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Part 4 of 8 Solving 0 C'(x) leads to 5√x² + 1 = 10x which can be re-written as x²+1= Submit Skip.(you cannot come back)

Part 4 of 8 Solving 0 C'(x) leads to 5√x² + 1 = 10x which can be re-written as x²+1= Submit Skip.(you cannot come back)

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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4.6 Q9

Tutorial Exercise 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 4 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 from the refinery should P be located? Part 1 of 8 If x measures the distance from the point directly across the river from the storage tanks to the point P where the underwater pipe reaches the opposite shore, then there are m = 3.4 refinery M Part 2 of 8 Part 3 of 8 3 We have m X C'(x) = -5 + P 11 x z²+1 We need to minimize the cost (measured in $100,000) of the pipeline, C(x)= (No Response) 5 X+10V Part 4 of 8 Solving 0= C'(x) leads to 5√²+1 = 10x which can be re-written as x² + 1 = km of pipe underwater. Doo storage tanks (No Response) √x²+1 Submit Skip (you cannot come back) 20 - (No Response) 20 10. z -10√x² +1. 4-x km of pipe over land and
Transcribed Image Text:Tutorial Exercise 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 4 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 from the refinery should P be located? Part 1 of 8 If x measures the distance from the point directly across the river from the storage tanks to the point P where the underwater pipe reaches the opposite shore, then there are m = 3.4 refinery M Part 2 of 8 Part 3 of 8 3 We have m X C'(x) = -5 + P 11 x z²+1 We need to minimize the cost (measured in $100,000) of the pipeline, C(x)= (No Response) 5 X+10V Part 4 of 8 Solving 0= C'(x) leads to 5√²+1 = 10x which can be re-written as x² + 1 = km of pipe underwater. Doo storage tanks (No Response) √x²+1 Submit Skip (you cannot come back) 20 - (No Response) 20 10. z -10√x² +1. 4-x km of pipe over land and
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