# Thread: differenciation

1. ## 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?

i got 6-√3 km from the refinery .. help me whether its right or nt please!

2. ## Re: differenciation

No, that's incorrect. Can you show us how you set the problem up?

3. ## 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?

4. ## Re: differenciation

I would draw a diagram first:

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

By Pythagoras we find:

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

We also know:

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

and so the cost function is:

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

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