solve-the-given-problems-by-finding-the-appropriate-derivative-a-computer-is-programmed-to-inscrib-3

Question

Solve the given problems by finding the appropriate derivative.
A computer is programmed to inscribe a series of rectangles in the first quadrant under the curve of $y=e^{-x} .$ What is the area of the largest rectangle that can be inscribed?

Stanley Enemuo · Accepted Answer

Ah. So we need to find the largest area at the right time that can be inscribed in the first conference under the curve. Why equal to eat the negative? So to find the largest area of a rectangle, we need to find them just value of Y and the value of X. So what&#x27;s wise its maximum? So finding the first derivative of Y is equal to eat the negative back D. Y E. S. They were the X. Is equal to negative E. To the negative X. So salmon, D Y. D. X equal to zero. To find the critical points, we have negative Make this easter and negative activity equals zero.

Nathan Mankovich · Answer

because this questions asking us on the area under this curve in the first and second quadrant. So what are the first and second quadrants? He can&#x27;t remember. It&#x27;s just quadrant one. This is squadron too, right? So what does that mean about X and Y? So this is adding a couple constraints. This is saying that ex Well, this can be. There&#x27;s no constraints on axe, right from water constraints. Some Why? Well, why has bigger than zero right for being in the first and second quarter? It&#x27;s right bigger than equal to zero. We&#x27;re trying to maximize the area of a rectangle. So the area of this rectangle let&#x27;s see. So I really want to tell you that I already know what this function looks like. But let&#x27;s say you have some idea immediate looks something like this, then this rectangle that we&#x27;re drawing just gonna be like this, right? So what&#x27;s the base of the rectangle? Two x when the height of the rectangle, What&#x27;s going to be e to the minus X word? Gosh, I hate it when you run out of it in there. She&#x27;s going to fix it e to the minus X squared uh, there we go. Cool. Right. So then the area of a rectangle, it&#x27;s gonna be based on site these times, right? Which is gonna be the same as the base of two eggs and the height of each line sex squared. And if we&#x27;re trying to maximize this when you take its treatment of seven equals zero. So take the derivative used product rule. So it&#x27;s gonna be too eats in mind. Sex squared times the derivative of E to the minus X squared, which is minus well, two x days. Right, then, just to be really clear here, you know, have a minus here, general to x E to the minus expert. It&#x27;s not infect her out when we can divide out by to Berlitz, since equal to zero can cancel out it to you. Then we can factor on year than my sex squared. Then you subtract two x squared. So right, we&#x27;re factoring this out. So there&#x27;s a one here, right? Boom. And then you notice that then you notice this is bigger than your equal to zero lips other way around. This thing you didn&#x27;t My six squares never zero. Right. So you say this speaking or equal to zero. And then what is this thing? You go? Zero decided. Eagle of Zero. So now we have two x squared. Well, this one equal zero. I smoked by both sides by negative here. Does it really matter? Just makes it nicer. So then we have X because no, If we factor out to here two x keep doing a factoring here in minus what has where? I guess it&#x27;s 1/4 times, 1/4 times x plus 1/4. I kind of skipped two step here. But the ideas in factoring Ana too. So then this means X it is plus or minus. So are we done? No. Let&#x27;s go back up to here. OK, so now we do the first year of it of test because we&#x27;re looking for a max. Well, something to consider here is that Well, plus 1/4 and minus 1/4 are the same thing, right? Because minus 1/4 is this part from close 1/4? Is this part So that drop is drawing the same rectangle? Um, so we just need to be verified that this is a maximum, so I just planted on a number line 2 1/4 year, and I&#x27;m gonna look at the sign of a crime. It&#x27;s the left, one to the right. So if you click an X equals zero right, that&#x27;s gonna be I&#x27;m gonna actually just look, zero doesn&#x27;t really you restrict my ex values, cause Yeah, I think it actually isn&#x27;t important in this case, because, right, if a negative expert is the same as a positive X value here and what does that make sense? Because, um well, if X equals zero, you just have an area of zero. If X is negative 1/2 it&#x27;s the same as X equals positive 1/2. Right. So I think it can restrict this. I can&#x27;t restrict this. So now if we&#x27;re 1/8 year, then let&#x27;s look at the derivative from one ace a crime of one ace. It&#x27;s definitely well, this thing is definitely going to be smaller in this thing here, right? Ah, I did make a mistake. Let&#x27;s go back. Okay. This is important. So And the fact of the Saudis skipped a step, right? You talked about it too. Then you have a 1/2 here and 14 times 1/4 gives you 1 16 That should be one over the square root of two square two. I&#x27;m gonna just do this thing. Is the square one over two? Okay, so this should be one over the square with two here. So then we can actually check this side of our derivative at 1/4 which is going to be, well, it&#x27;s gonna be bigger than zero. So this is gonna be increasing and no check the son of the derivative. That may be one that makes this thing less than zero. This thing&#x27;s always bigger than Jewish, so it&#x27;s gonna be negative. Betweens were decreasing. This is indeed a max and the questions asking, What&#x27;s the area this maximum. So you just calculate the area of one over swears to, and that gives us well two times one over the square of two. Sure. Just barely have enough room here. So it&#x27;s well, two over the square does too. Times e to them, minus one over the square of two squared. So that&#x27;s minus 1/2. That should give us the area. What? Our function is maximized. I believe that&#x27;s what the question was asking. Just in case it was asking for the height. You have the base of the rectangle, so I&#x27;ll just add another page here. So the base of it. So the rectangle here is of that area. Very, uh, is one over the square, two area. This is the area. The base of this rectangle is too over the squared of two. So this thing and the height of the rectangle is e to the minus. Good house. Cool. So that&#x27;s it?

Matt Just · Answer

Okay, So a rectangle is going to be inscribed under the arc of the curves for co sign point by sex tears flying calls for co sign herepoint by X, and then a rectangle it&#x27;s gonna be inscribed. Write that, Miss Cecil B minus accidents will be X. Okay, so the area even imagine maximized the area. It&#x27;s gonna be two x of the base times the height, Which is why? And we can write. This&#x27;s a function of X by just plugging in what lies in trance of X to X times for co Sign of to your point five X And what&#x27;s the domain? Well, where is this policy? So yeah. So this is just going where I actually told this Its going from minus pi to pie. So the domain of dysfunction is just zero to play and its effect zero. The area is zero. And if X is Pai, why zero said the areas Awesome. Zero. So we know that the maximum has to occur at a critical point inside this interval. Okay, so let&#x27;s take the driven ample for spot. This is Hey, ex co sign of Honore. This&#x27;s except for two. So the derivative or B eight co sign XO over too. Minus four X Sign X over two. Okay. And then we want to solve a prime equals zero. And so see what we have. Eight Kasai necks or two equals four sign of X over tea. All right. I think there&#x27;s a text there, and so we&#x27;re gonna have a lot of trouble. Uh, solving this analytically. Hey, because well, I mean, you could drive. It&#x27;s This is not an easy equation too soft, analytically. I mean, you could divide by co sign and you get tangent Times X is equal to something, so we just we just want to use some sort of graphing ah system to solve it. And we see that there&#x27;s one expel you approach. This equation holds an ex is approximately two point two one four. Okay, And then so we called it. The dimensions are too X and which is about four point for two eight. So this is the base of the rectangle and heights. It is for papacy Fork Kasai NW of except for two. So, to your point, Teo on four number two, which is about one point seven nine. Okay, and then see. I mean, we actually want the largest area. So what is a two point two four? Well, it is about seven point nine to three. So here the dimensions and there&#x27;s the maximum area.

Regina Hays · Answer

Okay. We are starting with an arch that is represented by why is equal to four co sign of 1/2 X on the close interval from negative pie two pi. Um and we want to know the largest read tangle that can be inscribed. Um, and it&#x27;s the largest rectangle with the largest area. So the first thing we do, let&#x27;s go ahead and draw a sketch of what that curve looks like from negative pi to pie. So this is negative pie, and this is pie. And so we know at native pie and pie crosses the X axis, and it zero is up here at four. And so the graph actually will Look, um, something like that. And we want it. We want to inscribe a rectangle in which we have the largest area. I guess I should say the largest area, Um and we want to know the dimensions of that. And so if we arbitrarily lists, that is X, um, then this is going to be that high or that value is going to be that x value and, um, the y value that is represented by co sign of 1/2 X. Because if I want to know what that function is at that particular X, which happens to be the top. Most of the rectangle is going. I&#x27;m going to evaluate that function at the X value. Okay. And so we do know area area of a rectangle is given by the length of that rectangle, which is two x is two X and the height, which is four co sign of 1/2 X. So there is my length of my with, um and then we know that I have the domain eggs can only go from negative pi to pie. Um, that that is it. That is gonna be the value of X. And so if I we write this, this is eight x co sign of 1/2 X. And I also know, um, this is my domain. I also know that my range can basically only go from 0 to 4 until my range is just going to go from zero 24 So a couple things we had to think about, um and so now my take the derivative of the area with respect to X, and so this is a product rule. So this would be eight co sign of 1/2 X plus, um, four X. And it&#x27;s not gonna be a plus. It&#x27;s gonna be a minus. Um, because we&#x27;re to kick taking the derivative of co sign, um, of sign of 1/2 X. Okay. Now, um, and so I need we need to be able to determine, um, where the derivative equals zero. Um, and I&#x27;m actually going to utilize a calculator. Um, and so to help me out. And so I&#x27;m gonna go and and do a graph my derivative, and, um, used a calculator to help me out. Somebody go ahead. I had been doing something else on Dismas. So I&#x27;m gonna go ahead and put this in. I&#x27;m gonna actually graft my derivative and find out where it crosses the x axis. Um, and so we&#x27;re gonna dio It was eight. No sign of 0.5 X minus four eggs. Sign of with Dr X SOS. We have him somewhere in there. Um, I need to kind changed my graph, so I know I&#x27;m going from negative. Hi. Hi. And it looks like he&#x27;s pretty low. Um and so let&#x27;s just go from negative one one. Probably not, because to I have to go further than that. Let&#x27;s do, um, pre and three, which is not going to be probably eight. Okay, um, and then Okay, so here we have it. And so we know right here and right here. Or where it crosses the, um, x axis. So I do know where my derivative equals zero now, is that negative? 1.7 to 1. And positive 1.7 to 1. So, um, this is where my derivative equals zero. So where my derivative equals, zero X is a negative 1.7 to 1. And X is positive 1.7 to 1. So which makes sense because we wanted it. We would know we would have to have one on the negative side and the positive side. And so and I know that both of these will result in a maximum value because we already know what the graph looks like. The graph is calm. Cave down. Um, and so that means these would result a maximum, um, values. And so now, if I do why, and it doesn&#x27;t matter which one I use. So, um, so the length of my right tangle my length of my right tangle is gonna be twice as values. So the length of my right tangle is 2.61 and the height of the right tangle now is going to be, um, my function, my original function or my regional arch function evaluated at 1.7 to 1. And so when I do that, we get four times co sign of 1.7 to 1 divided by two, which is going to give me about 3.44 So the dimensions that will result in a maximum area is 2.61 and 3.44 and the maximum area is equal to 8.9784 um, square units and these air units, and there we have it.

Ankit Gupta · Answer

the area off a rectangle in describe in one article of the graph here Why you close to close six? Here is the figure given So now this is the graph equals to the area equals toe to x course off X, where X belongs to from zero to from zero to pi Divide with two is the idea and now we have to used this graph and find the idea fun and approximate the area of the largest in describe the neck tingled. So now that we have to find a Mexican value off this so the maximum value will be if here, if this value x course x is maxim and we know they&#x27;re the X courses, maximum value is 0.86 So that means area will be the area with a twice off ex classics so that it equals two multiplied with 0.86 equals two. One point went to this is the unity square. So united the anything that is so unit and it&#x27;s great. So this is the makes a media that is 1.12 units where

Heather Zimmers · Answer

all right, We&#x27;re using a graphing utility for this problem. And so for part A, we go into our y equals menu, and this is where we type in the area function that we were given and were asked to look at it on the interval from X equals zero to pi over two. So for my window dimensions, I let my ex minimum be zero and my ex Max Mumby high over to And then I knew the area wasn&#x27;t going to be terribly large. So for the y values, I went from negative 0.5 to 2. And here&#x27;s what the graph looks like and we can see that the area is growing and then it reaches the high point, and then it is going smaller. And so what we&#x27;re going to do is find that Max among area right here. So let&#x27;s go into the calculate menu and choose maximum. And then we want something to the left press. Enter something to the right press enter and then something in between, and we can see that the maximum area is right about 1.12 That&#x27;s the Y. Coordinate of the point. The maximum area occurs at an X value of about 0.0.86 but the area itself would be the Y. Coordinate 1.12 Now for Part B. We&#x27;re going to look for the values of X that give us an area greater than or equal to one. So let&#x27;s go back into y equals. And let&#x27;s let why to equal one that&#x27;s going to give us a horizontal line at a height of one. And let&#x27;s find these points of intersection so we can go into the calculate menu. Choose intersect. First curve, Yes, second curve, Yes, and then move that cursor over closer to the first point. And so we would say that between X equals 0.61 and whichever other number we get, we have an area greater than or equal to one. So let&#x27;s find the other number. Calculate intersect. First curve, second curve. And now let&#x27;s move that cursor over closer to the other point, and we get 1.10 basically. So between 0.61 and 1.10 the area is greater than or equal to one

Problem

Solve the given problems by finding the appropria…

Problem 42 Easy Difficulty

Solve the given problems by finding the appropriate derivative.
A computer is programmed to inscribe a series of rectangles in the first quadrant under the curve of $y=e^{-x} .$ What is the area of the largest rectangle that can be inscribed?

Answer

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Basic Technical Mathematics with Calculus

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Allyn J. Washington, Richard S. EvansBasic Technical Mathematics with Calculus

Catherine R.

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Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus