Modeling With Systems Of Inequalities Homework
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Students will be working through the whole activity before the debrief. For the most part, they won't have much trouble with this activity. As you walk around and monitoring the groups, make sure to be checking that their inequalities are correct. It can be really frustrating for a student to spend a lot of time graphing only to find out they graphed the wrong inequality.
During the Check Your Understanding problem, students should write and graph inequalities that are in standard form. This is a great time to help students practice finding intercepts. Ask students, "What is the greatest number of minivans that they can get with this constraint? What's the greatest number of SUVs?" This will help them to graph much more efficiently. They also have to decide the scale on the axes so finding the intercepts first is very helpful.
Hello and welcome to another E math instruction common core algebra one lesson. My name is Kirk weiler and today we're going to be doing unit number 5 lesson number 8 modeling with systems of inequalities. I'd like to remind you that if you'd like to get a copy of the worksheet that goes with this lesson and in the company homework, just click on the video's description. As well, don't forget that at the top right hand corner of all of our worksheets we have a QR code that will allow you to use a mobile phone or a tablet to scan that QR code and bring you right to this video. Anyhow, let's get into a practical problem that involves modeling with systems of inequalities. Let's take a look at exercise one. John mose yards for his father's landscaping business for $10 per hour, and also works at a bakery for $15 per hour. He can work at most 52 hours per week during the summer, and he needs to make at least $600 per week to cover his living expenses. All right, let's make sure that we really have all of this down. Okay? He's going to work first landscaping business for $10 per hour, work at a bakery for $15 per hour. He can work at most 52 hours, 52 hours per week, apparently I don't care about the word week. And he needs to make at least $600 per week to cover his living expenses at least. Letter a if John works 14 hours mowing and 30 hours at the bakery, does this satisfy all the problems constraints, right? Now there's really two constraints. He can work at most 52 hours, and he needs to make at least $600 per week. So I'd like to get a pause the video and see if those working hours satisfy all the constraints. Let's take a look. Well, that's something that's pretty easy. Let's see if he's worked too many hours. So 30 plus 14, right? That gives me 44. He can work at most 52 hours, so that works. Let's see if you made enough money. Let's see. He's going to work. 14 hours mowing, and he's going to get paid $10 an hour to do that. So he's going to make a $140 mowing. Right? Then he's going to work 30 hours at the bakery. Times $15 an hour. That gives me $450. Okay. Add those together and we're going to have $590. And he needs to make at least 600. So no. One of the two conditions is one of the constraints is met, but not the other one. Well, let's try to model this, okay? Letter B says if X represents the hours John spends mowing. And why represents the hours he spends at the bakery? Write a system of inequalities that describes the scenario. Well, one thing that's pretty easy is that first condition. When I add X to Y, it's got to be less than or equal to 52. You can work at most 52 hours. So when I add the hours that he spends mowing to the hours he spends baking, right? Then that's got to be less than or equal to 52 hours. Okay, so that's kind of the equivalent of this condition. Now, how about this? Well, X represents the number of hours mowing, so ten times X will represent the amount of money he makes mowing. Then if I add to that, 30, no, not 30. One mistake. If I add to that 15, $15 per hour, times the number of hours he spends at the bakery, that's got to be greater than or equal to 600. This is what's known as a system. Of inequalities. Right? Letter C says, if John must work or walk, apparently. That's not right. If John must work a minimum of ten hours for his father, will he be able to make enough money to cover his living expenses. Show the work that leads to your answer. Well, I'd like you to play around with that a little bit. If John must work, not walk, but if John must work, a minimum of ten hours for his father will he be able to make enough money. Play around with that. All right, well let's see. Let's see, he's going to work ten hours for his father, right? Times $10 per hour. He's going to make a $100 just for his father. Right? Now how much time does that leave him? While he can work at most 52, so subtract off the ten hours, he's got to work for his father. He's going to be left 42 hours that he can work at the bakery, right? So if he takes 42 hours and we multiply that by $15 per hour, all right, what am I going to end up getting? I'm going to end up getting a value that I don't have written down for some reason. Let's do 42 times 15. And we're going to get 600 and $30. Well, obviously, that means that we're going to make $730 in this scenario. Now he's got to make at least 600. It's got to be greater than or equal to $600. And so yes, he does make enough. Okay? So he can work that minimum amount of time that his father needs him to and still make the $600 he needs to as long as he can work the other time at the bakery. All right, we're going to keep working with this problem because it's a very rich one. Let me clear out the text. All right, that's all gone. Let's take a look at the next part of the problem. Letter D says, graph the system of inequalities with the help of your calculator if needed on the axes below. Use the space below to think about how to graph these lines. So here's that system of inequalities that we had, right? So what we want to do is somehow graph them. Now, at this point, the expectation is you've played a little around a little bit with graphing, systems of inequalities. So one thing I would do is I would immediately put this X on the other side. It would become negative. And I would end up having this. Now, remember, when we graph an inequality, what we do is we first graph the equation that the inequality is based on, which is going to be Y equals negative X plus 52. Now the other equation, that one's going to take us a little more work. So let me put that thing down here. We're going to get 15 Y is greater than or equal to negative ten X plus 600. Divide by 15, divide by 15, divide by 15, all right? And we're going to get Y is greater than or equal to we could reduce the negative ten fifteenths. It does reduce nicely to negative two thirds. All right. And then 600 divided by 15 is 40. Okay? So I think I am going to use the calculator right now to help me out because I'm going to have it graph these two lines. Y equals negative two thirds X plus 40. And Y equals negative X plus 52. So let's open up the TI 84 plus. All right. Let's go into Y equals. Okay. And Y one, I think I'm going to put in the negative X plus 52. And in Y two, I'm going to put in negative two thirds X plus 40. All right. I have to think a little bit about the window. But remember, X represents the hours working at mowing. Why represents the hours of the bakery? Neither one of them can exceed 52, probably most of them are well short of it. So why don't we go into our window? Let's put X-Men in as zero. Let's put X max in at, I don't know, let's say 60. Let's do Y min also 50. Sorry, Y min also zero. And let's do Y max also 60. Okay? Once we have them all in, let's hit graph. See how they look. All right. Now it's kind of important to remember which ones which so here we've got the actually, let me use. The prefab line maker here. Do the red one and maybe label that. And why not? That'll help. Y equals negative X plus 52. And I think I'll go with a blue line for the other one. This, maybe not the best in terms of my spacing. Y equals negative two thirds X plus 40. All right, actually, let me go back to red. Now remember, what I'm really doing is I'm shading Y is less than or equal to negative X plus 52. So that's going to be. Everything down here. All right. Y is greater than or equal to negative two thirds X plus 40. That's going to be everything above it. Right? So where is the solution to a system of inequalities? Well, it is now this triangular region. Right here. It's the overlap, right? Wherever the shading overlaps is the solution to our inequality. So that is the solution to our inequality. Every point within here makes our system true. All right? Satisfies those conditions. I was kind of interesting. Put that guy back. Letter E John's father needs him to work a lot at the landscaping business. Show the point on the graph that corresponds to the greatest number of hours that he can work. While still covering his expenses. All right, so it's got to be a point somewhere in here. Points out here, right? He's mowing too much. Points in here are fine, right? If I'm in here, let's say, like a mowing in here, that's all right. So this has got to be it, right? Whatever point that is. Is the greatest amount that he can mow while still being in this region. All right? Um, letter F then says algebraically find the greatest number of hours, right? That he can mow, work for his father, and still cover his expenses. So how do we find that point algebraically? Pause the video for a second, think about that. All right, we can find that point algebraically. By solving the system of equations, equations, not inequalities, given by these two. So in other words, I can solve for when negative X plus 52 is equal to negative two thirds X plus 40. All right. Kind of up to you, but I think I'll add a positive X to both sides. That'll leave me with 52 over here. Now negative two thirds plus positive one would be a positive one third X think about that for a second plus 40. All right, I'm going to rewrite that up here so that we can see it a little bit better. 52 equals one third X plus 40. I can now subtract 40 from both sides. And I'll get 12 equals one third X so I could multiply both sides by three sorry about the tight space. And I find that the most that John can possibly mow is 36 hours. So if he mows any more than 36 hours, if he's out here, then he can not make enough money in order to satisfy that condition of making $600. All right. I'm going to get rid of that. Copy down anything you need to. Okay. Now, sometimes the hardest thing to do with systems of inequalities is simply do the modeling itself. So the next problem, all we're doing is writing down systems of inequalities, but we're not going to solve them. We're not going to graph them. We're not going to find their intersection regions or anything like that. Let's just read. But array, Frank is putting together a bouquet of roses and daisies. He wants at least one rows. That's important. Any time you see things like at least underline it, he wants at least one rows and at least two more daisies than roses. Okay? Rose's cost $4 each daisy's cost $2 each. Frank must spend $40 or less on this bouquet. If our represents the number of roses, he buys, and D represents the number of daisies, write the system. Okay. Well, let's try to convert that at least one rows. That's pretty easy. So the number of roses are must be greater than or equal to one. Okay, he wants at least two more days than roses. So the number of number of days he's going to have is going to be at least. That's going to be greater than or equal to. The number of roses he has plus two, right? Two more, two more days than roses. Take the roses, add two, get the daisies. Right? Now, how do we do that last piece? Frank must spend $40 or less. Well, let's calculate how much he's going to spend. $4 times the number of roses is how much she's going to spend in roses. Plus, $2 times the number of daisies, how much he spends in daisies. Must be less than or equal to $40. There's our system of inequalities. Nobody said you could only have two, and we've got three. Let's take a look at one that has something to do with a diet bar. Let her be. A diet food company is attempting to create a non carb brownie. Yum. Composed entirely of fat and protein. The brownie must weigh at least ten grams. So it's got a way at least ten grams. But I have no more no more than a hundred calories. It's got to have a hundred calories or less. Fat has 9 calories per gram, protein has four calories per gram. If X represents the weight in grams of protein, so X is the protein. And why is the fat? We're going to write the system. All right, well, it's got a way at least ten grams. All right, so if we take the weight in protein and the weight, in fact, an add them together, that's got to be at least greater than or equal to ten grams. It's got to be greater than or equal to ten grams. I can have more than a hundred calories. Well, how would we get the calories? Well, we'd take 9 calories per gram, a fat, and we'd multiply its weight in fat. Let's see, what was its weight in fat? Ah, that's why, right? And then we take four calories per gram for the protein. And multiply it by how many grams of protein we have, which is X and that's got to be less than or equal to 100. All right. So only two inequalities this time. But still a system. Okay? All right, I'm going to clear these out. Let's take a look at the last problem. All right, exercise number three. A drama club at a local high school is trying to raise money by putting on a play. They have only 500 seats in the auditorium that they are using and are selling tickets for these seats at $5 per child's ticket and $10 per adult ticket. They must sell at least $2000 worth of tickets to cover their expenses. All right, so that's an important constraint. At least $2000 worth of tickets. What do we have? X represents the children's tickets, Y represents the number of adult tickets sold. Write a system of inequalities that models the situation. All right, well, we have only 500 seats, so we can't sell more than that. So if we take the number of children's tickets, add to it the number of adult tickets, that's got to be less than or equal to 500. We can't sell more than what we have. But we have to make $2000. So if we take $5, which is the cost of a child's ticket, and we multiply it by how many childs tickets we have, then we add to that, let's see, what is it? $10 per ticket times the number of adult tickets we have, which is why that's got to be at least $2000. So there's our system of inequalities. Right? Letter B, using technology sketch to the region in the coordinate plane that represents solutions to the system of inequalities. All right, well, it's kind of like what we did before. Let's take that first inequality. Let's move the X to the other side. It's going to become negative there. So it's going to be negative X plus 500. Right? Likewise, the other one, a little more work. We're going to get ten Y it's greater than or equal to negative 5 X plus 2000. All right, I can divide everything by ten. And we'll get Y is greater than or equal to. You could reduce that and becomes negative one half, or you could leave it. This then I think it's advantageous to reduce it. 200. That's very helpful to have our system of inequalities kind of rearranged like this. Because now we can put the equation Y equals negative X plus 500. And Y equals negative one half X plus 200. Into our calculators. So I'm going to open up the TI 84 plus snap. All right, it's all open. Let's go into Y equals. Okay, we're going to clear out our two equations from before. Okay. In Y one, I'm going to put in negative X plus 500. And Y two, I'll put in negative one half times X plus 200. Okay. Let's go into our window. Now remember, we're talking about the children's tickets in the adult tickets. So again, there's at most 500 tickets. We don't need to make our axes any bigger than 500. We'll go a little bit bigger just to fit everything. But let's do X-Men zero. X maxx let's go 600 again. Why not? Why men? Let's go zero. And Y max, let's go 600 again. And now let's hit graph. Okay. Now again, I want to graph and I want to do some shading at the same time. So let's again go with sort of two different colors. Let's start off with blue, yeah, looks good. We'll do the first one with that. So. Let's come up with its equation. Y equals negative X plus 500. Now, of course, what we've got is everything less than this. I'm not going to shade anything down here because these are non viable solutions down here. Let's now change our line to red, all right? And our other one kind of looks like this. And then that's why it's greater than. Well, let me throw the equation on first. We'll put it up here. It's oops. It's color coded. So Y equals negative one half X plus 200. But now we're shading above because it's Y is greater than. All right. So our viable region again, maybe I'll outline that in the screen. Our viable region is this. That's our solution set. That's our solution set. Okay? Now, let her see, says, if the students want to sell exactly 500 tickets and make exactly $2000, how many of each ticket should they sell? Why is this answer not realistic? Okay? So at this point, we want to have exactly 500 tickets sold. And we want to make exactly $2000. So let's solve this system now of equations. And let's do it by the method of elimination. Do you remember in the method of elimination? Let's do the method of elimination. What I'm going to do is I'm going to choose to eliminate either X or Y all right, I think I'll eliminate X and I'll do that by multiplying both sides of this equation. By negative 5. And that will give me negative 5 X -5 Y equals negative 2500. Then my other equation, I'm not going to change 5 X plus ten Y is equal to 2000. When I add these together, the X is eliminate, I end up getting 5 Y equals negative 500. Divide both sides by 5, and I get Y equals negative 100. Now, of course, that's not realistic because we can't sell. Can't sell a negative. Amount of tickets. It's a non viable solution. Can't sell a negative amount. Just can't do it. All right. So I'm going to clear out all of this and then we'll wrap it up. Great. So systems of inequalities actually show up a lot in the real world. They oftentimes serve as constraints and engineering problems or economic problems. Just being able to write the system is actually quite helpful. Apart from graphing it and thinking about viable solutions and things like that. So that's where a lot of the practice will come on both the homework and I wouldn't be surprised to see it tested as well. All right. Well, until next time, let me remind you that this has been an eMac instruction common core oops common core algebra one lesson. My name is Kirk weiler, and it's all next time, assuming that I can keep the camera on me. Keep thinking. And keep solving problems. 2b1af7f3a8