Math problem confusing, need help please?

Question

A drilling rig, a km offshore, is to be connected by pipe to a refinery on shore, b km down the coast from the rig. If underwater pipe costs w dollars/km and land-based pipe costs l < w dollars/km, what values of x and y give the least expensive connection?

Answer 1

You haven't said what x and y are, so I can't tell you the ultimate answer. ;-)

To minimise cost we want to drive one pipeline under water from the rig to the shore at some point in between the rig and the refinery, and then another pipeline on the land from that point to the refinery. The point we choose depends on the ratio of l and w. If l is very small compared to w, we will have the underwater pipe heading straight to the nearest point on shore. If l is close to w we will drive the underwater pipeline pretty well straight to the refinery.

Caution: Make sure you view the rest of this answer in a font that clearly distinguishes 1 and l.

Suppose we drive the underwater pipeline to a point r km along the shore from the rig (i.e. (b-r) km from the refinery). Then the length of the underwater pipeline is √(a^2 + r^2) and the length of the land pipeline is b-r, so the total cost is
C(r) = w√(a^2 + r^2) + l(b-r)
C'(r) = w(1/2) (a^2 + r^2)^(-1/2) . 2r - l
= wr (a^2 + r^2)^(-1/2) - l
= 0 if l (a^2 + r^2)^(1/2) = wr
Squaring both sides we get
l^2 (a^2 + r^2) = w^2 r^2
<=> (w^2 - l^2) r^2 = l^2 a^2
<=> r^2 = l^2 a^2 / (w^2 - l^2)
<=> r = la / √(w^2 - l^2).

Note that when l is very small compared to w, the denominator tends to w and we get r = la/w which will be small compared to a. So if a and b are comparable then r will be small, i.e. we go pretty much straight to shore and then along the shore, as expected.

If l is close to w then this expression goes to infinity - since r is physically bounded by b we need to make sure we use b if the indicated value is larger.

Note that if the ratio of w and l is close enough to 1 the cheapest option is to go directly to the refinery. You can verify yourself that this will be true if w/l ≤ √(1 + a^2/b^2).

So the underwater pipeline will go to a point
min{b, (la / √(w^2 - l^2))} km along the shore from the rig in the direction of the refinery. (You can also say that it is to a point max{0, b - (la / √(w^2 - l^2))} km from the refinery.)