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?

- anonymous

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- amistre64

are we assuming that the sea floor is flat?

- anonymous

yes

- anonymous

this starts out to be a minimization problem but then the question asks something different. confusing.

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## More answers

- amistre64

10sqrt(13) * 1000 then

- amistre64

the minimal distance between 2 points on a plane is a stright line

- amistre64

its a pythag problem that has no real world application lol

- amistre64

it seems to be asking; how much to build a pipeline along the legs of the triangle tho doesnt it

- anonymous

not to me but my reading comprehension might be off.

- amistre64

1 if by land; 2 if by sea..... yeah, they want the minimal cost compared to yada yada yada

- anonymous

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."

- anonymous

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"

- amistre64

we have to determine at what cost the land pipe and the sea pipe are qual

- amistre64

yeah, below being a pictorial term

- anonymous

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

- anonymous

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?

- amistre64

25980.76 for a straight run to the oil rig

- amistre64

it is; but they want some variation of that;

- anonymous

30 miles down shore AND 20 miles out to sea from a refinery

- anonymous

i take that to mean east 30 and then 20 north, for example. but of course i could be wrong.

- amistre64

60000 if we build it at a right angle

- amistre64

52426.40 if we go with a 45 at 10 feet from the start

- anonymous

##### 1 Attachment

- amistre64

C(x) = s(1500) + l(1000)

- amistre64

how to relate s an l? by angle perhaps?

- amistre64

sin(t) = l
cos(t) = 30 - cos(t) = s

- amistre64

if this was speed; then the fastest speed would be?

- anonymous

30mph

- amistre64

along which route :)

- anonymous

i dont understand your question?

- amistre64

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

- amistre64

its a matter if finding the route by land and by sea that is the cheapest to build

- amistre64

30(1000) + 20(1500) = 60 000 ; so just building it up the 30 then 20 is bad

- anonymous

is sqrt supposed to mean log?

- amistre64

log means log
sqrt means square root

- anonymous

what are you sqrting 11 down?

- amistre64

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

- amistre64

ideally, when the land pipe cost matches the sea pipe cost we have the cheapest cost

- amistre64

when s(1500) = l(1000) we balance out and have the cheapest cost for total pipe

- anonymous

o i get it ......

- anonymous

from the refinery to the oil rig would be the hypotenous of the other sides?
multiplied to get the cost?

- amistre64

yes

- anonymous

thank you soooooo very much for your help

- amistre64

can you get it from here?

- anonymous

yes i can thanks

- anonymous

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

- anonymous

thanks

- anonymous

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

- anonymous

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

- anonymous

i got it lol

- anonymous

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

- anonymous

ok

- anonymous

good luck.

- anonymous

thx

- anonymous

welcome

- anonymous

lol

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