An oil refinery is located 1 km north of the north bank of a straight river that
ID: 1410062 • 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 6 km east of the refinery. The cost of laying pipe is $500,000 per km over land to a point P on the north bank and $1,000,000 per 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 the answer to two decimal places.)
Explanation / Answer
where R is the refinery, O will be the x-axis origin, P is the point on the north bank,
and x= distance from O to the storage tanks. [Note, we could have put R at the origin,
but the algebra is a little simpler this way]
The cost C(x) of the pipeline as a function of x is:
C(x) = distance along north shore * pipeline cost over land +
distance under the river * pipeline cost under land
The distance along the north shore is 6-x
The distance (by Pythagorean theorem) under the water is sqrt( 3^2 + x^2)
So,
C(x) = (6-x)*500000 + sqrt(9 + x^2) *100000
[You should graph this]
To find the value of x where C(x) is minimized, we set dC/dx = 0,
[Reminder - use the chain rule to differentiate the second term]
Differentiating and simplifying, we get
dC/dx = C'(x) = -500000 + 100000x/ sqrt(9+x^2) = 0
100000x / sqrt(9+x^2) = 500000
100000x/500000 = sqrt(4+x^2)
Squaring both sides, we get
value of x after getting value
So the distance from the refinery to point P is 6-x =? km