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Find the dimensions 0l the open rectangular box of maximum volurne that can be made comers and folding up Ihe sides. Then find the volumte sheet of cardboard 19i by...

Question

Find the dimensions 0l the open rectangular box of maximum volurne that can be made comers and folding up Ihe sides. Then find the volumte sheet of cardboard 19i by 11 in by cutting congruent squares from The dimensions box of maximum volumo (Round ta Ihe nearest hundredth needed. Use commz ~cpaiale ans Bl? needed ) The maximum volume (Round the nearest hundredth a3 needed )

Find the dimensions 0l the open rectangular box of maximum volurne that can be made comers and folding up Ihe sides. Then find the volumte sheet of cardboard 19i by 11 in by cutting congruent squares from The dimensions box of maximum volumo (Round ta Ihe nearest hundredth needed. Use commz ~cpaiale ans Bl? needed ) The maximum volume (Round the nearest hundredth a3 needed )



Answers

Maximizing the Volume of a Box An open box with a square base is to be made from a square piece of cardboard that measures $12 \mathrm{~cm}$ on each side. A square will be cut out from each corner of the cardboard and the sides will be turned up to form the box. Find the dimensions that yield the maximum volume.

So this problem is your me similar to the one from before and that we're going to be maximizing the volume for a box. But this time the difference is that Um it's gonna be used using rectangular piece of cardboard. So we're going to have one side that's 30" long, meaning we're going to subtract Um two X. From it as well and then the other side is going to be 20" wide. So we're gonna have this and the high is still going to be X. So most of it's the same, it's just the dimensions that change slightly. Um So we want to find the dimensions of the box that give maximum volume now. So we're going to zoom out and see where we can maximize volume which is located right here, Maximizing volume occurs when X is equal to about 3.9- four. And at that point the volume is going to be 1,056.306. So with having maximum value in there, that's going to be our final answer.

All right in this problem 40. And we have a rectangle, a carport. And also we are cutting out a square whose side lengths these x from each corner of this rectangular cardboard. And then we're gonna try to figure out the largest possible volume of the new newly created shape. The first of all, let's try to Skase the situation. So we have this rectangular cardboard, as you can see. Okay. And I would like to cut out a square from the each corner with side length X. Okay, as you can see. Okay, so that's why the new dimensions after this cut out, the new measures will be X Ah five minus two x and also ate minus two X. Because we know that this entire length was equal to eight and this entire with was equal to five. Okay, so the volume of this shape if I multiply them X Times five minus two x times eight minus two weeks. And I would like to maximise this expression. Mike Smyth, this bomb. If I would like to maximize his volume, I need to focus on the Primex and set of equals here. But before that, I would like to simplify. Let's try to simplify. If I distribute X into the Prentice's, we're going to get five X minus two X squared times eight miles to x and let's try to multiply these two binomial Using the foil, we get 40 X and minus 10 X Square and minus 16 X squared and plus four X. Okay, so let's try to simplify. Let's try to combine like terms so we get 40 X and minus 20 X squared and plus for excuse. And that's gonna be equal to volley function. And I would like to maximize his volume function. So we need to figure out what is the first derivative of this volume function and set of the equals zero. So let's do that. So 40 minus 52 X plus 12 x squared, it's gonna be equal to zero. So we ended up with quadratic function, so if I try to divide each expression by four, we get that three X squared minus 13 X plus 10 is equal to zero. And if you tried to solve this quadratic equation for X, we're going to come up with X equals one and also X equals 10/3. Okay, so this X equals 10/3 cannot be the S because up because it produces negative volume. So when you said when you take this executes 10/3 and substantive back into this volume function, it's gonna produce a negative number for about, you know, that volume cannot be negative. But But if right when it comes to using X equals one, it's gonna produce a positive evolving bound. So let's try Teoh. Set this. Let's try to substance X equals one into the volume function. So we get V of one is going through my large possible volume off this newly created shape, which is 40 times one minus 20 times one squared, plus four times one cube. The result is gonna be cool to just 40 minus 26 plus four. Answer will be 18 feet. Cute. Okay, for the first problem, our four part B. In this case, we have a square cardboard, as you can see whose side lengths is X Great. Yeah. Excuse sidelines is l. Okay, so in this case again, we're trying to cut out as square from each corner whose side length is X. As you can see x x x x as this. We know that this is a square cardboard whose length who's one of the side length is out. So basically each of these dimensions are going to be equal to, um, this new dimension will be l minus two x and this new dimension will be l minus two x and the volume of the newly created shape will be x times l minus two x times l minus two x and I would like to I would like to again maximize this volume with the given the mashes. Okay, so let me use different colors. So we get if I distribute this X into apprentices, we get, um x Times l minus two x squared l minus two x So let's small to play thestreet by no means using foil. So we get X times L square minus two x squared minus two x squared l and plus four x cubed. Okay, Are in this case, if I would like to find the maximized volume that I need to focus on this the bottom I need to take the first urban of and set of equal to zoom. So when I do that we get. So we get, um, 12 x squared 12 X squared minus eight. L X and plus L squared is equal to yours. As you can see that this is a quadratic function. If I would like to find the solution of this quadratic function, I can apply the quadratic formula that is ab scribble to all kinds of quadratic functions. So let's do that. So actually gonna be a culture negative? Negative eight l plus minus under the square it. So we get negative. H l squared minus four times 12 times l squared and the bottom by two times 12 which is 24. Let's try to simplify this expression. So we get a Dell plus minus 64 l squared six for l squared minus 48 l squared under the square, divided by 24. And from there we get, you know, plus minus 64 minus 48 with 16 and we're gonna stick to the L Square under the square with divided by 24. So a dell plus minus this 16 else scored is going to come out come out of this radical form as four l over 24 okay. And from there. If I sold situation for l actually is gonna be called either l over six and X equals Al over to. So from there, if I stop stewed l over six into the volume formula, remember that that the volume is gonna be equal to AL over six times l minus two times AL over six times l minus two times L over six. Okay, so let's try to simplify this expression. So we get al over six times, uh, to al over three times to El over three. And once you multiply all of these three things together, we're going to get the volume. Actually, the largest possible volume, uh, will be four times l tripped over 27.

We know that we have a box open at the top that's made from a square piece of cardboard and we're going to cut out the corner which will be length X. Um so we know that the cardboard measures 40 cm on each side and we want to find dimensions that give maximum volume. So you see volume is equal to length width times height. Where we see that the length Is going to be 40 cm minus the side X minus the other side. X. That's giving minus two X. The length and the width of the same since this is a square base. So we're going to have again 40 -2x. And then the height when we fold up aside, we know that they're going to be a height of X. So we do that and now we want to maximize the volume. So that's going to occur right about here. Let's figure this out by letting XP smaller values so maybe negative 10 positive. Can we do this? And we end up seeing that the maximum will occur at about 6.667 centimetres as our X value

I need to minimize the volume of this box. How? Volume is length, times with times height and I have a restriction. This lidless box can Onley use two square meters of cardboard. So the surface area of my box is gonna be the base. No top because it's lidless and then two sides. Why Times e and two sides X times see? So if I take that equation and I solve for Z, that gives me too minus X y over two X plus two. Why? And I'm a substitute that back into my volume function which will give me volume in terms of two variables instead of three. So if I plug that value in frizzy and get rid of any parentheses, what I have is to x y minus x squared y squared over two. Why plus two x Okay. Now, to maximize my volume, the maximum value will happen at a critical point. So I need to take the partial derivatives with respect to X and why ex first, If I take this partial derivative, I get to why. Plus two x times to why minus two x y squared minus two x y minus X squared. Why squared times two over to why plus two x quantity squared. And I am going to simplify that. I'm gonna get rid of parentheses combined as much as I can, and what I end up with is two y squared times two minus two x y minus X squared over two. Why plus two x quantity squared. Okay, now on to the partial derivative with respect Toe. Why, If I do that, that gives me to why. Plus two x times two X minus two X squared. Why minus two x y minus X squared y squared times two all over that same denominator to why plus two x quantity squared. And with just like the previous one, I'm going to simplify as much as I can. I'm going to expand things out, get rid of parentheses, factor out common factors. And what I end up with is two x squared times two minus two x y minus y squared over to why plus two X quantities squared. So I need to find out where this thes partial derivatives equals zero. Well, I could say X equals zero y equals zero that works, but that means I have a collapsed box. So for this particular problem, that is not going to give me a solution I need So let me look at the other factors. I want two X Y Yeah, who is? I'm sorry. Let me erase that for a second. I'm coming to look right here and right here. There's so many letters. Let's just make sure we've got the right ones. I want tu minus two x y minus X squared to be zero and two minus two x y minus y squared to be zero, which means that X squared equals y squared or X equals. Why I don't have to worry about positives or negatives. This is a actual box. Both e values will be positive. So if I plug in XY Quinn, why into this equation? What I get is too minus two x squared minus X squared equals zero. That's putting it into this equation right there. Tu minus three X squared equals zero or X equals the square root of 2/3. And Aiken rationalize that square root of 6/3. Well, since why an X are equal, they put this over here. I've got a little bit of rum. There's my ex. There's my why. And if I go back up to Z this formula right here that shows how X and y and Z related. If I plug in those values for X and Y, my Z turns out to be square root of 6/6. So those are my dimensions. I have a base that's a square. Ah, square root of 6/3 on either side and a height a square root of 6/6 that maximizes my volume.


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