Differential Calculus: Maximum and Minimum Problem and Solution
An oil refinery is located on the north bank of a straight river that is 2 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 $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? (Round your answer to two decimal places.)
1 year ago Report Abuse Colorado...
Best Answer - Chosen by Voters
This is a min-max calculus problem, where we want to minimize the cost function:
We need a drawing of the situation: see https://docs.google.com/drawings/d/1PvkU…
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( 2^2 + x^2)
So,
C(x) = (6-x)*200000 + sqrt(4 + x^2) * 400000
[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) = -200000 + 400000x/ sqrt(4+x^2) = 0
400000x / sqrt(4+x^2) = 200000
400000x/200000 = sqrt(4+x^2)
Squaring both sides, we get
4x^2= 4 + x^2
x = sqrt(4/3) = 1.15
So the distance from the refinery to point P is 6-x = 4.85
1 year ago Report Abuse Colorado...
Best Answer - Chosen by Voters
This is a min-max calculus problem, where we want to minimize the cost function:
We need a drawing of the situation: see https://docs.google.com/drawings/d/1PvkU…
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( 2^2 + x^2)
So,
C(x) = (6-x)*200000 + sqrt(4 + x^2) * 400000
[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) = -200000 + 400000x/ sqrt(4+x^2) = 0
400000x / sqrt(4+x^2) = 200000
400000x/200000 = sqrt(4+x^2)
Squaring both sides, we get
4x^2= 4 + x^2
x = sqrt(4/3) = 1.15
So the distance from the refinery to point P is 6-x = 4.85