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# An oil refinery is located on the north bank of a straight river that is 2 km wide. A pipeline is to be constructed from the refinery to storage tanks located on the south bank of the river 2 km east of the refinery. The cost of laying pipe is $\$ $400,000 /km over land to a point$ P $on the north bank and$ \800,000 /km under the river to the tanks. To minimize the cost of the pipeline, where should $P$ be located?

## 0.845

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##### Top Calculus 2 / BC Educators   ##### Michael J.

Idaho State University Lectures

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### Video Transcript

in this problem, we want to build a pipeline between the factory and refinery Refinery is located across a two km river, and it's also two km east along the shore of the river from the factory. So what we do is we build a pipeline from the factory to a point P. On the coast and then across the water to directly to the refinery. And we want to locate the location P. That minimizes the cost of the pipeline. We also know that the cost of building along the coast, and I'm going to call L1, the pipe that's along the coast. The cost for our one is 400,000 dollars per mm per kilometer. I'm actually gonna call Helen Keller dollars per kilometer, and L two is \$800,000 per km. And so we can make an equation for cost. That is C. Is equal to 400 times L1 and that's the length of the papal on the shore plus 800 don't you? And we want to determine the minimum of C. So we're actually, we're going to take a derivative eventually. But first we're going to try to get this all within regarding just a single uh variable. So we can take a derivative with respect to that variable. And in order to do this, we're actually going to use this X, which is the amount the distance um eastwards between the doctor and the refinery minus the distance of the pipeline, Because both L two and L one are quite easily related to this distance here, and that is L one here, L one plus x Is going to equal two km as you can see. And using Pythagoras, pythagoras theorem, you know, the L two is equal to the square root of X squared plus y squared. And because why is the sector is the component of lT that goes directly across the river? We know that, you know that why is two kilometers long? So we can say that L two is equal to the square X squared plus two square, which is for Now, we're gonna plug in these new equations for L one and L two And we get C is equal to 400 two minus X plus 800. And the square root of X squared plus four. And we could not take the derivative with respect to X. But we're going to simplify attorney at first, I'm going to call say it's um 800 -400 x. What? 800? This is going to be X squared plus four to the power of 1/2. Okay, now taking the derivative. So, well, that's true. The derivative of C with respect to X Going to equal to the derivative of 800 with respect to acts which is 0- the derivative of 400 x. With respect to X, which is -400. And then the derivative of this thing with respect to X. And I'm going to use a changeable for that. So first power is derivative with respect to X squared plus four of 800. That square plus four. One of the two. And the second part is the derivative with respect to X. So that's it. They live with respect to X of expert plus four. And this year is going to be just using power rule is 800 times the power which is 1/2 times expert plus four. And the power you subtract one from the power. So this is negative 1/2. Multiplied by the different in this respect with respect to X. Which is to its power with your ex. So this in total is equal to negative 400 plus 800 times a half times X squared plus four to the power of negative 1/2 times two X. And I'm gonna simplify this bit this um two And this one over to cancel our and I'm gonna put this in the bottom seeing as it's a negative exponents. So that's equal to negative 400 plus, I would say 800 x over square root of X squared plus four. And this is how do we live three of that was a bit fast. That should be all the steps that you eat though. So hopefully that is clear. Um Now when we're trying to minimize costs, we're going to find the critical point which is going to be when the derivative is equal to zero. So we're gonna solve for zero equal to this negative 400 plus 800 X over a square root of X squared plus four. We're going to multiply everything by spirit of X squared plus four so can be equal to negative 400. The spirit of X squared plus four 800 X. And I'm going to switch it to the other side. I'm also going to divide it by 400. So you can say good of X squared plus four equal to 800 X over 400. Now I'm going to great everything First. I'm actually going to simplify this. 800 over 400 is to over one. So this is Spirit Ex scrapers boy Is equal to two x for everything that's X squared Plus four is equal to four X squared. And then I'll move X squared to those sides. So it's four is equal to four X squared minus X squared we'll be done or is equal to three X squared for over three equals x squared and you can say X is equal to the square root of four or 3. So let's just step by step simplifying that and screw it. Or four with three is about 1.1 55. Nothing got back into the diagram. This length here is one point 155. And since we're looking for L well you know that X plus L one is equal to two kilometers. So I'm going to say L one plus 1.155 Deep of the two, And we can sell it for L one is equal to zero eight for right, So the Point P is located 0.845 km east out of the factory. I hope this was helpful. Thank you.

KS

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Derivatives

Differentiation

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##### Top Calculus 2 / BC Educators   ##### Michael J.

Idaho State University Lectures

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