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Least Cost for Laying a Pipeline


Date: 06/21/99 at 00:04:03
From: sandi
Subject: Trig and Calculus

I am halfway through working on a problem and am hoping you can point 
me in the right direction where I am stuck.

An oil platform is at point O, at sea, 10 km from the nearest point P 
on a stretch of straight coastline. An oil refinery is at point R, 16 
km along the coast from point P. It is necessary to lay a pipeline 
from the platform to the refinery, consisting of straight-line 
sections. Laying pipeline underwater is 4/3 times as expensive as 
laying pipeline on land. This ratio is the relative cost.

    1\
    1 \
    1  \
 10 1   \
    1    \
    1     \
    1      \
    1_______\______
    P        X     R
    <-----16km----->  

If the pipeline reaches the coast at point X, distance x km from P in 
the direction of R, find an expression for the cost C (in dollars) of 
laying the pipeline in terms of x, given that laying pipeline on land 
has a fixed cost of A dollars per kilometer.

I've worked out this expression to be:

     C(x) = (4A/3)[sqrt(100 + x^2)] + A(16 - x)

Find the least expensive route for laying the pipeline. I think that 
this is when x = 11.34?

This is what I'm stuck on:

Compare the cost of the solution you have found with the cost of a 
pipeline directly from O to R and with a pipeline from O to P to R.

Also, if the cost of the underwater pipeline changes in relation to 
the cost of the coastal pipeline, explore how the least expensive 
route changes. In particular, see if there is a value of k (the 
relative cost) at which the direct OR route becomes the cheapest 
solution, or one for which the route from O to P to R is the least 
expensive.

I am hoping that you can please give me some hints about how to go 
about solving this. I am fine with theory, but with an application 
problem such as this one, I don't know where to start. If you know of 
any similar problems that might be able to help also, I'd appreciate 
if you could let me know. Any help you can give would be very much 
appreciated.

Hoping to hear from you soon,
Sandi


Date: 06/21/99 at 07:10:51
From: Doctor Floor
Subject: Re: Trig and Calculus

Dear Sandi,

Thanks for your question!

Let us consider the general case, where at sea, the costs are p (p>0) 
times as much as on land. 

Your function C(x) then becomes:

     C(x) = pA sqrt(100+x^2) + A(16 - x)

Of course we should only consider x in the interval [0,10].

To find possible minimum costs we have to use the derivative C'(x). 
This is given by:

     C'(x) = pA [2x / 2sqrt(100+x^2)] - A
           = pA [x/sqrt(100+x^2)] - A

(To find the derivative of sqrt(100+x^2) I used the chain-rule.)

A possible minimum should occur when C'(x) = 0. If we don't find a 
solution x in [0,10], then we should check which of x = 0 or x = 10 is 
the least costly.

                        C'(x) = 0
      pA[x/sqrt(100+x^2)] - A = 0
             px/sqrt(100+x^2) = 1
                           px = sqrt(100+x^2)
                       p^2x^2 = 100 + x^2
                 (p^2 - 1)x^2 = 100
                          x^2 = 100/(p^2 - 1)

This gives you possibly one solution for x in [0,10].

When p = 4/3, as it was in your first example, then you find that 
x > 10, and you just have to check whether x = 0 or x = 10 gives the 
desired result.

When p = 5, for example, you find that x is approx. 2.04. Then you 
should still check whether this gives less costs then x = 0 or x = 10, 
because from C'(x) = 0 we might find a maximum as well.

If you need more help, just write us back.

All the best,

- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
College Calculus

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