minimizing cost: An offshore oil well is 1 mi off the coast.

[maria3128]

An offshore oil well is 1 mile off the coast. The refinery is 2 miles down the coast. If laying pipe in the ocean is twice as expensive as on land, what path should the pipe follow in order to minimize the cost?

I have no idea where to start...the first thing i did was have X equal the distance from the point of shore closest to the point where the pipe comes ashore. Then the equation

y^2=1^2 + x^2 where y equals the length of pipe

I then had C(x) = 2 sqrt (1 + x^2) + (2 - x)

I then though I take the first derivitave of this but I'm not sure if I am even on the right track. Thanks!

 

[soroban]

Re: minimizing cost

Hello, Maria!

You're doing great!

An offshore oil well is 1 mile off the coast.
The refinery is 2 miles down the coast.
If laying pipe in the ocean is twice as expensive as on land,
what path should the pipe follow in order to minimize the cost?

I have no idea where to start . . . you're kidding

The first thing i did was have \(\displaystyle x\) equal the distance from the closest point on shore
to the point where the pipe comes ashore . . . good!

Then the equation: \(\displaystyle \:y^2\:=\:1^2\,+\,x^2\) where \(\displaystyle y\) equals the length of underwater pipe . . . yes!

I then had: \(\displaystyle \: C(x) \:= \:2\sqrt{1\,+\,x^2}\,+\,(2\,-\,x)\) . . . correct!

I then thought I take the first derivative of this . . . right!

Exellent work . . . nice going, Maria!

 

[maria3128]

Ok, I took the first derivitave and got

2[x/sqrt (1 + x^2)] - 1

I set this equal to zero and got x = 1/2

I tried plugging this back into my equation C(x) = 2 sqrt (1 + x^2) + (2-x) and I ended up with C = 3.736

My problems are (1) I'm not sure if this is the correct answer and (2) if the math is done right, how does this necessarily answer the question of how the pipe should be laid? Thanks so much.