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5 . We are designing flier with 50 square inches of printed text, surrounded by 4-inch margins on top and bottom, and 2-inch margins on the left and right sides. (S...

Question

5 . We are designing flier with 50 square inches of printed text, surrounded by 4-inch margins on top and bottom, and 2-inch margins on the left and right sides. (See the picture on the next page: Find the dimensions of the flier that minimize the amount of paper used.

5 . We are designing flier with 50 square inches of printed text, surrounded by 4-inch margins on top and bottom, and 2-inch margins on the left and right sides. (See the picture on the next page: Find the dimensions of the flier that minimize the amount of paper used.



Answers

A rectangular page is to contain 50 square inches of print. The page has to have a 1 -inch margin on top and at the bottom and a $\frac{1}{2}$ -inch margin on each side (see Fig. 16 ). Find the dimensions of the page that minimize the amount of paper used.

Okay, so we have a, um, rectangular poster. So we have a right tank of the poster. Um, that has a total area of 50 inches squared. And, um, what we have is we want to, in case, um, that 50 inch squared. Um, we want a print it with a four inch margin, um, at the top and bottom. So we're gonna have a four inch margin at the top and bottom. Well, if I could draw a straight line, that would be perfect. Um, And so, um, this right here is four inches, and this right here is four inches and a two inch margin. Um, at the left and right. So Well, these are two inches. Okay, Um And so what we want to do is, um what are the overall dimensions? Um, to minimize. So we want to know what the overall dimensions are to minimize, um, the amount of paper used. Okay. And so, um, what we know is that, um we have this. If we label this X and this why we know that that 50 inches squared is X times y. And we also know that the new area the new area is going to be X plus four. Because from here from here now to here is the X Plus four. And so that new, um, length is X plus four, which means now that, um, this new hi is why plus eight. And so this is why plus eight. And you know, any time we want to minimize or maximize that is implying to take the derivative. And so we cannot take the derivative at this point with two variables. Um, my area formula dependent upon two variables. So what I need to do that's where this one comes in handy. I need to solve for one of my variables. And so I may say, why is equal to 50 over X, And so wherever there's a Y in here, I'm going to substitute 50 overact, so this becomes X plus four times 50 over X plus eight. Now, I'm gonna be a little lazy, have to be honest with you, And I would rather, instead of using the product rule for the derivative, Um, because eventually I'm gonna have to multiply all out. Anyways, I'm gonna go ahead and and multiply this out. And so this is gonna be 50 plus eight x plus 200 over X plus 32. Um, which is going to simplify too? So that new area formula is simplifies down to 82 plus eight x plus 200 over x. And now I'm set up to, um Now I'm set up to Excuse me, Teoh, take the derivative. Um, now, also, what I probably need to go back in here is also put some I know that, um X Husby bigger than zero. And why has to be bigger than zero? Who's, um I think those were the only stipulations right now. And so the derivative of that new area with respect to X is eight minus 200 over X squared. And I might go ahead and put this over a common denominator over X squared. And so I know the eggs now could not equal zero. And so now we want to take we want to know we're that derivative does actually equals zero. And so we know for a fraction. That's typically means where that numerator goes to zero. So I'm gonna set the numerator equal to zero, and I'm gonna solve for X. And so X is going to be equal to five inches now. When I take the square root of both sides, I should have a plus or minus. But we're talking about a real life poster, and so we know that's not gonna be a negative inches, and so X is five inches. And so, if X is five inches now, I can go back to my poster, um, and be able to determine. And so that means also tell me that if X is five inches, why is 10 inches? But if I go back, I want to know the overall dimensions to minimize the paper use. So that means that this is going to be a five but five plus four because X was five and I have four inches. So this is going to be a nine inch by an 18 inch um poster, which will minimize the amount of paper I would need to use

This problem, we're too. Minimize the amount of paper used for a poster where we have 50 square inches of printed area boundary of four inches on the top and bottom and two inches on side. Well, the area in the middle here will be our printed area. Would let one side B X and the other Why link times with with B 50. Now that's just for the inside box. Now we've got to take into account the border. So this border, what that be the wit and this be the link. Well, as we look at the inside as well, this length right here would be represented as line. This would be for work in this one before. And then you think about X. That would be X with two on this side and two on this side. So the link l was why waas eight. The wit w would be X plus four and the areas of the poster link times with and well, we have it in terms of X. And why weaken soul for why Here in the given equation, why would he for 15 divided by X and so our area in terms of X is going to be. Why 50 over X waas eight times quantity exports forward multiplying down, out and we'll get 50 waas 200 over X plus he'd x plus 32 so a X will be 82 waas 200. Let's go ahead and write X to the negative one Waas a Dex. Then we take the derivative to maximize. That will be 200 times negative one x to the negative too. Plus eight we'll have negative 200 over x Weird plus eight equals zero. I had 200 over x squared to both sides, divided by eight multiplied by x Weird and take this route There will be 200 divided by eight is 25. So exes X squared is 25. So ex get their X squared will be 25 so x will be five. And why will be 10? And what we really want is the length and the whip. The length is why plus eight, Why being 10 we'll have the length being 18 inches and the whip being X plus four x was five. So with would be nine inches

All right. So let's go ahead and solve this problem. Now the question asks for right, What is the minimum minimum amount of paper you can use? And so here is your piece of paper right here. Um, with the blacks being, um, its borders and then the green here represents, um the 50 square inches of printing, right, Uh, four inch margins on top and the bottom and two inch margins on both the left and the right side. And so when you're looking to when you're looking to solve for what the minimum amount of papers use, what you're really asking is what is the minimum right area that you can have, uh, of paper that you can have. And you have to remember that the area is equal to the length times the width right of the paper which correspond to these right here, right the length. And so the trick to this problem is unique. We need to come up with equations. Um, that can give us an actual area where we can find the derivative of the area, uh, and therefore find the minimum and the maximum of that function, and therefore find the solution to the problem. And so the first step here, All right? And so I'm going again. I'm gonna put this as step one. Okay, I'm going to say and step one that as we know, L w right. This is little L Times little, W, which is the, um the length and the width of the portion of the paper with the print on it equals 50. Okay. And then from here, I'm going to say that, uh, l equals 50. Divided by W, which is equal to 50 to the negative first. Okay. And then therefore Right. Um uh, from here, I can say for the actual paper, write a piece of paper using l right is equal to right the length of the paper. Plus, you see, I have four inch margins right up top and bottom. It's l plus a total of eight, right? And then the width of the paper Big W is equal to right. A little w right. Plus. And because you have two inch margins on both the left and the right, that is equal to four. Okay, Now, from there. Okay, so now we can go to step two, so I'm gonna put Step two right over here. Okay. For step two. Okay, Um, let's go ahead and set up an equation for your area of the paper, right? If the area is equal to right big at all times, big w link times with, that's all you can say that the area is then equal to right, plus eight times w plus for now. Okay, We want to get this to where we have a single variable. Now you have to remember from here. Okay, um, this equation says l equals 50 uh, minus 50 w to the negative first. And so I'm going to replace l with this right here. And so therefore, I get hey equals 50 w attempts to the negative first, plus a times w plus for Okay. Now from here, then I can expand this right here. I can expand this whole thing and bring it down to okay. A is equal to, um 50 right? Plus 200 w, yeah. To the negative first. Okay. Mhm plus eight w plus 32. Okay, if I simplify this if I put the 50 and the 32 together, okay, I get a equals. Okay. And here's the now an equation for my area using a single variable 50 or sorry, I'm going to go back there. I'm gonna put 50 and 32 together, and I get 82 plus 200 W to the negative. First plus eight W Okay, there's my equation for the area. Okay? And therefore, now we can go to step three, okay? Where we can get the derivative of the area, okay? And then find the minimum value for W. Or the value for W Sahara is the minimum. So a prime Now. Okay, I'm gonna I think it's a little bit clearer. A prime is equal to remember. This is a constant and therefore that this is zero. This becomes negative 200 w the negative second plus eight. And in order to make it, uh, in order for it to be at a minimum, the derivative has to equal zero to set a primary zero. And therefore I get, um, negative. 200. They'll be negative. Two plus eight equals zero. And then from there, I can say eight is equal to 200 over w squared. And if I do the algebra there, therefore Okay, I get W is equal to five. Okay, now, to check my work, okay. To make sure that, uh, this is actually a relative Max. Okay, I can go up here and say a double prime right is equal to Yeah, negative or positive. Sorry. 400 x to the negative three. Mhm. Sorry. And therefore, this is always right. Greater or less than greater than zero. Right. Therefore, the graph is at a concave down, and then this must be a relative. Men. Okay, birth of minimum. So this is a relative minimum. So now from here, right, I can say therefore, great. And now we're going into step for which is the final step. Okay, Um, all right, Big w is equal to right little w plus four and therefore is equal to five. Uh huh. Five past four, which is equal to nine. Okay. And then l big l rate is equal to l plus eight, which is equal to 50 over a little W plus eight, which is equal to 50 over five since w equals five from here, plus eight equals 18. And so your answer is for your dimensions. Big w equals nine, and the length is equal to 18. And this is both and inches

The following problem, there's a printed page with two inch margins of white space on the sides and one inch margins of white space on the top and bottom. So we know the area of the printed portion is 32 and just squared. So we want to find the dimensions of the page that the least amount of paper is going to be used. Um So what we have is that in this case since there's two of margins on the sides. Mhm. Um that's going to be 4" on the side, you can be 2" top and bottom combined. We know that the printed area is going to be 32 and just squared though. So we know that the length, whatever the length of the printed pages plus um Top and bottom is going to end up giving us too. So l. Plus two Times W. Plus four. That's can you give us the area of the page? But we know that the area of the printed portion is going to be 32. So we have 32 equals l. W. So we're going to solve for one of these W. And then we can and put it into the other portion


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