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Question -

An offshore oil well must be connected by pipe to the refinery. The oil-well is 4 km off-shore. The refinery is on the coast, 6 km from the nearest point of land to the oil well. It costs $s$ dollars per kilometer to lay pipe along the shore, but $u$ times as much per km to lay pipe underwater. The pipe may be laid either straight to the refinery, or to an intermediate point on the coast then along the coast.

1) Give the total cost as a function of the distance, $x$, between the point directly opposite the oil well and the place where the pipe comes ashore. This should be a formula involving $x$, $s$ and $u$

2) Then find the $x$-value of the critical point of this cost function. This point will be a function of $u$.

Working -

1) For part (1), I used Pythagoras Theorem and found that the total cost as a function of the total distance $x$ was given as $C(x)= su\sqrt{16+x^2}+s(6-x)$.

The first part of the function, $su\sqrt{16+x^2$}$, is the total cost of the pipe laid underwater while the second part, $s(6-x)$, is the total cost of the pipe laid on land.

2)To find where the critical points occurred, I first differentiated this function, $C(x)= su\sqrt{16+x^2}+s(6-x)$, and got back $\frac{dC}{dx} =\frac{xsu}{\sqrt{16+x^2}} -s $.

I then equated that to equal zero and got that the critical point occurs at $x=\sqrt{\frac{16}{U^2}-16}$ When I put this into the program which asked the question, it comes back incorrect but the cost function, $C(x)= su\sqrt{16+x^2}+s(6-x)$, is given to be correct.

Have I done something wrong in my differentiation and finding the critical points or is my cost function actually incorrect. Does anyone have an directions I can try instead. Please feel free to edit my question for clarity. Thank you!

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  • $\begingroup$ You didn't solve $c'(x) = 0$ correctly. $\endgroup$ – copper.hat Sep 20 '15 at 7:38
  • $\begingroup$ Any suggestions of what I might be doing wrong. Did it several different ways but keep get this same answer even though I also know that it's incorrect. $\endgroup$ – Joe Speedmen Sep 20 '15 at 11:30
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We can assume that $s > 0$, and factor $s$ out of consideration, so I will assume that $s =1$ to simplify notation.

Presumably we can assume that $u>1$.

Then we have $c'(x) = {ux \over \sqrt{x^2+4^2}} -1$.

If $c'(x) = 0$, we have ${ux \over \sqrt{x^2+4^2}} =1$, squaring gives $u^2x^2 = x^2+4^2$. This gives $(u^2-1) x^2 = 4^2$, and then $x = {4 \over \sqrt{u^2-1}}$ (the negative solution can be eliminated because it has a higher cost).

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