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Sheet of paper 68 cm-by-80 = cm made into an open bor (i.e: there'$ no top). by cutting x-cm squares out of cach corner and folding up the sides. Find the valu...

Question

Sheet of paper 68 cm-by-80 = cm made into an open bor (i.e: there'$ no top). by cutting x-cm squares out of cach corner and folding up the sides. Find the value of x that maximizes the volume of the box: Give your answer in the simplified rndienl forn:the

sheet of paper 68 cm-by-80 = cm made into an open bor (i.e: there'$ no top). by cutting x-cm squares out of cach corner and folding up the sides. Find the value of x that maximizes the volume of the box: Give your answer in the simplified rndienl forn: the



Answers

A box with no top is to be built by taking a $6 "$ -by- $10^{\prime \prime}$ sheet of cardboard and cutting $x$ -in. squares out of each corner and folding up the sides. Find the value of $x$ that maximizes the volume of the box.

So I'm gonna use a flat sheet of material to make a box. And to do that, I'm gonna cut out squares from the corners so as it would be folded up, it's gonna look like this. So, like, this side right here is this side pulled it up into space, and it is open top, which is good. That's what we're going for it. So with that figure in mind now, my goal is to find a volume that is maximized. The sheet of material that we're starting with is originally 12 by 16. So all the way end end 12 by 16 and then we're cutting out squares. So I know that the side of the square is gonna be the same eight seconds. Call it X. And it would be like XX on each of those little squares. The formula for volume is length, times with times height, and I can't go through and maximize that until I know just one variable here for length, for with for height. So Michael actually would be toe label. Are images with what to left behind. Like what is this length on the final product? What is this? With what is this height, so the length on the final product comes from this part right here. That's remaining well, if it's the stuff remaining. It was originally 16 but we took away two of those squares, so it's 16 minus two X, each of those squares not cut out. And then the wit is this part here where it was originally 12. But two squares got cut out. The height might be the toughest one to see. It's when it gets folded up. So notice when he gets folded up. Like if I just focus on this one side with the width, look at the pieces that are facing up with the X tall. So actually the square determined how tall the final product is for the box. So if I feel in that volume with length and with and height, then I just have one variable. Before I go through and take the derivative for this equation, I'd like to rewrite it, so I'm going to go ahead and expand the whole thing and just get it, like, kind of in a polynomial form, instead of how it is right now, with the factors so like 16 times 12 is 1 92 the negative 24 x negative, 32 x And then I still and I have a plus four x squared. Well, almost forgot that. And then I still have an X on the outside. So then combining like terms and bringing in that exits 1 92 x minus. This is 56 x squared plus for execute. Now that it's in a simpler form, it's a great point to take that first derivative. So this is just gonna be power rule for the derivatives. Here, this is 1 12 X and then just becomes 12 x squared. Now, I'm not feeling really confident about this, factoring out nicely. So I'm going to graph this equation for the first or evidence and then see where this graph cross zero because that's going to tell me where the critical numbers are because I want that first derivative T equals zero. So I'm gonna type this and as a graph and see where it crossed zero to help me find for those critical numbers are. So I found that those excess happened at 2.263 and X equal 7.70 now to see which of these maximizes the volume. I'm going to check the first derivative. I was gonna do a first derivative test here, using the prime and picking some convenient numbers around those critical numbers. So, for instance, if I pick zero, then the result of the first derivative would be 1 92 which is positive now. If I plug in three as the expel you, then I get a really big number. We make sure you didn't type in anything wrong. I did type in something wrong. Let me try that one more time. Here we go, still with number, but it's negative. It's negative. 36. And then if I plug in something like a eat to that first derivative, then I get 64 which is a positive number. The first derivative test helps me see the shape of the graph, where it's increasing and decreasing, so because it's increasing to decreasing on this first critical number, this is creating a maximum value, and I want the maximum volume. So if I were to say you know what, what measurement do I need to cut out of the squares? It would be this first critical number. If you want to maximize the volume, need to cut out a square that is approximately 2.263 inches for all four corners of this box.

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.

The solution to the question is length and breadth of her rectangular sheet is 45 centimeter and 24 centimeter respectively. Not considered. X centimeter be the side of each of the four squares cut off from each corner. So the dimensions of the open box formed by folding flaps. After getting office squares are 45 minus two weeks into 24 minus two X. And X centimeter. Now, consider said the north the volume for the open box. So the volume will be 45 minus two weeks into 24 minus two weeks in two X. So it will be equal to four X cubed minus 1 38 X square +1080 X. Now moving further days that by day X will be equal to 12 X square minus 2 76 X +1080 And the square said by dx square will be equal to 24 X minus 2 76. Now moving further after this as dessert by D. X. Must be equal to zero. So 12 X squared minus 2 76. 6 +1080 equals to zero. This will give you X squared minus 23 X plus 90 equals to zero. From here. You will get your X as five or excess 18. Now X equals 2 18 is to be rejected because X equals 2 18. Length will be 24 minus two. X equals two waiting. That is it will be minus 12 which is not possible. So X will be five in the turning point. Now at X equals 25 days squares. Set by the excess square will be equal to 24 into three minus 2 76. That is minus 1 56 which is the negative. So that is minimum had X equals 25 That is side of each square. To be cut off from each corner of maximum volume has five centimeters. Thank you.

We have the number eight, uh, number nine, part A. So there is a cardboard, which is the shape of a squired by six by six inches. Okay, six inches and by six inches. So Okay, so there are four Squires, its side x cut from the center from each corners. And now this remaining a Squire is being made a box with no top. Okay, so the resultant length will be six minus two weeks. Religion with it would be six minus two works, and the height will definitely be X. So the volume of the box we even let us say will be six minus two works into six minus two x into X. Now it is being also it is also being told that this these four, uh, small box small pieces are They are joined to make a box in no top or bottom, so their volume will be equal to this will be like you, so their volume will be equal to X Q. So we need to maximize mm v equal to V even plus V two, that is six minus two x whole square into X. Let's xq. Mm. Okay. So we need to maximize this. We need to maximize this. So let us open the back and it will be more convenient. Esquire plus four x Esquire minus 24 x into X plus x Q. That is for XQ minus 24 x Esquire just 36 x plus x Q. So vehicle to five X cube minus 24 x Esquire 36 x Okay, so to maximize the maximize or minimize and we need to find the critical point so we dish, let us suppose various expect v X because X. This is the function of the is the function of X only. So with a sex we had to quit equal to zero. So 15 X squared, minus 24 in 2 to 4 10 48 x plus 36. This should be called to zero. If you divide both sides by three will be getting five x Esquire minus 16 X plus 12 equal to zero. Okay, so we have two factories. This this will be characterized. Ads five x Esquire minus 10 X minus six x Australia equal to zero so five x x minus two minus six X minus two equal to zero so X minus two and five AC minus six should be called to zero, which means acts should be equal to two. And here, five X minus six. So x will be equal to six by five. Okay, so there are profitable points we need Now we need to find where it is maximum and where it is minimum. So let us travel. Differentiate all differentiate this this quantity again. So we'll be getting vegetable dash X equal to 15. 30 x minus 48. Mhm. Mhm 30 X miles 48. Now we have to find the value of this at X equals two first, this will be 60 minus 48. That is 12. So this is positive. So we have minimum value here. Minimum Okay, it is good. And zero. So minimum and visible. Dash X at X equal to six. By five. It will be 39 13 to 6 by five minus 48 at its 36 miles. 48 that is minus 12, which is less than zero, so maximum. So we have to find only maximum value. So we have Okay, so for X equal to six by five inches, that is 1.2 inches. It will have maximum volume. Okay, but be we have to solve the same problem, starting with four inch by six inch box. Okay, this is four inch. This is six inches. So we will be simply writing vehicle to four minus two x into six minus two x into x Bless xq. So the X Let us open the back it 24 minus 12 bucks minus eight x last four x square into X. Let's xq So this will be if you want to play with X 24 X minus 12 X squared minus eight X squared plus four x Q. Let's xq so v x will be equal to Okay, five x Q minus 20 X squared plus 24 x mhm. Okay, okay. Okay. So we have to just either we have to characterize it Odd. No problem. I wanted to factories because for maximum minimum, we have to find the critical point first. So we have to differentiate and should equal shoot equal to equal to zero. So 15 x Esquire minus 40 X plus 24 equal to zero. Okay. No, we have two factories. It so let us. Okay, okay. Factories and seems to be very tough here. So we'll be using the quality formula to get the value of acts so x will be equal to minus bee. That is 40 plus minus B Esquire minus 40 Squire minus four into 15 into. See, that is two and 2 15. So this will become 40 plus miners 1600 minus 14. 40. Do I buy 30? So the C 40 plus minus 160 divided by 30. Yeah. Okay, so this would have become square root of 160 years. 12.65 approximately 60 plus 40 plus minus 12 point green, 65 divided by 30. So we have two critical points. Access 40 plus 12.65 That is 52.65 by 30 and 40 minus 12.65 30 27 point 35 by 30. So let us just converted in decimal 215 50. That's equal to 1.755 and 27.35 jihad 30 zero point 91 two. So there are two critical points. Okay, So to get an idea of where it is maximum and where it is minimum. We have to find V double Dash X. So basically, that has differentiate this. So this will be 30 X minus 40 minus 40. So video X at X equal to 1.755 will be 13 to 1.755 minus 40. So 13 to 1.755 minus 40. That 12.65 It's just positive. So we have minimum here. Okay. And now readable. Dash X at X equals two. Is it a 0.9 12 will be equal to 13 to 1.7. I'm sorry. That went to 0.912 Okay, we required 912 and minus 40. 30 and 2.912 My 40. That is my industrial 0.65 which is negative. So maximum is here. So we have, in this case, the value of X approximately equal to 0.912 inches to get maximum video. So these are the answers. Thank you so much.


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