Thursday

July 24, 2014

July 24, 2014

Posted by **piper** on Thursday, April 1, 2010 at 5:50am.

• Straight, two-inch coated pipe must be used, at a cost of $ 1.50 per foot.

• A maximum of one elbow joint may be used. The elbow joint may be fabricated with any angle measure and will not be included in the cost.

• On normal terrain, installation cost is $ 1.20 per foot.

• Installation in the wetland area requires the use of a special hoe, at an additional installation cost of $ 60 per hour over and above normal installation cost.

• In a 10-hour day, this special hoe can dig approximately 300 feet of trench.

Write up your investigation as a report to your supervisor, who should be able to understand your report without reference to this sheet. Demonstrate that you and your partner(s) have considered the costs of what you believe to be several viable pipeline routes connecting the existing well at A to the new well at B. Discuss your selection for the route that incurs the least cost and mathematically justify that your selection will give the least cost of all possible routes from A to B. Be sure to explain why minimum distance may not yield minimum cost.

Hints:

• Since the wetland separating A and B is very irregular in shape, simplify your work from the beginning by making some (slightly exaggerated) assumptions. You are to assume that the entire wetland is in the shape of a rectangle. How far to the east will your rectangle go? How far to the west? To the north? To the south?

• Reduce the number of paths you consider before calculating anything. Should you consider a path around the swamp to the north of A (with only one elbow joint)? What about to the south of B? Can you really go due south of A only on normal terrain?

• Do some measuring and calculating of costs of various paths, but be sure you consider the case where the pipeline is laid some distance southeast of A on wetland terrain and then turns due south on normal terrain to B. Perhaps this is where calculus comes in?

If you type in this :Engineering Applications in Differential and Integral Calculus* onto google on page four you will see the map. Please help

the title of this lab is designing a pipeline with minimum cost

- MATH (CALCULUS) PLZ HELP -
**Damon**, Thursday, April 1, 2010 at 7:02amRoute 3: I think the hypotenuse is wrong in the text

d = d1-d3

cost = 2.7 x + 4.7 [(d-x)^2 +d2^2 ]^.5

dc/dx = 2.7 +4.7(1/2)[(d-x)^2 +d2^2 ]^-.5 *2(d-x)(-1)

now set that to zero and solve for x

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