $l = 100$(all shore then across).
• May 2nd 2009, 02:50 PM
sk8erboyla2004
For this did you use the distance formula and what else
• May 2nd 2009, 03:09 PM
Jester
Quote:

Originally Posted by sk8erboyla2004
For this did you use the distance formula and what else

For the cost function? Yes, the distance forumla and the cost per length.
• May 2nd 2009, 03:18 PM
sk8erboyla2004
so take the derivative and solve for 0
• May 2nd 2009, 03:45 PM
Jester
Quote:

Originally Posted by sk8erboyla2004
so take the derivative and solve for 0

Yep!
• May 2nd 2009, 06:06 PM
sk8erboyla2004
Ok whew I took the derivative which took a few good lines of work here is my final solution for the derv.

If this is correct I will continue on and solve this for 0

$10+\frac{40l-400}{2\sqrt{(100-l)^2+900)}}$

??? I am so unsure of this
• May 2nd 2009, 10:19 PM
mr fantastic
Quote:

Originally Posted by sk8erboyla2004
Ive encounter this problem which is very confusing for me

Tim needs to run a pipeline to his camp that is 100 feet west along the shore and 30 feet north across a lake from his starting position. It cost $10 per foot to run pipe over land and$20 to run pipe over water What path should Tim take to minimize his cost? What his minimum cost ?

I so if I draw a mental picture you a height of 30 feet on the y axis and 100 feet on the x axis

I know I can use the distance formula somehow but i am still lost

There is a similar problem in this thread: http://www.mathhelpforum.com/math-he...s-problem.html

Most of these sorts of optimisation problems have been asked and answered several times in the forums over the years. It's always worth doing a search of key words ....
• May 2nd 2009, 10:50 PM
sk8erboyla2004
Thanks however is my differentiation correct?
• May 3rd 2009, 06:09 AM
Jester
Quote:

Originally Posted by sk8erboyla2004
Thanks however is my differentiation correct?

Looks goods so far but cancel before solving for $l$.
• May 3rd 2009, 09:40 AM
sk8erboyla2004
Cancel what?
• May 3rd 2009, 09:52 AM
Jester
Quote:

Originally Posted by sk8erboyla2004
Ok whew I took the derivative which took a few good lines of work here is my final solution for the derv.

If this is correct I will continue on and solve this for 0

$10+\frac{40l-400}{2\sqrt{(100-l)^2+900}}$

??? I am so unsure of this

From you post

$C' = 10+\frac{40l-400}{2\sqrt{(100-l)^2+900}} = 0$

so

$10 \left( 1+\frac{2l-20}{\sqrt{(100-l)^2+900}} \right) = 0$

so

$1+\frac{2l-20}{\sqrt{(100-l)^2+900}} = 0$
• May 3rd 2009, 10:13 AM
sk8erboyla2004
damn thats gonna be hard to solve :?
• May 3rd 2009, 10:41 AM
sk8erboyla2004
Quote:

Originally Posted by danny arrigo
From you post

$C' = 10+\frac{40l-400}{2\sqrt{(100-l)^2+900}} = 0$

so

$10 \left( 1+\frac{2l-20}{\sqrt{(100-l)^2+900}} \right) = 0$

so

$1+\frac{2l-20}{\sqrt{(100-l)^2+900}} = 0$

Ok If I solve for l I do

$\frac{2l-20}{\sqrt{(100-l)^2+900}} = -1$

${2l-20} = -1(\sqrt{(100-l)^2+900}$

$2l-20 = -100+l-30$

$2l-l = -130+20$

$l = -110$

? ehhh
• May 3rd 2009, 11:47 AM
Jester
Quote:

Originally Posted by sk8erboyla2004
Ok If I solve for l I do

$\frac{2l-20}{\sqrt{(100-l)^2+900}} = -1$

${2l-20} = -1(\sqrt{(100-l)^2+900}$

$\color{red}{2l-20 = -100+l-30}$

$2l-l = -130+20$

$l = -110$

? ehhh

The step in red isn't right. I'll need to square both sides.
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