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

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- May 3rd 2009, 12:20 PMsk8erboyla2004
Square about side because you getting ride of the root on the other correct?

- May 3rd 2009, 12:37 PMJester
- May 3rd 2009, 01:16 PMsk8erboyla2004
Starting off at my mistake

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

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

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

$\displaystyle l(4l-81)=270$

$\displaystyle 4l=351$

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? - May 3rd 2009, 01:50 PMJester
Sorry, when I said it (**) was correct it isn't (should be 4000 instead of 400). Let me provide some details

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

then

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

Setting this equal to zero gives

$\displaystyle

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

$ so $\displaystyle 100-l = \pm \frac{30}{\sqrt{3}}$

Thus $\displaystyle l = 100 \pm 10\sqrt{3}$ where we choose the negative case as the postive gives to much. - May 3rd 2009, 02:14 PMsk8erboyla2004
- May 3rd 2009, 02:40 PMJester
- May 3rd 2009, 02:53 PMsk8erboyla2004
Just plug C

$\displaystyle 100-10\sqrt{3}$

into the original cost function? - May 3rd 2009, 02:59 PMJester
- May 3rd 2009, 03:28 PMsk8erboyla2004
what made you plug in the distance formula for 20l why not 10l or it would not have mattered?

- May 3rd 2009, 03:33 PMJester
- May 3rd 2009, 03:39 PMsk8erboyla2004
- May 3rd 2009, 03:43 PMJester
- May 3rd 2009, 03:47 PMsk8erboyla2004