# differenciation

• Apr 24th 2013, 09:22 AM
yugashan
differenciation
an offshore oil well is 2 kilometers off the coast. the refinery is 4 kilometers down the coast. laying pipe in ocean is twice as expensive as layiing it on land.
what path should the pipe folow for the minimum cost?
(Lipssealed)
i got 6-√3 km from the refinery .. help me whether its right or nt please!
• Apr 24th 2013, 09:36 AM
MarkFL
Re: differenciation
No, that's incorrect. Can you show us how you set the problem up?
• Apr 24th 2013, 10:07 AM
yugashan
Re: differenciation
oh, sorry.. i got 7.47..
c= land path + ocean path
d=√(2^2-x^2)
L=4-x
c(x)= 2√(2^2-x^2)+(4-x)
differenciation= ((2^2-x^2)^-1/2)-1
when dc/dx=0 , x=-1
c(-1) =7.47
is it right?
• Apr 24th 2013, 10:47 AM
MarkFL
Re: differenciation
I would draw a diagram first:

Attachment 28122

$\displaystyle O$ represents the under ocean distance and $\displaystyle L$ represents the over land distance. Let $\displaystyle k$ represent the cost per unit length of the over land pipeline.

By Pythagoras we find:

$\displaystyle x^2+2^2=O^2\,\therefore\,O=\sqrt{x^2+4}$

We also know:

$\displaystyle x+L=4\,\therefore\,L=4-x$

and so the cost function is:

$\displaystyle C(x)=2kO+kL=k\left(2\sqrt{x^2+4}+(4-x) \right)$

This is the function you want to minimize. What do you find when you equate the derivative to zero?