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A window is in the form of a rectangle surmounted by a semicircle. If the perimeter is $30 mathrm{~m}$, find the dimensions so that the greatest possible amount of ...

Question

A window is in the form of a rectangle surmounted by a semicircle. If the perimeter is $30 mathrm{~m}$, find the dimensions so that the greatest possible amount of light may be admitted.

A window is in the form of a rectangle surmounted by a semicircle. If the perimeter is $30 mathrm{~m}$, find the dimensions so that the greatest possible amount of light may be admitted.



Answers

A Norman window has the shape of a rectangle surmounted by a semicircle. If the perimeter of the window is 30 ft, express the area $ A $ of the window as a function of the width x of the window.

Yeah, okay, here's a norman window, it's a rectangle with a half circle on the top and we know that the perimeter is 30 ft and we want to find the dimensions of the window that lead in the most light. So that means we want to maximize the area. So we need a formula for area of this. It will be easy with the area of a rectangle plus area about half a half a circle. But we've got to name it cleverly. So let's Let's call the radius here to be x radius of the semicircle. So that means this side is two x. And then let's call these sides. Why? Since the vertical? And then if the radius is X, we need the perimeter of the circum half the circumference of a circle. So uh circumference is two pi R of a circle. So in this case circumference is two pi X but we only have half a circle, so one PX Okay, so then the perimeter is going to be this. Why? Plus pi X. That's why plus pi X plus Y plus two X. Okay, so that's our constraint constraint, Y plus pi X plus Y plus two X. Has to Equal 30. Okay, or two Y plus two plus pi X equals 30. And then we're trying to maximize area. So a equals well it's a rectangle. So it's areas linked times with so two X. Y plus the area of half a circle. So remember the area of the circle is pi r squared. So in our case pi X squared and we need half of that. So half the area is half pi X square. Okay, So to find the maximum of anything, you take the derivative and you set it equal to zero and then you solve and then you answer the question Okay, But we have a problem here in our problem is we have too many variables. We have X, Y and X squared. So we need to get one of them out of there and that's what the constraint is for. Okay, so now you decide do you want to solve it for Y or do you want to solve it for X? And then after you do that, which one do you want to plug in? Well, I'm solving for Y. I get to Y equals 30 minus two plus pi X. and so why is 15 -2 Plus pipe X oops. Two plus pi x oops. Mhm. Two plus pi X over two. Mhm. So now my area formula is two times X Times Y. Which is 15 -2 plus pi X over two plus one half by X square. So if I multiply two extra here, I get 30 X plus two plus pi X squared plus one half pi X squared. So I have 30 X plus two X squared plus pi X squared plus one half pi X squared. So one plus a half. That's three halves. So now I have 30 x Plus two x squared plus three halves, pi X squared. All right, That's a so now I'm gonna take the derivative with respect to X. So it's 30 plus for X. I've made a mistake. Right here, That's 30 x two plus pi X squared. And so that is 30 X minus two X squared -909ared. So that is 30 X minus two X squared -1 X half pi X squared. Excuse me? The way I knew that I made a mistake is because I was going to get all positives here and I wasn't gonna be able to set it equal to zero. Alright. Here we go again. Sorry about that. All right. Going to take the derivative derivative of 30 x. 30 minus for X -1/2 pie times two X. So 30 -4 x -9 X equals zero. So 30 equals 4 -9 times X. 04 plus pi So x needs to be 30 over. Okay, sorry four plus pi And then remember we had why was Why was 15 -2 plus pi X over two. So why is 15 two plus pi Over two times 30/4 plus pi. Okay that two cancels with that and give you 15. So now it's 15 times 1 -2 plus pi Over 4-plus pie. I don't know why I'm going crooked there. Okay, so um that's 15. Going to get a common denominator four plus pi That's four plus pi minus parentheses. T two plus pi Over four Plus pipe. So 15 times to over four plus by or 30/4 plus one. Hi so the dimensions are this Okay. X. is 30/4 plus pi. So make the base 60/4 plus by and then make the verticals. Whatever. Why was I forgot already? Um 30/4 plus. Why four Plus by Army. Okay. If you don't like that, change into decimals. Okay so the trick was first you have to get a good constraint and you had before that you had give everything good names. I could have called the base X and then the radius would have been one half X. And then I would have a bunch more fractions and clearly that would not have been good. Okay, so name everything correctly, write your constraint. Figure out what you're going to maximize takes to take the derivative saturday equal to zero. Super fun time.

Find the dimensions of a rectangle with perimeter 100 m whose area is as large as possible. Large. Okay, so that means we're trying to maximize something. A thing we're trying to maximize is the area of ah rectangle. Alright, so let's get us a rectangle gets, um, labels going here. Let's call it X and y So we're trying to maximize X times. Why? So to maximize anything, you take the derivative set it equal to zero. But we have too many variables here. So notice we have a constraint. And the constraint is that the perimeter of the rectangle has to be 100. So constraint perimeter, which because 100 and the perimeter of this rectangle is X plus y plus x plus y so two X plus two y equals 100. So now what you're going to do with the constraint is you're gonna either solve it for X or solve it for why gonna plug it in the area formula so that the area formula e house one variable in it. So let's just go ahead and solve it for why? So we get to y equals 100 minus two x bye bye to So why is 50 miners X So then our area formula in place of why we're gonna put 50 minus six area is X times 50 minus X, which is 50 x minus X squared. Alright. To maximize it, take the derivative said it equals zero. So 50 equals two x so X equals 25 and then why equals 50 minus x, which equals 25. All right, so what we just found out is that the rectangle of maximum area, whose perimeter is 100 is actually the square whose sides are 25. And so it's it, uh, you could see its perimeter is 100 then it's area is 625 or 25 squares.

Given that we have this rectangular window pane with the semi circle up top, we could calculate the total area of this and starting with the area of the rectangle on the bottom, notice that two X is the whole width and that makes X the radius for the semi circle up top because it's halfway across. But the area of the rectangle on the bottom would be two times x times. Why plus, and then the area of the circle up top a semi circle is pi r squared but then divided by two. And so it would be one half I and then our. In this case, the radius is ex soap I X squared. Okay, Next, what we want to do is take into account the amount of light that's coming through and the amount of light on the the in the top part. Let's start with is half of the amount of light that would come through on the bottom part so we could multi multiply this first one by basically a factor of one and then divide the second one by a factor of two, and those green parts will then take into account the light, Um, here as well. So let's just take a minute to rewrite that. So we have. Well, actually, I'm going to pause here because, um, we're about to take the derivative. And so well, let's just go ahead and do that too. X. Why? Because what I was about toe say before that is we want to then calculate the derivative of this to find a max. And to do that, we're going thio need to come up with the expression for the perimeter. So we'll go ahead and do that now. So the perimeter is just everything, of course, added around the outside. So that's two Why plus two x waas and then Thisted op part. So this is half of the circumference. The circumference is two pi r. So it's two pi r divided by two and the twos cancel. Let me just write that out too high. R r s X divided by two. Since it's a semi circle and that's where these twos then cancel. Okay, let's look back at the area. The area is in terms of X and y. We'd much prefer if it was only in terms of one variable and maybe Constance, of course. And so that's where we're going to solve for two y and the perimeter equate equation eso It's gonna be p minus. Let's get why by itself. So we're going to subtract two X and then he's go away so that we're gonna mind us Ty X and then all of that is equal to y. But then divide that to over and so it's going to be all of that divided by two. And that's what we're going to go out and replace that by with over there. So then we've got the area is equal to now two times x times. Okay, why is mhm minus two X minus two? Why, Nope. Sorry. Minus high X, all divided by two. Just leave it like that. All right. And then at the end of that, plus from 1/4 hi X squared. So just pausing here because this is kind of the important formula that we're getting to. This takes into account the area and the light that passes through, and this is what we want to maximize before we do, though, it's going to be a mess to derive this. So that's where we're just gonna go ahead and distribute this two X real fast first, and then we'll derive so two x p divided by two. That would just be X times p the perimeter and then subtract two x times Negative two x negative four X squared, divided by two That would just give us negative do expert and then two x times Negative pi X Divided by two that would get us negative pie x squared the twos canceled and then finally plus 1/4 Hi X Square. And what's nice about this is that these last two formulas our terms, I should say, are like terms. Yes, we could go ahead and combine a negative one with a positive 1/4 and those would combine to a negative 3/4 pie expert. Now we're ready to derive. And so let's go down just a little bit. I didn't get rid of these two just to make sure we don't think they're still part of the formula and see, Let's go ahead and use. Maybe. Well, let's just stick with black, actually, okay? A prime is equal to derivative with respect to X. Here is just going to leave us with a P minus four x and then at the end, What will that give us? That will give us negative Three halves Pi X. Okay, we want to set. That's equal to zero. And, um, we want to know. Okay, Basically, if we just imagine this is a line here, we know that it looks like it's gonna have a negative slope. And so we know somewhere it's going to go from positive to negative, which would tell us that it's a maximum since it's a crime, since it's the derivative graph. So that's good. So that should give us a max. And, of course, the step Before that, we had negative X squared. So we know that that formula for A is just a downwards parabola. So as much of that analysis is, we could do ahead of time. We know that we're moving in the right direction and that we will actually have ah, um a Max on this case. So let's just factor out a negative acts here. Let's go Negative X and four minus three pot 3/2 pi. Oh, except that would then turn to a plus because we factored that in negative, and that's equal to zero. So this will finally allow us to solve for what X is actually Because we could add that over and then divide both sides by the four plus three over to fine. And that is all now equal to X because we just added that whole expression year over to the other side and then divided. Okay, Finally, we could just rewrite that just a little bit. And so you might see this because I have looked up some other ways that people have kind of written this in and another form just multiply the top and bottom by two. So if you're thinking well, that doesn't look exactly like what I've seen. Maybe in the answer key or something you could also of course, right it multiply that by to multiply that by two. So that would be eight plus three. Pie three pie is equal to X. Okay, so X is now just a function of the perimeter, which is fixed, and so that is good to G O. And then if we then plug that into the UAE equation, which again we could use, um, the perimeter equation appear, for instance, to go ahead and plug that into. Then we can go ahead and sulfur. Why? So I'll just jump to that and that will give us why is equal to simplifying that would be pie. What's for that's p and all of that divided by six pie plus 16. And so now this is completely a function of whatever you want to make the perimeter. But this will maximize the amount of light that we're trying to let in here. Just a couple of checking. Find the proportions that will admit the most light. So those air it so that would be our Y value. That would be our X value, and that's it.

It's this problem. We're looking at a window specifically a Norman window, which is a rectangle topped with a semi circle. Now for this window, we wanna maximize the light coming through, so we want to maximize the area of our window. Our constraint is the perimeter. So let's put some variables on here and see what we can find out about our window. Let's let the hype B X and the width be. Why? Because the semi circle is sitting right on top of the rectangle. Thio radius is half of why or why over to. So let's look at our Doesn't matter which we look at first. Let's look at the constraint first. The constraints are perimeter. That's what we're told is fixed. And if I go around this window, my perimeter is two x plus. Why plus half of a circumference? A circumference is pi to pie are half of that is going to be pi r or in this case, pi times y over to There's my perimeter and we're told that this perimeter is going to equal 30. Now. What about our area? What's the area of the window? Well, the area is going to be the area of the rectangle, plus the area of that semi circle. The area of the rectangle is just length, times width or X y and the perimeter are I'm sorry. In the area of that semi circle is half of pi R squared or pie times. Why over two squared. And if I get rid of these parentheses by area is gonna be X y plus pi over eight. Why squared? Yeah, So what I want to do in order to maximize the area, I'm going to take the derivative of the area, but I have one step I need to do. First. I need to get the area in terms of a single variable I have right now both X's and y's. So let's see if I can get rid of the X. So come back over to our perimeter. Let's solve this equation for X. So I have two. X equals 30 minus. Why, minus pi over two. Why divide everything by two excess 15 minus one half? Why minus pi over four. Why? And I can plug that into my area equation, So that's going to give me Here's X times. Why, plus pi over eight y squared. Okay, let's before we dio the calculus, let's do a little bit of algebra. We get rid of my parentheses, clean it up a little bit, the needle we make this piece, the easier the derivative is to find. So I'm going to get 15. Why? Minus one half y squared minus pi over four wives squared plus pi over eight y squared and I can combine those last two terms. I could give them a common denominator and I get 15. Why? Minus one half life squared. And if I do a common denominators gonna be eight So this first one's gonna be two pi over eight. I could just fix it like that, which means I'm gonna have minus pi over eight y squared. Okay, Now let's take our derivative Derivative is going to be 15 minus. Will bring that to down as we minus. Why minus pi over, uh, over four consent. If I that a little bit pi over four. Why now? In order to find the maximized, the optimal solution here to find the maximum area. I am going to take this derivative and said it equal to zero. And I'm just gonna come down just a little bit where I've got a little more room. So setting this equal to zero, this is equivalent to I'm gonna push the wise to one side and keep the constant on the other. I have Why, plus pi over four. Why? Okay, Factor out a y from both of these, and I can simplify. So this is why equals 15 divided by one plus pi over four. And if I put this thio into a calculator to getting approximation, I get why equals approximately 8.401 Okay, so now what is my value for X? And I'm just gonna come over here onto the side where I've got some more room and I'm going to plug in that value of why? Into my boxed equation for X, so X equals 15 minus one half. Why? Minus pi over four. Why squared? And that gives me a value of X. It's approximately 4.201 And those dimensions are going to give me my maximum area. If I plugged that all the way back into my area equation, I'll come back up here to remind you here's our area equation. right here. Plugging those values in gives me an area of 63.8 square feet. So these dimensions are going to give me my maximum area.


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