An oil refinery is located 1 km north of the north bank of a straight river that
ID: 2886550 • Letter: A
Question
An oil refinery is located 1 km north of the north bank of a straight river that is 3 km wide. A pipeline is to be constructed from the refinery to storage tanks located on the south bank of the river 7 km east of the refinery. The cost of laying pipe is $300,000/km over land to a point P on the north bank and $600,000/km under the river to the tanks. To minimize the cost of the pipeline, how far downriver from the refinery should the point P be located? (Round your answer to two decimal places.)
Explanation / Answer
R
________________
|
|_
|__..........river wide = w = 1 km
|___
|____
----------------------------
..........P...............T
<--p---->|<-----(8-p)-->
note that :
R = refinery
P = point of pipe on north bank
p = horizontal distance of P from of refinery
T = location of tank
note that we can find the length of pipe under the river (RP)
using pythagoras method
RP^2 = w^2 + p^2
RP^2 = 1^2 + p^2
RP^2 = 1 + p^2
RP = (1 + p^2)^0.5
The cost of laying pipe is $200,000 per km over land to a point P on the north bank and $400,000 per km under the river to the tanks
the cost of pipe (C) can be written as
C = 200,000(PT) + 400,000(RP)
C = 200,000(8-p) + 400,000((1+p^2)^0.5)
cost is minimum when dC/dp = 0
C = 200,000(8-p) + 400,000((1+p^2)^0.5)
dC/dp = 200,000(-1) + 400,000[0.5(1+p^2)^(-0.5)*2p]
note :
if C = 8-p
dC/dp = -1
if C = (1+p^2)^0.5
dC/dp = 0.5(1+p^2)^(-0.5)*2p
you know how to get it, don't you??
dC/dp = 200,000(-1) + 400,000[0.5(1+p^2)^(-0.5)*2p]
dC/dp = 200,000(-1) + 400,000[p(1+p^2)^(-0.5)]
to obtain minimum value dC/dp = 0
0 = 200,000(-1) + 400,000[p(1+p^2)^(-0.5)]
200,000 =400,000[p(1+p^2)^(-0.5)]
0.5 = [p(1+p^2)^(-0.5)]
0.5(1+p^2)^(0.5) = p
square both left side and right side
0.25 (1+p^2) = p^2
1+p^2 = 4(p^2)
3p^2-1 = 0
p^2-(1/3)=0
(p^2 - 0.333)=0
(p + sqrt0.333)(p - sqrt0.333) = 0
p = - sqrt0.333 or p = sqrt0.333
p = sqrt 0.333 because length cannot be minus
p = 0.58
so, P must be located 0.58 km east of refinery