Expert: Scotto - 11/22/2008

Question
A pipeline must be laid from point a to point b. Point b is located 31 miles east and 40 miles north of point a. horizontally the first 21 miles east will be under water and the other 15 miles in the easterly dirrection and on land. A perfectly verticle shoreline seperates the water and the land. At what point in the shoreline should the pipeline meet the shore in order to minimize the cost? Pipeline on land costs $74951.35 per mile and in the water 2x as much, at $149902.70 per mile. and how much will the whole project cost?

Answer
Points: A,B
Point B: 31 east, 40 north of point A
Horizontally: 21 east water, 15 east land
Cost: x on land, 2x by sea, C=$74,951.35

Hit shore line y miles north of horizontal.
Water distance: √(y²+21²); cost is 2C√(y²+21²).
Land distance: √[(40-y)²+15²]; cost is C√[(40-y)²+15²].

Total cost: C{2√(y²+21²) +√[(40-y)²+15²]}.

To do to solve:
1) Multiply the (40-y)² out, group by term, add 15² to constant.
The result is √(1825-80y+y²]

2) Take the derivative.  Remember, the derivative of √f(y) is
f'(y)/[2√f(y)].

3) Set equal to 0.  Divide both sides by C (0/C is still 0).

4) Subtract one of the squareroots from both sides.

5) Square both sides of the equation.

6) Convert to quadratic equation by putting all of the terms for y², y, and the constant on one side.  Add together where appropriate.

7) Find the distances for each.

8) Determine the cost for each.