   Chapter 3.7, Problem 51E

Chapter
Section
Textbook Problem

# 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 $400,000/km over land to a point P on the north bank and$800,000/km under the river to the tanks. To minimize the cost of the pipeline, where should P be located?

To determine

To minimize:

To minimize the cost of the pipeline where should P be located

Explanation

1) Concept:

Absolute minimum- let c be a number in the domain D of a function f then f(c) is the absolute minimum value of f on D if fcf(x) for all x in D

2) Given:

Oil refinery is located on the north bank of a straight river that is 2 km wide

Pipeline length from refinery to storage banks is 6 km east from the refinery

The cost of laying pipe in millions of dollars is 0.4

The cost of laying pipes under the river, in millions of dollars is 0.8

3) Formula:

Differentiation formula-ddxx=12x·dx

4) Calculation:

Use the given information to draw a rough image

Let P be located 6-x km far to the oil refinery

Oil refinery is located on the north bank of a straight river that is 2 km wide

Pipeline length from refinery is 6 km east from the refinery

The cost of laying pipe in millions of dollars is 0.4

The cost of laying pipes under the river, in millions of dollars is 0.8

Therefore the cost of laying pipes on the land, in millions is 0.4(6-x)

Therefore the cost of laying pipes under the river, in millions is 0.8x2+4

Therefore the total cost of laying pipes in millions of dollars, will be

C=0.46-x+0.8x2+4

To find the minimum cost, differentiate the cost function

C'=0.4-1+0.812x2+4·2x

Simplify

C'=-0.4+0.8xx2+4

To find critical points, set C'=0 and solve x

-0.4+0.8xx2+4=0

-0.4+0.8xx2+4+0.4=0+0.4

Simplify

0.8xx2+4=0.4

Multiply by x2+4 both sides

0.8xx2+4·(x2+4) =0.4·(x2+4)

Simplify

0.8x=0.4x2+4

Divide by 0.4 on both sides

0.8x0.4=0.4x2+40

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