# Differential Calculus: Maximum and Minimum Problem and Solution

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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 km

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