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 $\displaystyle x$ = distance from point A to point P

let $\displaystyle C$ = cost in thousands of $as a function of$\displaystyle x\displaystyle C(x) = \sqrt{x^2 + 5^2} \cdot 100 + (8-x) \cdot 75$find the value of x that minimizes the cost 3. That helped a whole bunch. But how did you figure out C(x) ? Also the problem says that B is 8 miles from A why do you have an 8 - x from P to B ? After all this to find the minimum will I have to integrate C(x) ? If so how would I go about finding the values of my upper and lower limits ? 4. ohh never mind the 8 - x. 5. C(x) = (water distance)(cost of water pipe) + (land distance)(cost of land pipe) A to P = x P to B = 8-x -------------- add'em up A to B = 8 you use derivatives to find maximums and minimums, not integrals. 6. Thank You very much. 7. Ok I sometimes have a problem graphing a function on my calculator so i was just wondering if the critical point is 0.15. If anyone might be able to confrim it that would be great. 8. Originally Posted by wair Ok I sometimes have a problem graphing a function on my calculator so i was just wondering if the critical point is 0.15. If anyone might be able to confrim it that would be great. calculator? put that thing away until the very end.$\displaystyle \displaystyle C'(x) = \frac{100x}{\sqrt{x^2+25}} - 75 = 0\displaystyle 4x = 3\sqrt{x^2+25}\displaystyle 16x^2 = 9(x^2+25)\displaystyle 7x^2 - 225 = 0\displaystyle \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.