Question

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?

Answer 1

are we assuming that the sea floor is flat?

Answer 2

yes

Answer 3

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

Answer 4

10sqrt(13) * 1000 then

Answer 5

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

Answer 6

its a pythag problem that has no real world application lol

Answer 7

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

Answer 8

not to me but my reading comprehension might be off.

Answer 9

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

Answer 10

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."

Answer 11

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"

Answer 12

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

Answer 13

yeah, below being a pictorial term

Answer 14

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

Answer 15

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?

Answer 16

25980.76 for a straight run to the oil rig

Answer 17

it is; but they want some variation of that;

Answer 18

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

Answer 19

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

Answer 20

60000 if we build it at a right angle

Answer 21

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

Answer 22

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

Answer 23

how to relate s an l? by angle perhaps?

Answer 24

sin(t) = l

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

Answer 25

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

Answer 26

30mph

Answer 27

along which route :)

Answer 28

i dont understand your question?

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

is sqrt supposed to mean log?

Answer 33

log means log

sqrt means square root

Answer 34

what are you sqrting 11 down?

Answer 35

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

Answer 36

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

Answer 37

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

Answer 38

o i get it ......

Answer 39

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

Answer 40

yes

Answer 41

thank you soooooo very much for your help

Answer 42

can you get it from here?

Answer 43

yes i can thanks

Answer 44

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

Answer 45

thanks

Answer 46

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

Answer 47

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

Answer 48

i got it lol

Answer 49

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

Answer 50

ok

Answer 51

good luck.

Answer 52

thx

Answer 53

welcome

Answer 54

lol