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

2. Originally Posted by sk8erboyla2004
Square about side because you getting ride of the root on the other correct?
Yep - that's right.

3. 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?

4. 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

$
2(100-l) = \sqrt{(100-l)^2+900}\;\; \Rightarrow\;\;4(100-l)^2 = (100-l)^2 + 900
$
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.

5. 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

$
2(100-l) = \sqrt{(100-l)^2+900}\;\; \Rightarrow\;\;4(100-l)^2 = (100-l)^2 + 900
$
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

6. 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).

7. Just plug C

$100-10\sqrt{3}$

into the original cost function?

8. Originally Posted by sk8erboyla2004
Just plug C

$100-10\sqrt{3}$

into the original cost function?
Yep. That's what I did.

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

10. 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?

11. 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

12. 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.

13. 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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