A company needs to run an oil pipeline from an oil rig 25...

A company needs to run an oil pipeline from an oil rig 25 miles out to sea to a storage tank that is 5 miles inland. The shoreline runs east-west and the tank is 8 miles east of the rig. Assume it costs $50$ dollars thousand per mile to construct the pipeline under water and $20$ dollars thousand per mile to construct the pipeline on land. The pipeline will be built in a straight line from the rig to a selected point on the shoreline, then in a straight line to the storage tank. What point on the shoreline should be selected to minimize the total cost of the pipeline?

Key Concepts

Cost Function Formulation

This concept involves modeling the total cost as a function that combines the different expenses incurred over various segments of a project. In problems like this, different components (e.g., underwater vs. on-land construction) have separate per-unit costs, so the overall cost is represented as a sum of functions, each weighted by its corresponding cost rate.

Geometric Distance Calculation

When planning routes or connections, distances often must be computed based on geometric relationships. The Pythagorean theorem is typically used in these scenarios to calculate straight-line distances between points when their positions are given in a coordinate plane. This step translates physical positions into mathematical expressions that can be embedded within the cost function.

Calculus-Based Optimization

Optimization involves finding the best solution, such as minimizing or maximizing a quantity—in this case, the total construction cost. By expressing the cost as a function of a variable (like the chosen point on the shore), calculus provides techniques for identifying the value that minimizes the cost function.

Application of Derivatives

Finding the minimum of a function typically involves taking its derivative with respect to the decision variable and then setting it equal to zero to locate critical points. The derivative reflects how the cost changes with respect to small changes in the chosen point, allowing for the determination of a point where the cost is at a local minimum.

Transcript

00:02 Hello, so here we have the diagram of the whole thing, right? so let's just say that the oil rig, that's, this is the sea and this is land, right? now, let's just say that the oil rig is 25 miles in the sea and the, well, the thing that has to be constructed from the land is five miles inside the land, right? and this whole distance, it's eight miles from the sea to the, i mean, from the rig to the place that has to be constructed.

00:31 Of the pipeline.

00:32 So now let's just take, let's just say that the distance from the oil rig to the shore that has to be constructed with the line is l1.

00:43 And the distance from that to the place that has to be constructed in the pipeline with is l2.

00:50 Right.

00:51 And since we know this is 8, so let's just take this distance here to be x.

00:56 So this will be 8 minus x.

00:58 And using pythagoras, you can say that l1 is 6 under root 625 plus x square and l2 is under root 25 plus 64 plus x square minus 16x, which is 89 plus x square minus 16x.

01:15 And here, let's just say that the cost is px.

01:18 That's a cost function.

01:20 So the px cost function would be 50 into l1 plus $20 into l2 because $50 is a cost.

01:27 $50 is cost per mile for the pipeline in c and 20 is for land.

01:33 So let's take l1 squared, it includes 625 plus x square.

01:37 L2 square is 89 plus x square minus 16.

01:41 This is the equation that we have right now.

01:44 Now to minimize the cost, what we do is we differentiate px with respect to x, which gives us 50 dl1 by the x plus 20 dl2 by the x.

01:52 Now using implicit differentiation, we can differentiate both of these...