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DM
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Problem 76 Hard Difficulty

A rain gutter is to be constructed from a metal sheet of width $ 30 cm $ by bending up one-third of the sheet on each side through an angle $ \theta $. How should $ \theta $ be chosen so that the gutter will carry the maximum amount of water?

Answer

$\theta=\frac{\pi}{3}$

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MN

Mikyla N.

June 16, 2021

6. Let f : R ? R be a differentiable function and P(a, b) a fixed point which is not on the graph of f. Suppose that Q(t0, f(t0)) is the point on the graph of f which is nearest to P. Show that the line through the points P and Q is perpendicular to the t

Video Transcript

So in this problem we're making rain guttering out of a 30 centimeter wide a piece of metal and we're bending up a third of the width on each side. And asked to figure out the angle theta that maximizes the amount of liquid that the gutter would carry. So what do we have? We have guttering. It looks something like this. Right? And we're looking for this angle feeder across here. Okay. We know that this is 10 cm. This is 10 cm. This is 10 cm. And so this height up here, right? This is some H heights. We're not sure what it is just yet. It will be dependent on that angle wanted. All right. So we know all the area inside of this trapezoid here. Right. We're trying to maximize that area to give us the most amount of water in this in this rain gutter. Well, we know that the area is Well this inter rectangle that's length times width. So that's 10 h. Plus we got two triangles, one on the left and right side. In a triangle is one half the base times the height. Right, Okay. Well, what do I know about this based down here and the height of this thing? Well, I know that in relation to data, I know that sign of data is going to be what going to be a church over 10 and I know that coastline data will be. Yeah. Um No, we know that the base with here, right would be or the coastline today would be The base over 10. All right. So that means this is 10 H plus well two times a half this cancels. And the base times the width will be what? Well the base is going to be tin Cosign data. That's 10 co sign data. And the height is going to be 10 science data. So that means I have 10 h plus 100 co signed data scientist. Um Okay. So if I'm trying to maximize this area then what I should do is take the derivative derivative And set it equal to zero to find the critical points. Right. Well this would be maximized Well the derivative of this. Well a derivative. Um What we forgot to put in here is that right here? Alright let's put let's use this relationship right here And let's put in here at times 10 sign of data. Okay. That means I have 100 signed data. So the derivative of sine data is cosign data. So this is 100 co sign data plus 100. Okay. So I have two functions of data multiply together. So I use the chain rule on the derivative. And when I do then I will get well the duality of co sign is is minus sign data. Time signed data. So it's my squared sine squared data. And then the driver of signed data is co sign. So this is going to be plus go sine squared data, isn't it? Okay. Now settle this equal to zero. So what do we have? We have factor out the 100 everywhere times. What do I have? I have coastline data minus science grid data. Well, that's one minus coastline square data, isn't it? And then plus co sine squared data. Okay, So that means now I have zero equals well if I divide by 100 on both sides so that 100 goes away then I have zero is equal to everything in the parentheses. Okay, So that means I have co sign data. Let's see minus one minus of minus. That's plus. That's plus cosine squared plus cosine squared. So this is plus two co sine squared data. So let me rewrite this for a second coastline squared data Plus Co Sign Data -1. Now this kind of looks like X squared plus x minus one, doesn't it? And so this factors This is supposed to be a two right here. Sorry. This factors into two. Cool Science data minus one times. Co signed data plus one. All right. So this means then that co sign data is a half or Cool. Science data is -1, doesn't it? Okay, well, we know that All of our data angles are less than pie between zero and pi over two because if I were to 90°. Right. Okay. So that means that that means that our our only solution is this one Because cosign theta is equal to -1. Right? Is this one. Right? is a -9/2 which doesn't work for us. All right. So what does that mean? Well, that means then We have co signing data is 1/2 which occurs when theta is pi over three and pi over three is inside Of this interval, right between zero and pi over two. Okay, So thus or maximum, which is a critical point or extreme? Is that theta equals pi over three in the area? At Pi over three. When you plug it into our formula above You get 75 times a skirt of three, which is approximately 129.9 centimetres squared for the volume of the gutter. The key point here is we found this angle theta that maximizes the volume in this guttering.

DM
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