# Thread: Modeling Problem

1. ## 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

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

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

Now minimize C (i.e. $\displaystyle \frac{dC}{dl} = 0$ and find l). Don't forget to check the endpoints $\displaystyle l = 0$ (all river) and $\displaystyle l = 100$(all shore then across).

3. For this did you use the distance formula and what else

4. 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.

5. so take the derivative and solve for 0

6. Originally Posted by sk8erboyla2004
so take the derivative and solve for 0
Yep!

7. 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

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

??? I am so unsure of this

8. 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 ....

9. Thanks however is my differentiation correct?

10. Originally Posted by sk8erboyla2004
Thanks however is my differentiation correct?
Looks goods so far but cancel before solving for $\displaystyle l$.

11. Cancel what?

12. 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

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

??? I am so unsure of this
From you post

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

so

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

so

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

13. damn thats gonna be hard to solve :?

14. Originally Posted by danny arrigo
From you post

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

so

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

so

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

Ok If I solve for l I do

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

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

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

$\displaystyle 2l-l = -130+20$

$\displaystyle l = -110$

? ehhh

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

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

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

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

$\displaystyle 2l-l = -130+20$

$\displaystyle l = -110$

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

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