An oil refinery is located on the north bank of a straight river that is 1 km wide. ?

A pipeline is to be constructed from the refinery to storage tanks located on the south bank of the river 10 km east of the refinery. 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. To minimize the cost of the pipeline, how far from the refinery should P be located? (Give your answer correct to two decimal places)

thank you!

R

________________

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|_\

|__\..........river wide = w = 1 km

|___\

|____\

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..........P...............T

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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(10-p) + 400,000((1+p^2)^0.5)

cost is minimum when dC/dp = 0

C = 200,000(10-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 = 10-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

i hope you'll understand

it's so difficult to write equation in here :)