Calculus: The Box Problem

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Optimization
Maxima/Minima
The Box Problem Page [1 of 1]
All right. Well, welcome to our continuation of the classic calculus capers. There's a tongue twister right there.
Now it's time for the ever popular and much beloved making a rectangular box out of a sheet of steel. So, the question is you are given a rectangular sheet of steel, such as we have here. And what we want to do, in fact, let's take the measurements of this piece of steel here. I measure, I don't know if you can read that or not, but I measure sixteen inches here. So this is a sixteen by -- let's measure this one; I get thirteen, sixteen by thirteen-inch rectangular sheet of steel. Okay, and now we want to make this into an actual box that we can put things in.
So what are we going to do? Well, here's a method that we're going to use to make this into a box. We are going to actually cut off with special shears. I happen to have some right here, use these special shears that cut into steel. And we're going to cut out a small square piece out of each corner. So if I cut a square out of this corner, for example, I get a square like this. And that's now completely removed from here. And I remove the exact same size square from each of the other three corners, the exact same size square. So here, I get this one comes out, and here, and when you've done this as long as I have, it's real easy to cut. Let me get this. Okay, these are all the same size squares. And you'll notice that what remains is a sort of truncated rectangular piece. It sort of looks like this. It's almost like a very, very fat plus sign. And now to make this into a box, what we can do is actually fold up this sheet along these sides like this. But if I do that, what I actually get you see is a box. I'm doing this for you right now live. It's an awful day here, it's raining and thundering and I think I'm catching a cold. You see if I tape part of this, you can really see that it folds up perfectly and makes a nice box. See that? You can put things in it and put your stuff in it.
Okay, well, the question is how much should we cut off of this rectangular sheet in order to get the largest possible volume after we make the box? So the question is find out what is the size of the square we should remove from each corner so that when we fold up and make the rectangular box, that box will have as large a volume as possible? That's the question.
Okay? Well, let's quickly go through my method for solving all of life's problems. Number one - understand what is being asked. Can you even hear that? It is thundering! You would not believe - on days like this, what you want to do is you want to be inside doing calculus problems. So luckily we're here together doing calculus problems, let it rain outside. Let the rain hit the roof, let it thunder. We're warm and cozy making boxes. Okay, so let's go on.
Okay, so understand what is being asked. What are we asked to find? We're asked to find the size of the square we should remove, how much we should remove. Let me draw a little picture here. And I'm going to draw a very small picture of what I just showed you. We actually did it live. I'm going to draw a little sketch here. Here's the rectangular sheet of steel after I cut off the pieces. If you want to see what I cut off, I cut this off here. Those are the four identical, identically sized squares that I removed. That was the original sixteen by thirteen. What I'm going to do now is take this sheet and fold it along here and I'm going to make a rectangular box. Let me now show you what that rectangular box would look like. Here's a perspective drawing of that rectangular box after you make it, it would look like this. You see? I can open the box so you can actually see inside of it. And we want to make this thing as large as possible. So the question is, -- what we want is, find the amount we have to cut off here. Let me give that a name right now. We could call it x; you might want to call it x. I'm actually going to call it s, just to indicate that we can use any letter at all and also to show that, in fact, it stands for the side of the square. So, let me call this length right here s. Which means automatically that this is s that is s; this side is s, that's s. All these sides are s since they're all the same. So we want to find the value for s such that the volume of the box - what I mean by this is the box we make after we fold, after we fold up these sides here, and get this - that volume is maximized. Okay, well that's the question. That's the question.
I wonder if you can even the thunder? Do you think they're going to be able to hear the thunder? I'm asking the soundperson here. Wow, so you can. Well, that's great! So, you can enjoy this rainy day with me. That's great!
Okay, second phase, notice that behind my problem solving method, there is an obscured sun, so it's not surprising that we're getting this thunder. The second thing we have to do is determine what we know. So what is it that we know? Well, what we know here is that, well, we're given this size here of sixteen by thirteen. So we know that we have a sixteen by thirteen original slice of rectangular metal. I'm trying to actually maximize the volume. I should write down what the volume is, by the way. Let's write down what the volume is -- so what I know is the formula for the volume. Now what is the formula for the volume? The formula for the volume is - okay, this is something that we actually have to recap here because we have never talked about this. How do you find the volume of a rectangular box? Well, the answer is - what I do is it's base times width times height. So it's the product of three things, base times width times height. Okay, so let's do that right now and see what that is. Now you've got to be very visual here to figure out what the base, what the width and what the height is. So let's see if we can figure that out right now. Base, width and height.
Okay, let me just write down base, width and height. So volume, I'll call it v, equals the base times width times height. Now what are all of these things equal in this case? Well, that is the last part of our issue, finding a connection between how this is set up, the original picture, how all of this was given to us, and what we know, namely the formula for a volume. So now let's see if we can tie these things together, the method we use and getting this formula into one variable. Because this one now has three variables in it, seems like we're losing ground. Before in the other example, we had two variables, now we have three. So let's see if we can now reduce the number of variables by finding a connection between these things. And the connection is to return to this picture. Each of these sides is length s. So this is s. And this is s, and this is s and this is s. So when I fold them all up one of the lengths, one of the dimensions, we can actually easily see. It's the height. The height is plainly s. You can see it visually here or you can look at this picture and realize that when I fold this up, when I turn that up, this becomes the height. So, in fact, this is s. So the height is equal to s. Great.
Now, what about - let's do the base. So what about this length here? Well, what is that length right there? That's this length right here. How would I find that? Well, I knew that the original length of the whole thing was sixteen. So what is this new one? Well, all I've got to do is unfold it and notice that - well, it's this length right here. And when the whole thing was sixteen, I cut something off. I cut off how much? I didn't cut off just s. I cut off s on this side, but also s on this side. So this length inside here is actually sixteen minus one s minus a second s, sixteen minus two s. So, in fact, this length is sixteen minus two s. Sixteen minus two s because this whole thing was sixteen and then I cut off an s, cut off an s, so I have sixteen minus two s. So this is sixteen minus two s. And so that's the base, sixteen minus two s. And what about the width? Well, the width we can find in a similar way. This entire width we saw was thirteen, but the width of the actual box, which is this, is just the thirteen minus an s and minus another s, so it's thirteen minus two s. So this equals thirteen minus two s. So the volume I can now write in terms of just one variable -- s, and it equals sixteen minus two s times thirteen minus two s times s. So, I have volume in terms of s. And I want to now maximize the volume so I want to find out where is the high point. So I would take the derivative of volume with respect to the variable s and see when that derivative is zero, because that will give me a candidate for a max.
So now I'm going to take this formula and I'm going to differentiate it. I'm going put this over here for a second. So we have volume equals sixteen minus two s times thirteen minus two s times s and we have to differentiate it. Now, how would we take the derivative of that? Well, we could use the Product Rule. That would be completely great. Or you could be lazy and multiply this out and then take the derivative. Either one is perfectly fine. Since it's rainy outside and thundering, I think we should be lazy. Aren't you lazy when it's sort of rainy and sort of lethargic, at least I am. I'm always lethargic.
Anyway, let's try this right now. So we have to do sixteen times thirteen. Sixteen times thirteen is, I think two hundred eight. Then my inside term gives me a minus twenty-six and my outside term gives me a thirty-two, so that gives me twenty-six and thirty-two gives me a fifty, twenty-six and thirty-two gives me a fifty-eight minus fifty-eight s. And my last times the last gives me a plus four s squared. So, that's all times that outside s. All times that outside s. So I multiply through by that s. What I see here is two hundred eight s minus fifty-eight s squared plus four s cubed. So that's volume.
And now I can take the derivative pretty easily, domestic violence/ds equals two hundred eight minus and fifty-eight times two is one hundred sixteen s, plus twelve s squared. So there is the derivative of volume.
I want to now set that equal to zero to find candidates for max/min. So I'm actually setting this equal to zero to see where, in fact, things level off. Okay, now how are you going to solve this thing? Well, this looks really, really bad; it's a quadratic. First thing I can do is divide through by two to simplify this a little teeny bit, but it's not going to be that big of a help, let me tell you. So I have a hundred and four minus fifty-eight s plus six s squared equals zero. Let me write this out in a slightly different way, putting the s squared term first, that's what we usually do. Six s squared minus fifty-eight s plus one hundred four equals zero.
You can try to factor this, but you're not going to succeed. We actually have to use the Quadratic Formula here. By the way, what's the Quadratic Formula? Let me write it over here. If you have a x squared plus b x plus c equals zero, then there's two solutions for x. X equals negative b, plus or minus the square root of b squared minus four a c, all over two a. So we actually could use that formula right here. Here a would be six, b would be minus forty-eight and c would be a hundred and four. And if you plug all that in -- that formula, you would see that s equals, well, negative b, which would be negative negative forty-eight, which is forty-eight plus or minus the square root of b squared minus four ac, which turns out to be eight hundred and sixty-eight. You can check that. All divided by two a, which is twelve. Okay, so there's two answers here.
Okay, now let's regroup to make sure this is all okay, because maybe there's a mistake somewhere here. All right, so let's make sure that we're doing everything okay here. We're doing everything okay here, so if we work through this. Okay, here, this looks good. This looks good, looks great. This is here. This is here; this looks good to me. Okay. Maybe it's okay. Forty... Oh, when we divide through by -- a little mistake here. When we divide through by two, I think we don't change anything here. So the question is when I flip these two things, I shouldn't write a forty-eight here. I should keep it as a fifty-eight. That subtle suggestion was made by the staff, by the way. So I have a typo here. Instead of writing fifty-eight, I wrote forty-eight. Sorry about that. If you saw that and you were confused, good for you! This is happening on the live folks, we're not sort of doing this - there's no script here. I'm doing these things with you and I made a mistake. Okay, good, it's good to make mistakes. It's also good to catch them. So thank you for catching it if you caught it out there and thanks to the crew right here, live, for catching it right now.
Anyway, so these are the answers that we get, there's two answers here. Let's actually figure out what those two answers are. And I have my calculator right here to quickly just whip this out. There are two answers here and I want you to see what they are just in case you're not too familiar with this. There's the s equals fifty-eight plus square root eight six eight, all over twelve and then there's s equals fifty-eight minus square root eight six eight, all over twelve.
Let's see what each of those equals. The first one equals, let's see, turn this one. So I see eight six eight, I take the square root. I add to it fifty-eight and now I divide by twelve and I get seven point two eight eight stuff. And the second one, I get this answer is eight six eight. Take the square root and then I subtract fifty-eight, I take away fifty-eight and I divide by twelve and I get two point three seven eight. So look at this, now we have two candidates, which one do we pick? Which one gives us the maximum volume? Maybe both of them are maximum. This is like Jim's question earlier on, or how do we know what to pick?
Well, first of all if you think about the original problem, you will see that one of these answers is actually not going to be a good one. Let's take a look at this one for a second. If I cut off my square, each side having length seven point two, then when I go back to the original thing here, I'm taking off seven point two inches here and seven point two inches here. So how much are the sum total of these two things? Around fourteen point four. But what do I know about this entire length? Remember how long that was? Let me remind you how long that was. It was thirteen. So look what I'm doing, I'm saying hey, you're going to take off fourteen inches from something where you only had thirteen to begin with. That's impossible. So by the physical constraints of this problem, which of course, this function didn't know. This function's just plodding along. It doesn't know that there's a little box back here. It turns out this answer is impossible. I'm cutting off too big. I'm cutting off so much there's nothing left so the volume would be zero, so in fact that's the minimum. It would be volume zero, cut everything away. So this is not it. And so our answer must be this one.
And we can check to see that's a max by doing a little sign test here for the derivative, dv/dx. And I put this point in, two point three something. Oh, the actual answer, in fact, on a test or if you were talking to me, this answer might look good to you but that's an approximation. The actual answer is this one with the square root, because that's exact. This is just a numerical estimate, but this is the real answer, by the way. That's the one you want to say on the test, you don't want to actually, I don't think, say this rough estimate even though it gives you a sense of how big it is.
Now we'll take a point to the right and left and see what the derivative is doing. The derivative, by the way, is right over here in case you forgot. So let's take an easy point, let's take s equals zero. That's certainly to the left of two. Plug in zero here and zero here, I just see two hundred eight and that's certainly positive, so everybody in this region's positive because if not, we would see a change somewhere. We didn't see a change. Oh, that's actually, yeah, we didn't see a change. That's great.
Then here we see a zero. And now I have to pick a point to the right of two, but I've got to pick that point to the left of seven because things change around seven, we thought that was going to be a min. So let's pick something that's between two and seven, how about if we pick five. And I'm going to let you try this on your own. Plug in five into here. You're beginning to learn how miserable at arithmetic. But if you plug in five in here, take two hundred eight subtract off a hundred and sixteen times five plus twelve times twenty-five and see what that is, it will be negative. I'll let you try that.
So what we see is what's happening. Well, since the derivative is positive the function itself must be climbing. So it's climbing up and then right after that point, immediately after that point, we see it's falling down. So what happens here, this must be a summit. It must be a max. It's the only max we found. This we saw was a min, so this is the max. To get the maximum volume what we have to cut out is roughly two point three inches from each side and that would give us the maximum volume. So that answers the question. So that's what it would look like. Really, really neat.
Okay, well I hope you enjoyed that. You guys chew on this and, in fact, I invite you to try this problem on your own. You know, by the way, it's not a bad idea to just do a problem over after you see someone else do it or you work through it with someone else. Do it on your own, make sure you can work through all of the kinks. That's when you really understand for sure whether you see the ideas and the big picture. Try this, another calculus classic coming up.