Trigonometry Question

Question

A company needs to connect a physical data line to a building on an island located 1000m off the shore. The hub station that needs to be connected to the island building is located 3000m away from the shortest (1000m) point to the island. The cost to run the line on land is $3/m, and the cost to run it across the water is $6/m. One option would be to run the line directly to the island in the water the whole way. Another option would be to run 4000m across the land then turn 90º and run it directly to the island 1000m in the water. Or you could choose to Run a portion across land, then make a turn of θ° and run the rest across the water. Answer all of the following:

1) What is the cost to run directly to the island?

2) What is the cost to go across the land 3000m, and then turn and go in the water?

3) What is the cost function C(θ)= that represents the cost with respect to the angle you would turn to get other possible paths/costs?

4) What are reasonable bounds for the domain (θ)?

5) What is the angle that gives the minimum cost to run the line?

6) What is the minimum cost?

7) How much cable must be used (how many total meters) to achieve the minimum cost?

8) If the owner of the island expects to make $1,500/month for access using the new line, how many months before they recoup the cost of the project?

Answer 1

I will walk you through the most important question here which is #3. Questions 1 and 2 are self-explanatory and solving question 3 will make the rest of the questions more trivial. In this situation, we have a right triangle between the hub station, the closest point to the island which I will call the shore, and the island building. Let us say we decide to travel from the hub to the shore and then cut straight to the island partway through. I will call the point at which we change directions the bluff. We can call the angle between the (bluff-shore) and the (bluff-island), θ.

We can see that our cost function will have two parts, one is the part on land between the hub and the bluff while the other is the part between the bluff and the island. Thankfully we can use θ to solve for both. Let us begin with the distance between the hub and the bluff. This distance is the original 3000m between the hub and the shore minus the distance between the bluff and the shore. I encourage you to graph these points to follow along. If you do, we can see that we have an inner right triangle between the bluff, the shore, and the island. Since we know the distance between the shore and the island is 1000m, we can solve for the distance between the shore and the bluff. Using our known trigonometric identities, we see that tanθ = 1000/(bluff-shore distance). Solving for the bluff-shore distance, we get bluff-shore = 1000/tanθ. Thus the land distance the cable will travel over is 3000m - 1000/tanθ.

The second part of the cost function is a little simpler if you got this far. We can see that the cable distance through water is just the hypotenuse of the inner triangle between the bluff, the shore and the island. Using our trigonometric identities again, we see that the sinθ = 1000/(bluff-island distance). Solving for the bluff-island distance, we get that bluff-island = 1000/sinθ.

Now we know the land distance and water distance the cable will travel for any θ, we can put together the entire cost function. As the cost for running a cable through land is $3/m and the cost of running a cable through water is $6/m then we multiple the first part of the cost function by 3 and the second part by 6. All together, C(θ) = $(3(3000-1000/tanθ) +6(1000/sinθ)).

You can check your work by looking at the cost function if θ is the angle between (hub-shore) and (hub-island) i.e. the full water route. You can also check what would happen as we use the right angle approach when θ approaches 90 degrees. Don't solve for when θ = 90 degrees as tanθ will be undefined but you will get an idea what will happen.