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Suppose the refinery in Exercise 51 is located $ 1 km $ north of the river. Where should $ P $ be located?

4.87637 $\mathrm{km}$ east of the refinery

04:17

Wen Z.

01:25

Amrita B.

Calculus 1 / AB

Calculus 2 / BC

Chapter 4

Applications of Differentiation

Section 7

Optimization Problems

Derivatives

Differentiation

Volume

Missouri State University

Oregon State University

Harvey Mudd College

Boston College

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We're told to suppose the refinery and exercise 51 is located one kilometer north of the river and rest where the point P should be. So from exercise 51 we have this sort of set up where we have a refinery here which is one kilometre north of the river. Then we have our river which has a width of two kilometers not to scale here on the picture. Then we have some point p on this North bank on. And then we have a total of six kilometers. Yeah, away on the other shore, we have our tanks where we're storing material from the refinery. So we want to minimize the length of this path from the refinery to point P two tanks. Now this distance from the refinery horizontally from the refinery to the point P will call this X. This is going to be in kilometers. Yeah, yeah, yes, yes. So this length here is X and therefore this length here on the other side is six minus x. Yeah, you've done it? Sure. So using this graph and Pythagorean theorem for distance the pipe should cover. She'll call D is the square root of one plus x squared for X squared plus one plus the square root right, uh, six minus X squared plus two squared or four here. You you, That was Mm. And the cost function is a sum of both distances, each multiplied by its cost per kilometer. The cost function C as a function of X. This is 400,000 times the square root of X squared plus one plus and then the cost over the river or under the river is 800,000 per kilometer times the square root of expanding. This is X squared minus 12 X plus 40 the minimum cost. We take the derivative of the cost function and set it equal to zero. So see, prime of X. I'm not actually going to take the whole derivative here, but this simplifies to 400,000 X over the square root of X squared plus one plus 800,000 times X minus six over squared of X squared minus 12 X plus 40. Now, if we set this equal to zero to find the critical values, well, eventually we get mhm x times the square root of X squared minus 12 x plus 40 equals negative two times X minus six times the square root of X squared, plus four. If we square both sides, we get X squared times X squared minus 12 X plus 40. This is equal to four times X minus six squared times X squared plus one. To simplify this polynomial equation, you should get three x to the fourth minus 36 x cubed plus 108 x squared minus 48 x Yes, plus 144 equals zero. Now this is a core tech polynomial. You're probably going to use a solver or a graphing calculator to find the zeros, so we have to Real zeros that we care about X is about 7.1399 Notice, however, that our domain well, in our figure, it's implicit that X can only believe between zero and six. So we don't care about this zero. The other zero X that we find is 4.87637 which is insider domain. Now we use the first derivative test, So if you choose a point to the left of this say 4.8, you should find that C prime of 4.8 is approximately negative 20,000, which of course is less than zero. And that's C prime of a number greater than this. Let's say 4.9 is approximately 6386 which is greater than zero now. Since first relative changes from negative to positive, this means see changes from decreasing increasing. So it follows that C has a minimum at X approximately equal to 4.87637 they should 1000 and therefore the minimum cost is C at about 4.87637 Once again, if you use a calculator, can you plug this in? This is approximately three million, 826,000 358. I'm just going to leave it in this form. Obviously there's more and this is in dollars. And our answer is that. Point P is 4.87 six kilometers was east of the refinery

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