 Good afternoon. Good afternoon. I'm Carol Stern. And Chris Gann is an instructor of mathematics here at the School of Science and Math. And we are going to have an opportunity to learn about variations on the math problem. Chris put together this presentation for the Teaching Contemporary Mathematics Conference that was held here at the School of Science and Math earlier this winter. And he agreed to do the presentation again for a video conference. We are streaming this session live and archiving it. And at this time, we don't have any live video conferencing participants. But I will be over here monitoring to see if we have any participants with questions or comments through the live stream feature. So without any further ado, let me turn it over to Chris. All right, thank you very much, Carol. So today, we're going to look at a classic problem that I first was introduced to this problem as a calculus student. But I don't think there's any reason to wait that long into the sequence, the math sequence, whatever we might call it, to introduce this problem. The primary focus today, what I want us to think about and get our students to think about, is how to write a mathematical model. This is a very important skill. It's very emphasized in our. So we want to make sure that our students can, when given a real world situation, think about how we write a function that represents our phenomenon. So we want to begin, let me begin by introducing the problem itself. Given all the top or wider, it's not a bad idea to have a class to demo or pass out to all the students because they can actually follow along. So suppose you were given this piece of wire and you were asked to cut it somewhere along the wire. To bend the two pieces of wire into regular shapes, by that I mean a regular polygon, a square, a circle, an equilateral triangle, a hexagon. For our instance, for our first stab at this problem, let's consider the situation where I said, here's your one piece of wire, cut it wherever you like. I cut it like this. And now bend one piece into a circle and the other piece into a square. OK, let's do it. So let's bend this piece into a circle, try to. This is my attempt at bending something into a circle. There's our circle. And let's bend this piece into a square. Here's my freehand attempt at a square. Now I actually brought a ruler along and we should probably be making our circle and square more square. But for our purposes, for the first look at this problem, I've got a piece of wire that I've bent into a circle and a square. The question is, now how much total area do I have? Well, I can measure my circle and I can measure my square and I can calculate how much area. What we really want to find out is, is there a way I could have cut this into two pieces in such a way that I would have minimized or maximized the total enclosed area? The question is, where would I place this cut along my original wire if I wanted the total area to be either maximized or minimized in the end? So a lot of students, when faced with a question like this, just don't even know where to begin. There's a lot of head scratching, some sighing, and maybe some giving up. But other students start to write some things down and a good approach to begin this problem is to just try something, to actually see what kind of area we get when we cut this wire at random and bend the two pieces. And so this is where it's great to either have a pair of wire cutters yourself and to do this and measure the area, or if you have the resources to have all of your students cut their own wire wherever they like, measure the two lengths that they have, and then calculate the area of their two shapes by bending them into the shape and then measuring with the ruler and applying familiar area formulas. So well, let's see what happened when I cut my wire the way I did. So let me first measure how much wire I started with. And so my particular piece of wire was two feet. So I'm going to write some things on a whiteboard here that you should be able to see. So given a 24 inch wire and told to cut it into two pieces, bending one into a circle and the other into a square, can we maximize or minimize the total enclosed area? And I really think that experimentation is the way to begin this problem. So what I would recommend my students do is to cut their wire and measure what they've got. So I did. I cut my wire. And I'm going to allocate this bigger piece here to the square. And that's roughly, and I'm being rough here, 12 plus 16 inches, leaving me with, I would hope, 8 inches for my circle. And that's what it appears to be. OK, so I've got use 16 inches for my square and the remaining 8 for the circle. Let's take a minute and calculate how much total area we have. The students at the level that this is appropriate for, about middle school age to early high school age students, should already be familiar with the area of a square is equal to its side squared. So if we consider that I've got 16 total inches for my square and it has four sides, we clearly have four inches for each side. Interesting. So if I have 16 inches for my perimeter of my square, then what's my area of my square? Four squared, also 16. So the area of my square is 16 inches squared. But what about my circle? So I bend this into a circle I'm going to consider. Well, how big is this? This one is a little trickier, because what I know is that I have 8 inches total for the circumference of my circle. But you ask your students, what's the area of a circle? They don't tell you the area in terms of its circumference. They're going to tell you the area in terms of its radius. They'll say it's pi r squared. So they have to do a little deducing. If I tell you that it's got an 8-inch circumference, the circle does, then we need to first deduce its radius from that before we can figure out the area. So let's do a little bit of algebra. What we know is we have 8 inches for the circumference of the circle. And we want the circle's area. We know that circumference of a circle is 2 pi times its radius. And we also know that the area of a circle is equal to pi times its radius squared. But does area depend on circumference? So a little intermediate function writing. What we have essentially is a function for the circumference in terms of the radius, as well as a function for the area in terms of its radius. And what we can do with this is eliminate the radius and know what the area of any circle is given its circumference. So to do this, we just have to determine how the radius comes into play. That is, if I know what the circumference is, does that inform me what the radius is? Well, sure it does. If I know that the circumference is 8 and I need to know the radius, all I need to do is say 8 is equal to 2 pi r. So r is equal to 8 over 2 pi. In this situation, that's my radius. But more generally, the radius of the circle is its circumference divided by 2 pi. In this case, that's 8. But more generally, the radius, r, is the circumference divided by 2 pi. What this does for us is it allows us to substitute into our area formula and eliminate what its dependence on r. That is, if the area of any circle is pi r squared, then the area in terms of its circumference is pi times its circumference divided by 2 pi squared. And it's perfectly OK actually to leave it like this, although we can simplify it just a little bit more and say, if I know the circumference to get the area, I just need to square the circumference and divide by 4 times pi. So what we know is that if I'm given 8 inches to make my circle, that my circle has 64 divided by 4 pi square inches of area. So I'm going to reach for the calculator because I don't want to divide 64 by 4 pi and just take a guess at it. So let's say about 5.09, or we'll say 5.1. So here, my circle, now this is an attempt at a circle, if that were really a circle with 8 inches of circumference, its area would be 5.1 square inches. 5.1 inches squared. OK. So what I found is that with 2 feet of wire, 24 inches of wire, if I cut it and allocate 2 thirds of that wire to the square and the remaining third to the circle, 16 and 8, that my resulting total area is 16 square inches for the square plus 5.1 more for the circle. And I have 21.1 square inches total enclosed volume. So what I would ask my class, since I've been following along and cutting their own wires, is what did you get? How does your area compare to mine? I got 21.1 square inches. Who got more than that? If we were trying to maximize the total enclosed area, who did better than I did? And I'd get a show of hands and I'd ask the students, well, what were your two pieces of wire sized? Can we determine, based on what other students said, whether we should allocate more or less to the square or circle if we were trying to either maximize or minimize the result? So now that we have kind of explored the scenario and gotten our students to tell us how they did compared to how we did, we got 21.1 with our particular cut size. Who did better? Who did worse? Well, better or worse depends on whether we're trying to minimize or maximize the area. But we can still get a sense of, if I give more to the square, more to the circle, what happens? Well, this is going to lead to writing a function for the total area enclosed as a function of how much area I give to one shape or the other. So what I would ask my students to do now is we've simulated this as many times as I have students in class. But if I wanted to get a broader sense, if I wanted to collect a lot more data about what happens when I cut the wire into two pieces, we can simulate this on the graphing calculator. So one thing about asking a bunch of students in a classroom to cut their wire at different places, most of the students are going to naturally kind of tend to cut the wire into around 50, 50, or 1 3rd and 2 3rds. They're not going to do some of the more extreme cuts, such as I'm going to only give half an inch to my circle and 23 and 1 half inches to my square. And maybe we get to gain some insight by seeing what would happen if we did that. It's really physically hard to bend. So we want to try to simulate this on the graphing calculator. And the students get a kick out of this, too, because they get to do some stuff on the calculator that's a little bit different than what we typically do in class. It's going to involve using the stats button on the calculator. We're going to create some lists. So if you have a TI calculator, this would be a good place to follow along, I'm using a TI-83. A TI-84 would be appropriate to all of the syntax, but the calculator syntax would be essentially the same. The other calculators in the TI family could all be made to do this, too, but the syntax might look a little bit different. OK, so to start, we're going to go press the stat button in the top middle of your calculator. And we want to edit our lists. So what we want to simulate here is an entire classroom of students doing what we just did in real life, except now every student is going to have 20 attempts at cutting the wire at random places and the result of that happening. And we're going to make a graph. And every student's going to get a slightly different graph. So what we want to do is first consider, if I have 24 inches of wire and I say, cut that wire at some place randomly along it, then I get two pieces of wire, one that is going to be used from a square, and one that'll be used for the circle. So let's define L1 to be the amount of wire that gets allocated to the square. So if L1 is the amount of wire that gets allocated to the square and I really want to make this random, then I want to generate 20 random numbers that are between zero and 24 in L1 if I'm simulating 20 students doing this activity. So to define L1 to be a sequence of 20 random numbers between zero and 24, you want to first make sure that your cursor is up here at the L1 at the heading of the list, not down here in the first entry. So again, you don't want to be down here in the first entry. We can only put a single number at a time in there, but rather we want to be up here where the cursor is highlighting L1 so that we can define the entire list to be a sequence of 20 random numbers. So that's where we're going to start. Enter on L1 and your cursor will pop down to the bottom where we're defining L1. The next thing that we want to do is create a sequence of random numbers. So we need the sequence, which is most easily found in your catalog. So if you press second zero, it brings up your catalog of all of your different functions that are built into the calculator. And I want the sequence function. Now to get there quickly, I'm going to press S because the alpha's already lit up in the top right corner of your screen. And if you press S, it'll jump down to all of the functions that start with S that are built into the calculator. We want this one, Seq open parentheses, not to be confused with capital Seq, no parentheses. So make sure your students are on the right sequence. Make sure you've got the right sequence and hit enter. So now we're defining list one to be a sequence of things. Well, I don't just want it to be any sequence. I could, using the sequence function in the calculator, I could define arithmetic sequences or geometric sequences or all kinds of sequences, but I actually want a sequence of random things. To do that, I also have to ask the calculator to, I also have to use a built in function that the calculator has. So right back to the catalog screen, we've got sequence and then second zero again, bring back this catalog screen and now I want RAND for random. So go down to your Rs and we want to do RAND now. This RAND right here, R-A-N-D, if we hit, if we run it, and this actually, this would be illustrative to pop back to the home screen real quick. So let me leave this real briefly and show you what RAND does. When we bring up RAND on the home screen, and press Enter, this time I get 0.18. If I do it again, I get 0.16, 0.81, 0.877. Interesting, 0.72. What RAND gives me is a random number between zero and one. So if I want a random number between zero and 24, what students then should we do? And so I like to post this question to my students and make them think about it a little bit. If I have a built-in function that gives me a random number that's between zero and one, how do I then get a random number that's between zero and 24? And the answer is to do 24 times RAND. So let's see that on the calculator. Still on the home screen, 24 times RAND. Well, there's 15.6. Let me hit Enter again, 8.3, 21.5. By doing 24 times RAND, we are simulating choosing a random amount of wire to allocate to the square. So now, back to our stats. We're going to edit our list now because I left that screen, everything I had started to do is gone. So let me quickly jump back to getting a sequence. Random numbers that are between zero and 24. Now, so far I've asked for a sequence of random numbers between zero and 24. I have not asked my calculator for a particular number of random numbers. So now I want to be sure that I'm getting 20 random numbers. So to do 20, it's not actually as simple as just doing comma 20 here. I have to do comma x comma one comma 20. And I could actually have used any letter. I used x because it's convenient. There's a button for x. I don't have to press alpha. But what we're doing with this is we're creating an index for our sequence. And so by putting an x there, I'm saying I want x to be a counter that keeps track of how many random numbers I have. The next two slots there, the comma one and comma 20, initialize and end the values of x. So x is my kind of label for each of these random numbers. And so I have x going from one to 20. So this will give me 20 random numbers that are between zero and 24 as soon as I hit enter. Okay, so this populates list one with 20 random numbers between zero and 24 as we wanted. So student A said, oh, I cut 13.4. And student B said, well, mine is 21.77 for the perimeter allocated to the square. So then the next question is, well, what's list two gonna be? Well, how much wire do I have left? So let's define list two to be the amount of wire that we have left for each of these particular cuts. That is, if I used 13.4, if I'm the student who cut 13.4 and gave that to the square, then that means that I only have 10.6 left for my circle. How do I see that? Well, I can define L2 simply to be the values that are in L1 deducted from 24. That is, I have 24 inches of wire. I give some of it to L1 and how much is left? So if I define L2 to be 24 minus the amount that I'm giving to the square, I can see I've got this much left for the circle, about 10.6 in the second case, 2.23. And so now in each row in my list of statistics, I can see how much wire, what I essentially have here is 20 different random experiments where people cut the wire differently. And now we wanna see how much area we get. And let me remind you again, the whole point of all of this of what I'm doing here, simulating this on the calculator and having the students simulate it, is with the wire, is that it will help them build their intuition about how we actually write the mathematical model. Because the things that we're doing, the mental considerations that we're going through when we're thinking about how we define these lists are very similar. They really mimic the process of writing a function. But I think for many students, they're much more comfortable simulating this 20 random times to a 20 random simulations than they are actually writing the function. So, what we want to do now with our two lists, now that we have list one is the amount of wire that is for the square and list two is for the amount of wire that is for the circle. We want to list three to be the area of that square and list four to be the area of the circle. So how do we do that? Well, we want our students to consider, what did I have to do when I had the 16 inches for the square? And I wanted to know how much area I get for that. Well, I couldn't just take all 16. I needed to deduce what the side length was. If I have 16 inches for the perimeter of my square, what then is the length of one of the sides? 16 divided by four. So more generally, when I have a random amount of wire allocated to my square, I just need to divide that amount by four if I want to know how long one of the sides is. From there, to know the area of said square, I just need to square that result. So let's go ahead and define list three to be list one divided by four so that we're looking at the length of one side and then square that result. So that is, list three is list one divided by four squared. Now what I have in list three is the area of every possible square given the random cuts that I made. So I made all these 20 random cuts and then these are the size squares that I get. Now, list four is a little more complicated. List four, as you probably anticipated, is the area of the circle. And I have in list two the area of the circumference of the circle. So it might be a good idea to pop back over to the whiteboard and look at what we wrote down for the area of the circle in terms of its circumference. We said, how do we get to the area of a circle if we know its circumference? We kind of have to go, we kind of have to make us pit stop at the radius before we can really understand. That is what we think of when somebody walks up to us and says, what's the area of a circle? You immediately tend to say pi r squared. You haven't learned the expression for the area of a circle in terms of anything else other than its radius even though we can clearly define areas equal to circumference squared divided by four pi as it says right here, right? So if we chose to, we could say always teach in a geometry class that area of a circle is c squared over four pi. I don't know why all those silly people teach pi r squared. It's c squared over four pi. Of course we don't. So this, but this is of course, this is what we need to calculate the area of all of our different circles that we get. So let's go back over to the calculator and recognize that what we have in list two is c. So I need to square that. I need to square list two, divide it by four and divide it by pi. So let's define list four to be equal to my list two squared. That is my circumference of each circle squared. Divide that result by four and also by pi. Having a little memory issue on my calculator. When I asked it to do that, it said it was out of memory. So let's try doing that again. So that should work just fine. Just had to go and delete some matrices that are taking up too much memory in the calculator. Okay, so now we have list four, sorry for the technical difficulties. List four is now the area of all of our different circles. It was the circumference squared divided by four and divided by pi. Now of course this could be done over a multitude of lists. That is, I didn't necessarily have to jump straight to the areas from the circumferences. And if you find your students having trouble with that, what I would recommend is make an intermediate list that is the radius of the circle, right? We're more comfortable thinking about the area of a circle as pi r squared. And we know what the circumference is that's in list two. So we could have made a list and called it the radius list where we divided list two by two pi. Then we could have taken the entries in that list, squared them and multiplied by pi to get the area. Okay, but I wanted to use the result that we had already worked out. The area was c squared over four pi, but you might need to modify in your own classroom depending on how comfortable your students are eliminating variables from two formulas and expressing the one in terms of the other. Okay, so now what we want to do is see what is our total enclosed area. We have all of our squares, different areas in list three, and all of our different circles areas in list four. So our total enclosed area, list five, is just the sum of list three and list four. This is how much total area I get in these situations. Let's add those again. This time I think I've deleted enough that I shouldn't have that problem again. Oh, okay, now that was definitely my error. I don't know if anybody saw me type that into the calculator, but I just made a mistake on the calculator where I did what I warned you not to do. I typed list three plus list four in the first slot of list five. And the calculator objects to that with a data type error because you can't put an entire list of 20 numbers into just the first entry slot of a list. That is, it's really important that I'm defining up here list five itself to be list three plus list four, not the first entry of list five. Okay, so now here's all my total combined area. If I was trying to maximize area, I'd say these ones in the thirties are pretty good. If I'm trying to minimize area, wow, 20 is looking like about the least we can do. But let's get a graph of all of the different areas, total areas, given how much we're going to allocate to the square. That is, let's define a stat plot. So second y equals to be the relationship between the amount of wire that I allocate to list one, or to the square, which is in list one, compared to the total amount of area that I get when I put everything together. Now I can go ahead and just zoom stat to my data here. And there's the trend. The x-axis indicates how much area, how much wire got allocated to the square. And the y-axis is how much total area I end up with. And so what we can see scrolling around is that if I try to give a small amount of area to the square, I'm up here, a small amount of perimeter to the square, then I'm up here and I'm getting a fair amount of total area. If I jump over here and I'm giving, oh, I said that backwards. That is, when we're down here, we're using a small amount of perimeter for the square. You can see that my cursor is roughly on the trend. We haven't fit a curve to this yet, but my cursor seems to be about where another data point would fall. So if I had to estimate, I'd say, if I made, if I took my 24 inches of wire and only gave my square 3.12 inches, then I would expect roughly 36.5 total square inches of enclosed area. Whereas if I instead gave a lot of perimeter to my square and only a little bit to my circle, then I could end up with almost the same amount of total enclosed area over here. You can see I'm giving nearly all of the perimeter, all of the wire to the square, only leaving less than half an inch to the tiny little circle, but the result is still 36.56 square inches total enclosed area. But if I was trying to minimize the total enclosed area, then the solution is probably down here is what the trend suggests, roughly 13.8 to my square, all the rest to my circle and the result is 20.23 square inches of area. Okay, we can at this point try to fit a curve to this data and let's do that real fast. Depending on what other things you've been doing in class, you might find it appropriate to fit a curve to this data. Now, if your students look at this and guess that it's quadratic, it actually is in this case, but it's something that we don't always want them to do. We don't want them to look at anything that's curved like this and guess that it's quadratic. So I kind of steer away at this point from having my students just fit a parabola to it because we don't have any evidence that it's actually parabolic. Now, it turns out it is. So if we do fit a parabola to here, if we go to stat and calculate a quadratic regression through list one and list five, we get this regression here and actually I needed to store that when I did it. Let me run that regression again. That is, I wanted list one, list five, and then I want to store that into y one. So if you run a linear regression on list one and list five comma y one, that'll do the exact same thing except now it'll store my function for me in y one. So I don't have to type it in. If I graph this, that's not at all the right graph. Oh, because I didn't wanna run a linear regression. I wanted to run the quadratic regression that I ran the first time. Quadratic regression, list one, list five, y one. Last time, then this is gonna work. Okay, now we can see our quadratic regression stored in y one. I don't wanna have to type all those decimal places in by hand. So being able to directly store it is very useful. And now I can plot this and I can see that a quadratic really does provide an excellent fit to this data, this experimental data that we just came up with. And if I want to calculate the minimum on this curve, I can do that. I can press second trace for the calculate, choose minimum, set a left and a right bound and kind of guess where I think the minimum is on the calculator. And sure enough, it turns out the very best I could do would be to give 13.44 inches to the square and the resulting total enclosed area is only 20.16 square inches. Now by best, that is, if I was trying to minimize the total enclosed area, that's what I should do. You can see that the answer, the question of what should I do to maximize the total enclosed area does not have as obvious an answer. There's not a local maximum anywhere on this curve. Now, to finish up, let's come back over to the whiteboard and let's recreate the same ideas that we just did in the lists to write the actual function for our total enclosed area. And this is what we'll finish with today. This is the main point. Now that we have thought about how we define list one, list two, list three, list four and list five, in our calculator, this helps us come up with how we're going to write our mathematical model. That is, list one was my random stuff. List two was 24 minus L1. List three was the random stuff chopped into fourths and squared. List four was my circumference of my circle squared and divided by four pi. And then list five finally was list three plus list four. If we just mimic this process, we are going to get the area, the total enclosed area as a function of the amount allocated to the square. So when I define a variable, let x be the amount, let x equal the amount of wire for the square. The reason we call it x is because we're not specifying what the amount is. This is completely analogous to saying, let x be some random thing between zero and 24. In fact, we would say the domain of reality of this model is x is in zero to 24. So what we talk about as the domain for our function matches what we allow x to be in our random generation of, or the amount of perimeter in our random generation. So there's nothing too fancy about our variable definition. But what that means then is that 24 minus x is the amount for the circle, just like we have our list two defined to be. Well then the area of the square is equal to, I see. We had a question, how do you get the y1 into your calculator? Y1 into your calculator. Let me go through that very quickly before we proceed. That's a very important thing to be able to do. That y1 can be found, let me back off here onto the home screen. There's a button called vars. It's right to the left of clear. And so if you press vars, arrow one to the right for y vars, and then there's a list of function, parametric, polar, and on, off. But it's under function here. Now is your list of all your y equals, which is very convenient because I can just grab y1 now in my home screen, say, and press, well, I can grab y1 and say y1 of x, and it's going to, oh, I see what it is, nevermind. X is currently defined to be the 13.44. So when I'm typing y1 of x, it's evaluating my function at 13.44. But the important thing is, is that we can grab the functions themselves as they're stored in y equals, via this vars menu. Carol's it gonna cut off right at 415? Yeah, let's see. Okay. 420. Okay, great. So we should be able to finish up within the next five minutes. So this might go slightly past 415, but we're practically done. So did that answer your question about how you get the, have you found the vars now? It's great to know that somebody is currently watching. I didn't have an audience when we started. So glad to know that you're out there, Stephanie. Well, please let me know if there's anything else. Okay, but we are close to finished. So let's wrap up. Going back to the whiteboard, we were saying the area of the square then is x, since x is the same as L1, divided by four, and squared. That's two came up quite sloppy. Okay, x over four squared is the area of my square. What about the area of my circle? So that's a semicolon here. Area of my circle is what's in L4. That is, it was the 24 minus x. Then I need to square that. And divide by four pi. The same as I did, right? Because this gives me my circumference of my circle, squared, and then divided by four pi. And so this process I find with my students, this process of having them experiment on their calculator and generate these random numbers in their lists, facilitates the process of writing the function. Because now we can say my total area is simply the sum of these two terms, x over four squared, plus 24 minus x, all divided by four pi, and the numerator is also squared. This function is better than what we did with the calculator lists, because with the calculator lists, I only had 20 random numbers in L1. Whereas the function is all infinitely many numbers that x could be between zero and 24. We have not specified anything for x, except that it comes from the domain zero to 24. So this function that we wrote is a lot like the scatter plot that we have on our calculator, but it's actually infinitely many points. It is all of the points. And so what we're gonna see is when I graph this, it will go through the data points that we collected. So back to the calculator, I'm actually gonna clear out y one so that you see I'm not using the quadratic fit that I used before, but rather I'm just defining my function as x divided by four squared, plus quantity 24 minus x squared, then divide that by four and divide by pi. And there's my curve, connecting all of the randomly generated data points from my calculator. And it's really the exact same thing, except instead of 20 points, it's all of them, all of the points from x equals zero to x equals 24. And then just like we had done a moment ago, because like I said, I tend to avoid having my students actually hit the quad fit button because we don't want them assuming that everything that is shaped like a parabola actually is a parabola. That's a bad assumption to make. But now that I've created the model myself and I could see that it was quadratic, right? The x's were all squared in there. This gives me more evidence that it actually is a parabola. But anyway, even if I didn't find that it was a parabola, I could still ask the calculator to minimize my curve or to find the minimum rather on my curve. So second trace for calc, grab the minimum option here. You can see that I'm slightly to the left of the minimum. I could arrow around on the curve. This is just like being in trace where I'm gonna stay only on the curve. I'm gonna press enter when I'm to the left of the minimum. I'm gonna arrow over somewhere until I'm to the right of the minimum. And then it asked me for a guess. You don't really have to get that close, but that's close enough. And I'll just guess that the minimum's about down there. It thinks for a moment and says, ah, there it is. Give 13.442379 inches to the square. That's the way that we possibly, that's the way that we minimize the total enclosed area. Okay. But again, to summarize the whole point of this exercise that when I use it with my students here at NCSSM is to facilitate the idea of writing a function. When you ask students to write a mathematical model, I find a lot of times they feel like they don't know where to start. And so by helping them through this simulation by getting them to think about the process in a very step by step broken down manner, then put it all back together. It seems to make it a lot more clear in their minds. So this is something I use here in class and something I wanted to share with other people across the state. So thank you for attending or for watching my talk and whether you got to come in while it was still live or if you're just watching the stream someday. Thank you for taking your time to watch me present this problem.