Industrial costs A power plant sits next to a...

Question

Industrial costs A power plant sits next to a river where the river is 800 ft wide. Laying a new cable from the plant to a location in the city 2 mi downstream on the opposite side costs $180 per foot across the river and $100 per foot along the land.

b. Generate a table of values to determine whether the least expensive location for point Q is less than 2000 ft or greater than 2000 ft from point P.

Accepted Answer

Welcome back everyone. In this problem, a telecommunications company needs to install a fiber optic cable from a central office to a new development 3 miles away on the opposite side of a lake. The lake is 1200 ft wide. The cost of laying the cable under the lake is $250 per foot and $150 per foot along the road. Determine the least expensive point for the cable to reach the development if it is less than or greater than 3000 ft from the starting point along the road. A says it's 600 ft, B 900 ft, C, 1200 ft, and D 3000 ft. Now let's see if we can visualize what's going on here, OK? So we're saying that we have a building, an office, office building. It's good right here. OK, that's 3 miles away on the opposite side of a lake of a new development that they want to run a fiber optic cable to. So if we can imagine our lake being here, OK. And this is where it's going to run the fiber optic cable across. Let me bring this over here just to make some space, OK. And then we have our development. OK, along which it's going to run and now in red, let's say that this is our fiber optic cable, so it runs across the lake, OK. And then it runs to the development, right? And we know that there's a road that runs along that, OK? So let's say that this is our road. Now for our dimensions or for the information that we have here, there are a few things that we already know. For starters, we know that the lake is 1200 ft wide. So on our diagram, we could show that 1200 ft here. OK. So let's take this length or this width and let's say that's 1200 ft. And as such, then let's say that the distance along the road, OK, where the cable enters the lake is the distance X. OK, this is the distance along the road. And now, since we know that the development is 3 miles away from the opposite side of the lake and 3 miles is 15,840 ft. Then after the cable leaves the lake and goes to the new development, that distance is going to be 15,840 minus X, OK, because we know, as we said that the entire distance is 3 miles or 15,840 ft. Next, we know the cost of laying the cable under the lake is $250 per foot, and the cost along the road is $150 per foot, and we want to determine the least expensive point for the cable to reach the development. Now for the cable to reach the development, we're going to have to, or to find a cost, we're going to have to multiply the amount of money per foot by the length of the cable for each scenario, OK? So in other words, The cost function The cast function. See of X, OK, where X represents the feet is gonna be equal to the cost to lay the cable under the lake, $250 multiplied by the length of the cable under the lake. Let's call that L. OK. Let's call that L. and let me label that on our diagram. Plus the cost delay along the road $150 multiplied by the length of the cable on the road which we found to be 15,840 minus X. Now how can we express the length of the cable L in terms of X? What can we do? Well, let's think about it here, OK. Notice that if we were to put it here on our diagram, our lake is not drawn the best way, but we, the cable under the lake forms the hypotenuse of a right triangle with one leg being the lake's width of 1200 ft and the other leg being x the distance along the road. So that means that the length, OK, by the Pythagorean theorem, the length of the k squared is going to be the sum of the squares of the two other sides of the Pythagorean or the of of the right triangle it forms which would be X2 + 1200 squared. Which means then that l is going to be equal to the square root of X2 plus 1200 square. Now that we have L in terms of X, this implies then that the cost function is going to be equal to 250 multiplied by the square root of X2 + 1200 square plus 150 multiplied by 15 + 840 minus X. This is going to be the cost to lay the cable overall. No, we want to find the least expensive point. In other words, we would like to minimize this function. So to find out, oh, let's try to take a few benchmark lengths for uh the length of our cable X or let's take a few. Benchmark lengths for a benchmark lengths for the value of X, since it's the variable we're minimizing on and see which of them can give us the least possible value, OK. So, for example. If we let X be equal to 300, then the cost to lay is gonna be equal to 250 multiplied by the square root of 300 2 + 1200 square plus 150 multiplied by 15,840 minus 300. I know if you put that in your calculator, you should get the cost to be equal to $2,640233 OK? That would be our cost. No, we could do the same with the rest of values, and I'll allow you to do that on your own. But if you go ahead and do that and let's say we were to make a table. Of X and our total cost CF X, OK. Uh let's extend this table here for the values, 300, 600, 900, 1200, 1500, and 1800. OK. Then by substituting these values, we know when X is 300, the cost is 2,640,233, when X is 600, you should get 2,621,410. When X is 900, you should get 2,616,000. When it's 1200, you should get 2,620,264, it's 1500, 2631, 234, and when it's 1800, 2,646, 833. Now what do you notice here? Well, by analyzing the data in the table, we can conclude that the minimum cost occurs at X equals 900. Thus, the least expensive point for the cable to reach the development is 900 ft from the starting point along the road, which satisfies our condition of being less than 3000 ft. If we look back at our answer choices, that means B is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Optimization

Distance Calculation

Cost Function

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