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anonymous

  • 5 years ago

An oil rig is to be built 30 miles down shore and 20 miles out to sea from a refinery. A pipe line needs to be built from the refinery to the oil rig. It costs $1000 per mile to build the pipe on land and $1500 per mile to build it at sea. The planners of this oil rig want to have the lowest cost possible for this pipe. How much would it cost to build the pipe line all the way down the shore until it is directly "below" the oil rig and then out to sea?

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  1. amistre64
    • 5 years ago
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    are we assuming that the sea floor is flat?

  2. anonymous
    • 5 years ago
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    yes

  3. anonymous
    • 5 years ago
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    this starts out to be a minimization problem but then the question asks something different. confusing.

  4. amistre64
    • 5 years ago
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    10sqrt(13) * 1000 then

  5. amistre64
    • 5 years ago
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    the minimal distance between 2 points on a plane is a stright line

  6. amistre64
    • 5 years ago
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    its a pythag problem that has no real world application lol

  7. amistre64
    • 5 years ago
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    it seems to be asking; how much to build a pipeline along the legs of the triangle tho doesnt it

  8. anonymous
    • 5 years ago
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    not to me but my reading comprehension might be off.

  9. amistre64
    • 5 years ago
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    1 if by land; 2 if by sea..... yeah, they want the minimal cost compared to yada yada yada

  10. anonymous
    • 5 years ago
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    but it doesn't ask for minimum cost, even though it says "The planners of this oil rig want to have the lowest cost possible for this pipe."

  11. anonymous
    • 5 years ago
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    is says "How much would it cost to build the pipe line all the way down the shore until it is directly "below" the oil rig and then out to sea? 8 minutes ago"

  12. amistre64
    • 5 years ago
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    we have to determine at what cost the land pipe and the sea pipe are qual

  13. amistre64
    • 5 years ago
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    yeah, below being a pictorial term

  14. anonymous
    • 5 years ago
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    the shoreline is in a distance straight line from the refinery the oil rig is verticalto the shoreline and at angle between the shorline and refinery

  15. anonymous
    • 5 years ago
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    it looks to me as if it is just asking how much it would cost to build the pipeline 30 miles down the shore and then 20 miles out to sea. that is what the last statement says unless i am reading it incorrectly. is this for a calc class?

  16. amistre64
    • 5 years ago
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    25980.76 for a straight run to the oil rig

  17. amistre64
    • 5 years ago
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    it is; but they want some variation of that;

  18. anonymous
    • 5 years ago
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    30 miles down shore AND 20 miles out to sea from a refinery

  19. anonymous
    • 5 years ago
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    i take that to mean east 30 and then 20 north, for example. but of course i could be wrong.

  20. amistre64
    • 5 years ago
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    60000 if we build it at a right angle

  21. amistre64
    • 5 years ago
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    52426.40 if we go with a 45 at 10 feet from the start

  22. anonymous
    • 5 years ago
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  23. amistre64
    • 5 years ago
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    C(x) = s(1500) + l(1000)

  24. amistre64
    • 5 years ago
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    how to relate s an l? by angle perhaps?

  25. amistre64
    • 5 years ago
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    sin(t) = l cos(t) = 30 - cos(t) = s

  26. amistre64
    • 5 years ago
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    if this was speed; then the fastest speed would be?

  27. anonymous
    • 5 years ago
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    30mph

  28. amistre64
    • 5 years ago
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    along which route :)

  29. anonymous
    • 5 years ago
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    i dont understand your question?

  30. amistre64
    • 5 years ago
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    straight shot; 54083.26 10 down and 45degrees gets us: 52426.40 11 down and a distance of sqrt(19^2 + 20^2): 52379.34 12(1000) + sqrt(18^2 + 20^2)(1500) = 52360.87

  31. amistre64
    • 5 years ago
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    its a matter if finding the route by land and by sea that is the cheapest to build

  32. amistre64
    • 5 years ago
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    30(1000) + 20(1500) = 60 000 ; so just building it up the 30 then 20 is bad

  33. anonymous
    • 5 years ago
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    is sqrt supposed to mean log?

  34. amistre64
    • 5 years ago
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    log means log sqrt means square root

  35. anonymous
    • 5 years ago
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    what are you sqrting 11 down?

  36. amistre64
    • 5 years ago
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    the pipeline goes along the beach; and each foot costs so much money ($1000) in that instance, we went down 11 feet and bent the pipe to go to the rig out at sea; the sea pipe costs 1500 per foot

  37. amistre64
    • 5 years ago
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    ideally, when the land pipe cost matches the sea pipe cost we have the cheapest cost

  38. amistre64
    • 5 years ago
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    when s(1500) = l(1000) we balance out and have the cheapest cost for total pipe

  39. anonymous
    • 5 years ago
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    o i get it ......

  40. anonymous
    • 5 years ago
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    from the refinery to the oil rig would be the hypotenous of the other sides? multiplied to get the cost?

  41. amistre64
    • 5 years ago
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    yes

  42. anonymous
    • 5 years ago
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    thank you soooooo very much for your help

  43. amistre64
    • 5 years ago
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    can you get it from here?

  44. anonymous
    • 5 years ago
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    yes i can thanks

  45. anonymous
    • 5 years ago
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    i agree with amistre solution. however, since this is calc class you probably should (to make the teacher happy) have some sort of function to find the minimum of. so for the second part, and if you put "x" as suggested. then the distance from where the pipe enters the water to the oil rig is \[\sqrt{x^2+400}\] miles by pythagoras and the cost is $1500 per mile for a total of \[1500\sqrt{x^2+400}\] the distance from the refinery to where the pipe hits the water is 3\[30-x\] miles at a cost of $1000 per mile for a cost of \[1000(30-x)=30000-1000x\] and therefore your total cost will be \[C(x)=1500\sqrt{x^2+400}+30000-1000x\] this is the function you need to minimize

  46. anonymous
    • 5 years ago
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    thanks

  47. anonymous
    • 5 years ago
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    \[C'(x)=\frac{1500x}{\sqrt{x^2+400}}-1000\] set = 0 to find the critical point get \[x=8\sqrt{5}\]

  48. anonymous
    • 5 years ago
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    and don't forget that the distance down the shore line is not x, but rather 30-x or about 12.11 rounded.

  49. anonymous
    • 5 years ago
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    i got it lol

  50. anonymous
    • 5 years ago
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    i did this on the fly so you should check my work. sure about the derivative though, and fairly positive about my \[x=8\sqrt{5}\]

  51. anonymous
    • 5 years ago
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    ok

  52. anonymous
    • 5 years ago
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    good luck.

  53. anonymous
    • 5 years ago
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    thx

  54. anonymous
    • 5 years ago
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    welcome

  55. anonymous
    • 5 years ago
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    lol

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