An offshore oil well is located in the ocean at a point W, which is 5 miles from the closest point on the shore A on a straight line shoreline. the oil is to be piped to a point B in the shore that is 8 miles from A, by piping it on a straight line underwater to some point on the shore P between A and B and then on to B via pipe along the shoreline. if the cost of laying pipe is $100,000 per mile under water and$75,000 per mile over land, where should point P be located to minimize the cost of the laying pipe?

Help will be very much appreciated, I have no idea where to start or what to do.

2. Originally Posted by wair
An offshore oil well is located in the ocean at a point W, which is 5 miles from the closest point on the shore A on a straight line shoreline. the oil is to be piped to a point B in the shore that is 8 miles from A, by piping it on a straight line underwater to some point on the shore P between A and B and then on to B via pipe along the shoreline. if the cost of laying pipe is $100,000 per mile under water and$75,000 per mile over land, where should point P be located to minimize the cost of the laying pipe?

Help will be very much appreciated, I have no idea where to start or what to do.
how about starting with a sketch?

let $x$ = distance from point A to point P

let $C$ = cost in thousands of \$ as a function of $x$

$C(x) = \sqrt{x^2 + 5^2} \cdot 100 + (8-x) \cdot 75$

find the value of x that minimizes the cost

3. That helped a whole bunch. But how did you figure out C(x) ? Also the problem says that B is 8 miles from A why do you have an 8 - x from P to B ? After all this to find the minimum will I have to integrate C(x) ? If so how would I go about finding the values of my upper and lower limits ?

4. ohh never mind the 8 - x.

5. C(x) = (water distance)(cost of water pipe) + (land distance)(cost of land pipe)

A to P = x

P to B = 8-x
A to B = 8

you use derivatives to find maximums and minimums, not integrals.

6. Thank You very much.

7. Ok I sometimes have a problem graphing a function on my calculator so i was just wondering if the critical point is 0.15. If anyone might be able to confrim it that would be great.

8. Originally Posted by wair
Ok I sometimes have a problem graphing a function on my calculator so i was just wondering if the critical point is 0.15. If anyone might be able to confrim it that would be great.
calculator? put that thing away until the very end.

$\displaystyle C'(x) = \frac{100x}{\sqrt{x^2+25}} - 75 = 0$

$4x = 3\sqrt{x^2+25}$

$16x^2 = 9(x^2+25)$

$7x^2 - 225 = 0$

$\displaystyle x = \frac{15}{\sqrt{7}} \approx 5.67$

9. Thank you. When I fixed up my derivative i put the square root of x^2 + 25 on top. I neglected the sign of the power.