Calculus optimization problem need help ASAP?

Question

here is the problem..
a pipeline transporting oil will connect two points A and B that are 3 miles apart (hypotenuse) and on opposite banks of a straight river 1 mile wide. Part of the pipeline will run under water from A to a point C on the opposite bank, and then above ground from C to B. If the cost per mile of running the pipeline under water is four times the cost per mile of running it above ground, find the location of C that will minimize the cost (disregard the slope of the river bed)

Answer 1

Draw a sketch of the problem. Mark point D directly across the river from A. Mark point C between D and B at a distance x from D.

For simplicity, assume cost on river bank is 1 and cost under water is 4.

AC = sqrt(1+x^2)
BD=sqrt(8) ----> BC=sqrt(8) - x

Cost = y = 4sqrt(1+x^2) +sqrt(8) - x

dy/dx=4x/sqrt(1+x^2) - 1

Set dy/dx=0

Solve for x ----> x=1/sqrt(15)

C is about 0.258 miles downstream from A assuming B is downstream from A.

Answer 2

Let D be the point that is directly across the river from B. Then by the Pythagorean Theorem, solve for the distance AD.

Given

AB = 3
DB = 1

We have

AD² = AB² - BD² = 3² - 1² = 9 - 1 = 8
AD = √8

Let
CD = x

Then
AC = AD - CD = √8 - x

CDB is a right triangle with CB the hypotenuse.

CB² = CD² + BD² = x² + 1
CB = √(x² + 1)

Now we have the distance of the path from A to C to B.

Distance = AC + CB = (√8 - x) + √(x² + 1)

We want to minimize the cost of the path.

Cost = AC + 4CB = (√8 - x) + 4√(x² + 1)

Take the derivative and set it equal to zero to find the critical points.

dCost / dx = -1 + 4(1/2)(2x)/√(x² + 1) = 0

4(1/2)(2x)/√(x² + 1) = 1
4x/√(x² + 1) = 1
4x = √(x² + 1)

16x² = x² + 1
15x² = 1
x² = 1/15
x = 1/√15

AC = √8 - x = √8 - 1/√15 ≈ 2.5702282 miles
CD = 1/√15 ≈ 0.2581988 miles



If B is downstream from A then the point C is
√8 - 1/√15 miles downstream from point A and 1/√15 upstream from point D.

Answer 3

sqrt 1/15