Modeling Problem

**sk8erboyla2004** · May 3rd 2009, 01:20 PM

Square about side because you getting ride of the root on the other correct?

**Jester** · May 3rd 2009, 01:37 PM

Originally Posted by sk8erboyla2004

Square about side because you getting ride of the root on the other correct?

Yep - that's right.

**sk8erboyla2004** · May 3rd 2009, 02:16 PM

Starting off at my mistake

$(2l-l )^2= -130+l$

$4l^2-80l-400=-130+l$

$4l^2-80l-l=270$

$l(4l-81)=270$

$4l=351$

l = 87.75

now what would I put for

what path should he talk to minmize his cost and what was the min cost

$87.75?

**Jester** · May 3rd 2009, 02:50 PM

Originally Posted by danny arrigo

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).

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

Sorry, when I said it (**) was correct it isn't (should be 4000 instead of 400). Let me provide some details

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

then

$C' = 10 - \frac{20(100-l)}{\sqrt{(100-l)^2+900}}$

Setting this equal to zero gives

$<br /> 2(100-l) = \sqrt{(100-l)^2+900}\;\; \Rightarrow\;\;4(100-l)^2 = (100-l)^2 + 900<br />$ so $100-l = \pm \frac{30}{\sqrt{3}}$

Thus $l = 100 \pm 10\sqrt{3}$ where we choose the negative case as the postive gives to much.

**sk8erboyla2004** · May 3rd 2009, 03:14 PM

Originally Posted by danny arrigo

Sorry, when I said it (**) was correct it isn't (should be 4000 instead of 400). Let me provide some details

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

then

$C' = 10 - \frac{20(100-l)}{\sqrt{(100-l)^2+900}}$

Setting this equal to zero gives

$<br /> 2(100-l) = \sqrt{(100-l)^2+900}\;\; \Rightarrow\;\;4(100-l)^2 = (100-l)^2 + 900<br />$ so $100-l = \pm \frac{30}{\sqrt{3}}$

Thus $l = 100 \pm 10\sqrt{3}$ where we choose the negative case as the postive gives to much.

So l is the min cost and what should the path he should take be

**Jester** · May 3rd 2009, 03:40 PM

Originally Posted by sk8erboyla2004

So l is the min cost and what should the path he should take be

Lay the pipeline $l = 100-10\sqrt{3} = 82.679$ ft. along the shoreline then cross into the river. The cost is $C(100-10\sqrt{3}) = \$1,519.62$ . The endpoints ( $l = 0, 100$ actually give more).

**sk8erboyla2004** · May 3rd 2009, 03:53 PM

Just plug C

$100-10\sqrt{3}$

into the original cost function?

**Jester** · May 3rd 2009, 03:59 PM

Originally Posted by sk8erboyla2004

Just plug C

$100-10\sqrt{3}$

into the original cost function?

Yep. That's what I did.

**sk8erboyla2004** · May 3rd 2009, 04:28 PM

what made you plug in the distance formula for 20l why not 10l or it would not have mattered?

**Jester** · May 3rd 2009, 04:33 PM

Originally Posted by sk8erboyla2004

what made you plug in the distance formula for 20l why not 10l or it would not have mattered?

I'm sorry, I'm not sure of your question? Can you elaborate?

**sk8erboyla2004** · May 3rd 2009, 04:39 PM

Originally Posted by danny arrigo

I'm sorry, I'm not sure of your question? Can you elaborate?

I assume to cost function was

10l+20l

and you plugged in the distance formula for 20 l

**Jester** · May 3rd 2009, 04:43 PM

Originally Posted by sk8erboyla2004

I assume to cost function was

10l+20l

and you plugged in the distance formula for 20 l

The cost function is $C = 10 l + 20\sqrt{(100-l)^2 + 30^2}$ established earlier.

**sk8erboyla2004** · May 3rd 2009, 04:47 PM

Originally Posted by danny arrigo

The cost function is $C = 10 l + 20\sqrt{(100-l)^2 + 30^2}$ established earlier.

Yes I know but could you have done

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

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