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A steel pipe is being carried down a hallway $ 9 ft $ wide. At the end of the hall there is a right-angled turn into a narrower hallway $ 6 ft $ wide. What is the length of the longest pipe that can be carried horizontal around the corner?

3$\left(3^{2 / 3}+2^{2 / 3}\right)^{3 / 2} \approx 21.07 \mathrm{ft}$

04:11

Wen Z.

01:23

Amrita B.

Calculus 1 / AB

Calculus 2 / BC

Chapter 4

Applications of Differentiation

Section 7

Optimization Problems

Derivatives

Differentiation

Volume

University of Michigan - Ann Arbor

University of Nottingham

Idaho State University

Lectures

01:11

In mathematics, integratio…

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In grammar, determiners ar…

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Turning a Corner A steel p…

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For the following exercise…

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Okay, We have a steel pipe, and we're carrying it through building. And we're in a 9 ft hall and we make a turn into a 6 ft hall. And the question is, what's the longest pull that we could get around this corner? Um, if we were carrying it horizontally. All right, so we're trying to find the longest poll, so that means we're trying to maximize the length of the pole. All right, So since we have two things going on here, I'm gonna go ahead and extend that line and that line so that I have two triangles, so I'm going to call this one. Let's call this one l one. And this one l two. So we're trying to maximize the length, which is l one plus l two. So now we're gonna find some relationships between them, and I better have something to do with this angle theta. All right, so notice that we have two parallel lines cut by a trans verceles corresponding angles or congruent. So if that one is data, so is this one okay? Also, we have right angles here because thes air Ah, horizontal and vertical. And I know this is nine. And I know this is six. Okay? Because that's the length of the hallway. So in this first triangle right here, Ivan angle the opposite side and the high pot news. So I know that the sign of theta is opposite, which is nine over the high pot News l one. So L one is nine over the sign of data or nine times the co second of data. And then in this triangle appear I have an angle the adjacent side and the hypotenuse. So I have the cosine of data equal six over l two, or, um, sure did. Sorry. Well moving dribble there. H l two equals six over the cosine, which is six times the sequence of data. Okay, so now we have length equals L one, which is nine. Cosi can't data plus l two, which is six Seacon theta. So to find the maximum A, we need to take the derivative, set it equal to zero, and then we could answer the question. All right, here we go. Taking the derivative with respect to theta. So it's nine. The derivative of the co sequent is negative. Kosik in data co tangent data plus six times the derivative of the Sikh End, which is seeking data tangent data. And we want to set that equal to zero. Okay, I don't know about you, but I don't know anything very quickly about Cosi Kitco tangent seeking or tangent. I'm gonna change everything to sine and cosine here. Is that negative and nine Coast ticket one over the sine co tangent co sign over sign plus six times, one over the cosine times the tangent. Oh, I totally did those. Oh, no, that's backwards. That's right. Sign over. Go sign equals zero. Okay, so this one is nine. Cosign data over sine squared. Data equals six. Signed data over coastline square data. Um, cross multiply nine co signed Cube data equal six. Signed cube data can divide the six over here, divide the coastline cubed over to the other side. And the reason I know to do that is because I want to talk about tangents, not co tangents. So 96 that's three halves equals the tangent cube of theta. So then, um, I need thio. Take the cube. Er que brute of three halves equals the tangent of theta. Okay, so here's a picture of theta now opposite side cube root of three adjacent side Q Bert of To can remember our formula for the length Waas um nine co Seacon data plus six second data so we don't actually have to find out what data is. We'll just use this picture. It's nine times the co sequent. Oh, I'm gonna find the other side here. Um que brute of two squared plus Q brute of three squared square roots. Uh, que brute of four plus Cupid of nine. Ah, yeah, I'm not too crazy about this. I'm not going to do it this way. I've decided to get out my calculator because this is too messy. It's not worth my time to do it that way because there will be a lot of algebra and simplifying. Alright, so I've got the cube root of three houses, the tangent. So I'm gonna inverse tangent that on my calculator Inverse tangent que brute of three house that will be Thatta Can't remember your calculus so you gotta be doing it in Radiance three. Divide by two. Que brute! Oops! Three. Divide by two Q brute inverse tangent. I get data is 0.85 to 77 So now I got to Dio, because the question is, find the length I've got to do nine times the Corsicans of 0.85 to 77 plus six times the 2nd 60.85 to +77 which is nine divided by the sine because I don't have a co second button plus six divided by the cosine of 0.85 to 77 All right, let's see if I can get no 0.85 to 77 The sign Divide by nine. Reciprocal. 11.95 plus 0.85 to 77 Cosine divided by six. Reciprocal. So I have to do mine upside down and then do the reciprocal because I'm using my calculator on my phone. Okay, now I'm gonna add those together. 21.7 ft is the longest pipe that I can carry around that corner

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