optimization problems

warsatan

New member
Joined
Sep 12, 2005
Messages
36
An offshore oil well is 2 kilometers off the coast. The refinery is 4 kilometers down the cost. If laying pipe in the ocean is twice as expensive as on land, what path should the pipe follow in order to mimize the cost?

many thanks.:)
 
Hello, warsatan!

An offshore oil well is 2 kilometers off the coast.
The refinery is 4 kilometers down the cost.
If laying pipe in the ocean is twice as expensive as on land,
what path should the pipe follow in order to mimize the cost?
Code:
. . . . W
. . . . *                                  The oil well is at W.
. . . . | \                                The refinery is at R. 
. . . . |   \
. . . 2 |     \                       The pipe is laid underwater to B 
. . . . |       \                         then along the coast to R.
. . . . |         \
. . . - + - - - - - * - - - - - *
. . . . A . . x . . B . .4-x. . R
Let \(\displaystyle p\) = price for laying pipe on land (per kilometer).
Then \(\displaystyle 2p\) = price of laying pipe underwater.

From right triangle WAB, we get: .\(\displaystyle WB\:=\:\sqrt{x^2\,+2^2}\) km of underwater pipe.
. . This will cost: .\(\displaystyle 2p\sqrt{x^2 + 4}\) dollars.

Let \(\displaystyle x\,=\,AB\)
There will be \(\displaystyle 4 - x\) km of pipe laid along the shore.
. . This will cost: .\(\displaystyle p(4 - x)\) dollars.

The total cost is: .\(\displaystyle C\:=\:2p(x^2\,+\,4)^{\frac{1}{2}}\,+\,p(4\,-\,x)\)

And that is the function you must minimize . . .
 
Top