How Should the Pipe be Laid?

Date: 03/25/99 at 01:10:53
From: Kristin Babin
Subject: Minimizing cost from an oil rig to the shore

Dear Dr. Math,

Here are two calculus problems that I could not get. It seems as if 
one problem is giving extra information, and the other problem is just 
weirdly worded. Please help me with this if you can.

1. Oil from an offshore rig located 3 miles from the shore is to be 
pumped to a location on the edge of the shore that is 8 miles east of 
the rig (8 miles along the shore). The cost of constructing a pipe in 
the ocean from the rig to the shore is 1.5 times as expensive as the 
cost of construction on land. Determine how the pipe should be laid to 
minimize cost.

When I tried to solve this problem, I kept coming up with the 
Pythagorean theorem. I know the problem is not that easy, but my 
answer was the square root of 73 miles. Is that right?

2. A fence that is 8 feet high stands 27 feet from the wall of a 
building. What is the length of the shortest straight beam that will 
reach to the side of the building from the ground outside the fence?

Again, I used the Pythagorean theorem and got the square root of 793 
feet. I really don't think that is the answer.

Date: 03/25/99 at 08:03:29
From: Doctor Jerry
Subject: Re: Minimizing cost from an oil rig to the shore

Hi Kristin,

1. If x is the distance from (the point of the shore closest to the 
oil rig) to (the point on the shore where the pipe comes ashore), then
y^2 = 3^2+x^2, where y is the length of pipe in the ocean.

The length of the pipe on the land is 8-x, right?

So, the cost is C(x) = 1.5*sqrt(9+x^2) + 1*(8-x), where 0 <= x <= 8.

If you set C'(x) equal to zero, you'll find x = 6/sqrt(5).

Check C at this point, as well as at x = 0 and x = 8.

2. Let x be the distance from bottom of the fence to the point on the 
beam that rests on the ground. Let y be the distance on the building, 
from the ground to where the beam hits.  From similar triangles,

y/(27+x) = 8/x.

So, you can write y in terms of x. The length of the beam is
sqrt(y^2+(27+x)^2).  Differentiate this w.r.t x and set equal to zero.

- Doctor Jerry, The Math Forum
  http://mathforum.org/dr.math/