 Well, welcome to episode 26. This is Math 1050 College Algebra, and I'm Dennis Allison. This episode has to do with solving systems of nonlinear equations and then solutions to inequalities. We'll be looking at systems of nonlinear equations. We've been talking about systems of linear equations. Now we'll do nonlinear equations. Then we'll look at graphing solutions to systems of inequalities. And finally, a very important application of those graphs is something called linear programming that's used quite a bit in business and economics. OK, let's begin with an example and a problem that I think you've solved in previous courses like an elementary or intermediate algebra. Suppose I have this system of linear equations, x plus y equals 4, and x minus y equals 2. Now you may remember that in previous algebra courses, there were several ways to solve this. You could graph the solution, which I'm going to do in just a minute. But there were two other methods. There was something called the substitution method. And by the way, can anyone just sort of briefly describe how the substitution method works if you were going to solve this? Solve for either x or y, and then put that into the other equation. And then substitute into the other equation. And what that does essentially is it will eliminate one of the variables. And instead of two equations, two unknowns, you'll have one equation and one unknown. But then the third method is called the elimination method. And this is where you add or subtract equations to eliminate a variable. And we've been doing something like this with Gaussian elimination with back substitution, but in a much more abstract manner using matrices. Well, what I'd like to focus on for the moment is just how I would graph the solution of this system. So this should sound familiar. But let me just point out some shortcuts, I think, that you might find helpful that will come up in some later problems today. If I want to graph this first line, x plus y equals 4, I don't have to solve for y to get it in the form y equals mx plus b. But here's a shortcut that I use quite a bit, and that is to just locate the two intercepts and then connect them with a straight line. Where does this line cross the x-axis? This first line right here. You remember to find the x-intercept, you let y be 0. And so what did you say, Stephen? Four. Four, yeah. So this crosses the x-axis at 4, so I'm going to go over here to 4 and put a dot right there. And where does this line cross the y-axis? Four. Also at 4. Because to find the y-intercept, you let x be 0, and y is 4. So if you just plot those two points, and if you connect them then with a straight line, then that should be a rather accurate graph of x plus y equals 4. Now, you see the alternative would have been to solve for y. y equals negative x plus 4 from that equation. And then the slope is negative 1. The y-intercept is 4. So the y-intercept is 4. The slope is negative 1. You go over, down 1. And then sure enough, you'll get the same line. But I think in this case, in this example, I think intercepts are a little bit faster. Now let's graph the second line in the same way. In the second equation, Susan, what's the x-intercept of that line? Two. Two, exactly. Yeah, so you let y equals 0, x is 2. So it crosses the x-axis at 2. And Lene, what's the y-intercept for that line? Negative 2. Negative 2, exactly. So I'll go down here and plot negative 2. And so if I connect those points, I'm graphing the other line. I'm trying to do this fairly accurately because accuracy is everything in this sort of a problem when you're looking for the intersection of two graphs. I hope I wrote that big enough there. So here's the intersection. And you see this point is on the first line, so it should solve the first equation. And this point's on the second graph, so it should solve the second equation. Therefore, it should solve both equations. And what does it appear, or the coordinates, at that point? 3 and 1. Looks like 3 and 1. Yeah, so this is the ordered pair, 3, 1. And so we would guess that our solution is x equals 3, y equals 1, or many textbooks just write this as the ordered pair 3, 1 in the back of the book. Sure enough, 3 plus 1 is 4, and 3 minus 1 is 2. OK, now I'm not thinking that this is actually something that's new to you today. But now let me take this idea a bit further and let's take a system of nonlinear equations. And I want to try solving it, first of all, by graphing. So suppose we have this set of equations. Suppose we have y equals the absolute value of x plus 2, and y equals x. And I want to solve that system of equations for x and y. Now, if I graph both of these graphs, and we certainly know how to graph y equals x, and this is a transformation of a fundamental function, I think we can get a fairly accurate bead on what the solution is. So let's see. If I graph y equals the absolute value of x plus 2, this is the absolute value function shifted which way? Left. 2 to the left, exactly. So its vertex is right here at negative 2. And you remember the target points were to go over an up one, to go back an up one. So this is sort of ancient history for us now. But this graph looks like this. OK, so this is the graph of y equals the absolute value of x plus 2. On the other hand, if I graph y equals x, what does it look like? That's a 45 degree line right through the origin. It looks like this. Oh, I'm not making it very straight. Imagine that straight, if you will. So this is y equals x. Where do these two graphs cross? They never do. You know, these two graphs don't cross. Because this branch of the absolute value function has slope 1. I'll just write that on there. And the branch over here has slope negative 1. And of course, this graph has slope 1. So these two branches have the same slopes with a parallel, so they're not going to intersect. So there are no intersection points. And therefore, there's no solution for that. Now you see the alternative would be in intermediate algebra. You might have used the substitution and said x equals the absolute value of x plus 2. And you would have proceed by algebraic methods to solve this equation. This is where I'm substituting x for y up there. But instead of substituting, I'm using the graphical method to find that there's no solution for this. On the other hand, if I had chosen, what if I had chosen? Let me just take that one out and put in something that will have a solution for us. Suppose I were to change this to y equals negative x. Then now I think we do get a solution. Because y equals negative x is the 45 degree line coming down to the origin. So it goes like this. This is y equals negative x. And it looks like these two graphs intersect at a point. And although I don't think my graph is perfectly accurate, I bet we can surmise that that's the point negative 1, 1. Negative 1 and 1. So the ordered pair is negative 1, 1. So x is equal to negative 1. And y is equal to 1. Or we could write it negative 1, 1. Let's just check that and see if that really is a solution. If I plug in a negative 1 right here, this is the absolute value of 1, which is 1. Yeah, y is 1. And if I plug in a negative 1 right here for x, the negative of negative 1 is 1. So it does solve both equations. OK, if we go to the next graphic, we have a different system of equations that I can solve by graphing. Although trying to determine points of intersection in graphs can become more and more difficult to do. Let's try this in this problem. It says solve the system y equals x squared minus 2 and y equals 2x plus 1. So I'm going to write these here on the green screen. x squared minus 2 and y equals 2x plus 1. Let's try solving this graphically and see what happens. Can anyone tell me what the first graph will look like? It's going to be a parabola. It's a parabola, yeah. And what else about it? Shift to the right, too. No, it's going to be shifted down to. It's going to be shifted down to, right. If there's a negative 2 outside, it's shifted down. This actually provides a good review for the final, which isn't that far away, because you'll be graphing these transformations of fundamental graphs on the final exam. So let's see. If I take the parabola, shift it down to units, here's my new origin. The target points, they go over and up one and go back and up one. So those are my three target points. And when I draw the graph, now here's the rather delicate part. I'm not really getting a totally accurate graph for the rest of it. And so the accuracy of my answer depends on the accuracy of my graph. Now, if I graph the second equation, of course, this is a diagonal line. It crosses the y-axis at 1. And the slope is 2, or 2 over 1. So if I go over 1, I should go up 2. And therefore, if I connect these points, it looks like we have one solution right here. Anyone want to take a wild guess what that solution is? Negative 1, negative 1. Negative 1, negative 1. Yes, but you see, everything depends on how well I draw the graph. And I haven't been that careful with target points. But I think we can read that as negative 1, negative 1. How could I check that to make sure that is a solution? Plug it back into our equation. Of course, just plug it back in and see. If I put in a negative 1 for x, negative 1 squared minus 2 is negative 1. And 2 times negative 1 plus 1 is negative 1. So this is a solution. Now, the other point, the way I've labeled this looks like that could be directly under 3. And here's 1, 2, 3, 4, 5, 6. But you know what, this should actually be 7. So what that means is my scaling is off, or my graph is a little off. This should be the point 3, 7. Now, I'm not really concerned about that because I wanted to demonstrate to you the pros and cons of graphing. If you were to graph these on a graphing calculator, then the graphing calculator plots not just three points, but say on a TI-82, like I've been using in class previously, it plots 95 points from left to right. And it connects them with straight line segments. So this point would be much more easy to predict using a graphing calculator than using my hand-drawn graphs, or maybe your hand-drawn graphs. Let's just double check that answer. If I put in 3 for x, 9 minus 2 is 7. And if I put in 3 for x here, 6 plus 1 is 7. OK, so that means what we really need is an alternative to graphing if we really want to get accuracy. And we're going to be drawing these graphs by hand. We really need some alternative method. So let's try using the same example. Let's try using the substitution method and see how this would work. Who could maybe on the spur of the moment manufacture a procedure for the substitution method for this? Matt, what would you do? I would try to solve one of the questions in terms of another, for instance. Well, we have y. And then put like that into the other equation. Yeah, so if y equals this and if y equals that, then these things must equal each other. Or the way Matt puts it, he's saying, let's substitute this expression for y in the other equation. So that's going to become 2x plus 1 equals x squared minus 2. And you see essentially what the substitution method does is it eliminates one of the equations, but it also eliminates one of the variables. So instead of a 2 by 2, I have a 1 by 1 system. And we call it a quadratic equation in this case. So 0 equals x squared minus 2x minus 3. And we can solve that by factoring x minus 3x plus 1. Now one of these two factors is 0. If the first one is 0, then x equals 3. If the second one is 0, then x is equal to negative 1. But how do I get the y that goes with it with either one of those? Plug one of those x values back into either equation. Into either equation, because this is where the two graphs are supposed to intersect. So I should be getting the same y value in either case. So if I substitute 3 into the first equation, 9 minus 2 is 7. In fact, we saw that just a moment ago. And if I plug in negative 1 into either one of these equations, say the first one, 1 minus 2 is negative 1. So now I have more accurately my points of intersection. The solutions are negative 3, 7. And now I should say and, because these are both solutions, negative 1, negative 1. OK, there is yet another way to solve this system of equations. And that's to use the elimination method. And although this system, the way it's written, I would say, is more obviously substitution because you have y equals this and y equals that. How would the elimination method work? Multiply one of the equations by negative 1 and then add them together. Exactly. Or you could just subtract them. Or you could multiply the second equation by negative 1 and then add them together. And that would eliminate the y. And you'd be down to one equation and one unknown. You know, rather than doing that for this system, I think we've probably seen enough of this. Let's go to the next graphic and we'll see another problem where we're asked to use the elimination method to solve a system of two equations and two unknowns. Now, you notice these are not linear equations, but we can still use the same methods. The first equation says 2x squared plus y squared is 17. And x squared minus y squared is negative 2. And I guess graphing is not as reasonable as we'd like because we don't know how to graph the first equation. We don't know how to graph the second equation except by what primitive method? Yeah, making a table, plotting points, but that's probably not our favorite way to approach this. So we could use substitution, although I don't see any variables isolated, or I could use elimination. And I think that's probably the fastest way. In fact, that's the way we've been asked to. Let me switch to another marker. I think elimination probably is the most efficient way to solve this. So suppose I just add these right now. How many x-squares will I get? Three. Three x-squared. And the y-squares cancel off. And we get 15 on the other side. So I've eliminated a variable. I'm down to one equation and one unknown. It's not a linear equation, but we can solve it. x squared is equal to 5. And so x is equal to what? Is that the only answer? Plus or minus. Yeah, plus or minus. See, we're actually looking for every possible solution. So we'll take anything we can get, plus or minus square to 5. Now to get the y value, let's substitute plus square root and negative square to 5 up in here. And by the way, when you square that, you're going to get 5, like it said here. So either one of these, when I substitute it in back up here, I'll put a 5 for the x-squared. So this becomes 2 times 5 plus y-squared equals 17. So y-squared is equal to what? 10. Not 10. 2 times 5 is 10. 7, exactly. So y is going to be plus or minus the square root of 7. OK, well let me ask you this. How many ordered pairs do I have that are solutions for this system? Four. We actually have four. That's right. Because you see, with the square root of 5, I get a plus or minus square root of 7. And with negative the square root of 5, I get plus or minus the square root of 7. So we actually have four ordered pairs. So if I were to list the solutions over here, that's an L, isn't it? You'll have to take my word for it. The solutions would be the square root of 5, the square root of 7, the square root of 5, negative the square root of 7. But then there's also the negative square root of 5, square root of 7, and negative square root of 5, negative square root of 7. Now, of course, that's a lot of writing. So heck, why don't we just shorten it and write it this way? We'll say plus or minus the square root of 5, plus or minus the square root of 7. And when you see this written in a book or elsewhere, this means you take every possible plus or minus here and every possible plus or minus there, and that's a shorthand for all four solutions. The solutions didn't turn out to be integers. In fact, they turned out to be irrational numbers, but that's OK. They are the numbers that will solve both equations. OK, let's see. What is our next graphic? Oh, we have an example of a problem. It's the sheet metal problem. If you can bring up that graphic next, here we go. Here's an application of a system of equations that demonstrates how nonlinear systems of equations come up in real world applications. So let's try setting it up and then try solving it. The example says that a sheet metal company has a contract to cut 48 square inch rectangles from circular pieces of metal with radius 5 inches. And we're supposed to cut them as shown. That is that the rectangle is supposed to fill up the circle. And the radius of the circle is 5 inches. What should be the dimensions of the rectangle? OK, well, let me just draw that illustration over here. You can come to the green screen, and I'll draw this. Here's our circle. The radius is 5 inches. And we're going to cut a rectangle out of this. It could actually be a square, possibly. And the question is, what should be the length and what should be the width if the area of the rectangle was 48 square inches? 48 square inches, that would be the area inside here. Well, let's see. Now, the way I would approach this, and this is actually a rather common problem in applications of mathematics, I wouldn't choose x to be the length and y to be the width. What I would choose do is to choose this point on the circle, if you imagine that there's a coordinate system passing through here, I would choose this point up here to be the point x, y, which means this half length is x, and this, should we call that a half height is y. So that we go over x, we go up y to get to that point. So what would be the total length of the rectangle in terms of x? 2x times 2y. Yeah, this would be 2x, and it would be 2y tall. So the width would be 2y. Now, you might say, well, Dennis, I don't understand why you didn't. Just let x be the whole thing. Well, what's important here is that I can make a right triangle, and the base is x, the height is y, and what's the hypotenuse? Five. It'd be five. Yeah, because see, that's the radius of the circle. So this allows me to work the five into the problem. So let's see. One of my equations is that the area of the rectangle, that'd be 2x times 2y, is supposed to be 48 square inches. So 2x times 2y equals 48. I'm going to call that equation number one. And the other equation is in this triangle right here. Can anyone tell me the other relationship between x and y in a right triangle? In fact, I'll give you a hint, the Pythagorean theorem. x squared plus y squared equals 5 squared. Exactly. OK, so that's the other thing I'll use is that x squared plus y squared equals 25. Now, you see, when you look at this, we see we have two equations. We have two unknowns. And these are clearly independent equations. I mean, neither one of them is just a multiple of the other. But these two equations give me two totally different pieces of information about x and y. So I should be able to solve these. Now, how can we solve these? Well, why don't we take number one and at least multiply it out. So I have 4xy equals 48. And the second equation says, still, x squared plus y squared is 25. Now, why don't we divide by 4 in the first equation to make that a little bit simpler? So in equation number 1, I have xy equals 12. And still writing equation number 2, x squared plus y squared equals 25. This is a system of equations. Would you say this is a linear or a nonlinear system of equations? Nonlinear. Neither one of these are straight lines. This first thing has an x times a y. The second one has squares. And in a linear equation, you should have a constant multiple of x plus or minus a constant multiple of y equals a constant. In other words, the graph of a straight line. So these are nonlinear. So I need to find a way to solve this. I don't think graphing would be appropriate because we don't know how to graph the first equation. But we do know the second equation is a circle. In fact, we already have that one graphed. But I don't think graphing would be appropriate here. So I think we're down to substitution or elimination. I don't think substitution is going to work so easily either. Because here I have x squared, y squared. And here I have x times y. It would be a little complicated to try to make those terms look alike. So that leaves me only with substitution. Which means I need to solve for a variable somewhere. Why don't we solve for y in the first equation? Y is equal to 12 over x. And now substitute that for y in the second equation. And we have x squared plus 12 over x squared equals 25. Which says x squared plus 144 over x squared equals 25. Now at this point, this is called a rational equation. I think we've mentioned rational functions. This is a rational equation because I have a rational expression in it. What would you do to make this into a polynomial equation? Times by x squared. Multiply both sides by x squared, yes. This will be x to the fourth plus 144 equals 25x squared. Now I have a polynomial equation. We've seen this before where we have a problem and we solve it. And we change it to different forms until we get it to a form that we think that we can handle. We saw exponential and logarithmic equations turn into quadratic and linear equations. I'm going to erase my illustration over here so that I have room to continue. To solve this, I'm going to get everybody on one side. And if I could factor that, I think I could get out of this. Do you think you could factor it? Let's see, if we're going to factor it, what would you put at the beginning of the parenthesis? x squared. x squared, yeah, x squared and x squared. And there's a plus on the 144. What does that tell you about the signs? That's not going to happen. If a plus here means the signs are alike and a negative in the middle says something's negative, so they must both be negative. So if we factor it, it's going to have to be on this basis. Now there are two numbers whose product is 144 and the sum is 25. And I don't know if you're thinking of this, but I'm thinking of 16 and 9. 16 and 9, sure enough, make 25. And their product is 144. Now I could factor these further, but why don't we just step out and say x squared is 16, or x squared is 9? If you want to factor those separately, you can get four factors, but I think we'll get the same roots this way. This says that x is equal to plus or minus 4, or x is equal to plus or minus 3. Now I think two of those answers I can throw out immediately. Which two do you think they are? The negatives. The negatives, because we said that the length was going to be 2x, so we were assuming x was positive, so we're down to x equals 4 or x equals 3. Now you see there's an unwritten property here that we haven't mentioned, and that is that we were assuming x and y were positive, or at least non-negative. So that's referred to as a constraint. Let me just write that word down. We're going to see the word constraint later in this episode in another regard. But that's a constraint that wasn't written down, but because of this word problem, I knew that my measurements couldn't be negative. To get the corresponding y, I'll go back to the relationship I saw right here. y was 12 over x. That'll be 12 over 4 or 3. On the other hand, the corresponding y, when x is 3, is 12 over 3 or 4. So 2x, in this case, 2x would be 8. 2y would be 6. R, 2x would be 6. 2y would be 8. You remember those are the lengths and the widths. So to simplify this answer, what would you say are the dimensions of the rectangle? 8 and 6. 8 and 6. Yeah, 6 by 8, 8 by 6, same thing. So I would just say the dimensions of the rectangle are 8 by 6 or 6 by 8 inches. And you see, the reason we come up with two different sets of answers for this is if I go back to the illustration, we erased it to make room. But I think we can see in the illustration why we came up with this answer twice. In the illustration, we had this circle of radius 5. And when I drew my rectangle, I drew it longer left to right and shorter top to bottom. But I could have also have drawn it this way. And we would have gotten the same result. So this one is the one that's 8 by 6. This one is the one that's 6 by 8. I guess I've exaggerated a little bit, but those are closer together. But there are actually two ways you can cut the sheet metal. But either way, you get this solution. Let me solve one more problem now. We don't have a graphic for it, but I think it's an interesting problem nonetheless for a system of equations that I'd like to solve. And let me just describe to you how it goes. Suppose I have a sheet metal, a piece of sheet metal that's rectangular. And I want to roll the sheet metal up to make a cylinder. And let's say I'll put a line right down the middle. This is where the two pieces come together right there. And I'm getting this out of this book. Let me get some numbers here for you. I'd like for the rectangular piece of sheet metal to have an area of 1,200 square inches. That would be the area of the sheet metal. And I'd like for the volume of the cylinder to be 600 cubic inches. So the question is, with these two constraints that the area be this amount and the volume be this amount, what should be the dimensions of the sheet metal? Well, I'm going to let h be the height of the sheet metal and h be the height of the cylinder. And for the width of the sheet metal, I'm going to begin with the cylinder. Suppose the cylinder has a radius of r. Now, when I enroll that and lay it out, how long will this be across the top? If the radius is r over there. Two pi r. It'll be two pi r. And the reason is you see whatever the circumference is of the circle, by the way, the formula for the circumference of the circle is circumference is two pi r. If the circumference is two pi r, then when you enroll it, that's going to be the length of the sheet of metal. Okay, from this we can write a system of two equations and two unknowns to solve. But I think the method for solving it will be a little different than maybe what you've seen in here before. First of all, what product is going to equal 1200? That's two pi r. Okay, I'm going to write that as, I'm going to try to write that here if I get this marker to work. Two pi r h equals 1200. Okay, that's equation number one. What's equation number two? It's going to involve the volume of a cylinder. Does anyone know how to find the volume of cylinder, Stephen? Well, it's the area at the top times the height, which is going to be pi r squared times h. Okay, I'm going to say it this way. It's the volume of the base times the height. Stephen said the volume of the top. Either way, the volume is the area of the base times the height. That's true for any cylinder, whether it's a circular cylinder or a triangular cylinder or what have you. So that would be the area of the base, that's pi r squared times h. Okay, well the volume is 600. I'll put that in for v and our second equation says pi r squared h equals 600. Okay, I'd like to reduce this, but I'm not going to reduce it as much as you might first guess. I'm only going to divide out the two. So equation number one is pi r h equals 600. And equation number two, I'm going to write this way. I'm going to say r times pi r h equals 600. In other words, all I'm going to do is factor out the r. I'm not really reducing it, I'm just rewriting it. Does anyone have an idea why I wanted to write it this way? Because now you have a pi r h in the top equation and a pi r h in the second equation. Exactly, I'm going to use a substitution. I'm going to make a substitution. But the substitution isn't just for r or for h. I'm going to take out three things at once. Pi r h is 600. I'll put 600 right there in number two. We have r times 600 equals 600. So I'm using the substitution method, but in a little bit different way. So this says that 600 r is 600. So r is one. Now that would be in inches, one inch, because our volume was given in cubic inches, our area was given in square inches. But r is one. Okay, now once I know that r is one, I can go to either of these equations or either of these equations and I can get h. Let's go to equation number one in this form. Pi times one times h is 600. So h is equal to 600 over pi inches. You know what that says is if you want to get these dimensions, the width of the sheet metal should be two pi times one, that's a little over six, a little over six inches wide, two pi times one. And the height should be 600 over pi. Can anyone just tell us approximately how much that is, just sort of a ballpark figure? Around 200 inches. Yeah, it's about 200. Would you say it's a little bit more or a little bit less than 200? It's a little bit less because you're dividing by more than three. So it's slightly less than 200. So this piece of sheet metal, I drew it in this rectangular form, it should actually be much, much taller than it is wide because it's about six inches wide and it's about 200 inches tall. Even this isn't the right measurement. So this is gonna end up being a very tall, thin cylinder in order for us to get those dimensions. Well, mostly what I wanted to do is to show you how this substitution is made right here. You don't always have to get things down to individual variables. Okay, let's go to the next graphic and we're gonna shift gears now and look at graphing inequalities and this will lead us to something called linear programming. My guess is you've probably graphed an inequality like this first one before. The problem says to graph the solution for 2x plus 3y is less than six. Now, you know, let me put a little cloud over here. Here's what I'm thinking is that I wanna graph 2x plus 3y equals six. And when I draw that graph, well, I wanna make it solid or dotted. Dotted. Yes, why dotted, Steven? Because it's not less than or equal to. It's... Yeah, this is strictly less than. So equal to is not allowed and when I graph this, I'm finding the points where it's equal. So I'll make it dotted to indicate I don't want the points on the line but just to one side. Let's see, this says that 3y equals negative 2x plus six. So y is equal to negative two-thirds x plus two. Which means I could graph this line in slope-intercept form. But if I use my shortcut that I mentioned earlier, I could just find the two intercepts for this line and I think that would be quicker. So let's try doing that over here. Can anyone tell me where this line crosses the x-axis? At three. At three, right. Because you see when it crosses the x-axis, y is zero. So we have 2x equals six or x equals three. And where does this line cross the y-axis? Two. At two, yeah. Now I would say that's quicker than trying to graph this line with a fractional slope on it. But if you want to graph it this other way, that would be fine. And now I'm going to draw the line but we'll make it dotted across here. Well, if the solutions are not on that line, where are the solutions? Where would I find solutions for the original inequality up here? Where would they be? Below the line? Well, they're either above or below. If they're not on the line, they're either above or below. So what we're going to do is to pick a test point and if a test point solves it, every point on that side will solve it. The test point that I like to pick is the origin. If the line doesn't go through the origin, let's choose the origin because it's so easy to plug in, two zeros. If you plug in a zero here and a zero there, you get zero plus zero is less than six. I think we could agree with that. So that says the origin works. And if the origin solves it, every point over here will solve it. Now let's just look at that and see why. Suppose this point makes it less than six. I'll write less than six beside it. Suppose there's a point over here that makes it greater than six. Well, that would indicate that somewhere between here and here, there must be a place where the point makes it exactly zero because we've gone from less than, excuse me, not zero, but equal to six. Because if here were less than six and here were greater than six, then somewhere in between, somewhere in between, there should be a point where it's equal to six. But all the places where it's equal to six are over here. So there's no way we can change from less than six to greater than six. So what that tells me is all the points have to be less than six if one of them is less than six. So I would pick all the points on this side. And the way you normally see this written in a textbook is this is shaded in like this instead of drawing those arrows. You know, there's an expression in American history that goes like this. It says, as Maine goes, so goes the nation. Let me write this down here. This is the state of Maine. As Maine goes, so goes the nation. Have you ever heard that before? So goes the nation. Matt, what does that mean? You said you've heard that before. This quote. Do you know what the context of that is? Wasn't it in colonial period how people of Maine felt and decided that's how the rest of the nation? Well, you're close. It's not during the colonial period but it's actually during presidential elections. And it goes from the 18th century well into the 19th century. I'm thinking right about the time that Abraham Lincoln was elected. Maine would vote for their governor in September and then the presidential election would be several months later. And for many elections, however, Maine voted for governor for whichever party they voted for, whichever party won. That was also the party that won the national election. And so it became quite popular in newspapers to say as Maine goes in a gubernatorial election, so goes the nation in the presidential election. Now, if I'm not mistaken, that continued up until the 1920s or 1930s when finally Maine voted for a different party than the party that won the presidential election. And as I understand it, today they say as Maine goes, so goes Vermont. Like, I mean, do we really care how Vermont's gonna vote? That wouldn't be that surprising actually if Maine and Vermont are right next to each other that they would vote alike. But it doesn't seem to hold anymore in presidential elections. Well, in this case, as the origin goes, so goes the entire region. And if the origin had failed to solve the inequality, I would have chosen the other side and chosen all the points over there. So as one point goes, so goes all the points in that region. Well, this leads us to a little bit more complicated problem. Let's go to the next graphic. This one has several inequalities in it. And we're asked to graph four different inequalities and find the points that solve all four of these at one time. So let's take these one at a time and graph them. What would the first, if I made that an equation, what would the first equation look like? A circle. It'd be a circle. Yes, if you put an equal sign in there, that'd be a circle. What would be the radius of the circle? Square root of five. Square root of five, okay. So I'm gonna graph X squared plus Y squared less than five, which means I'm gonna have to make this into a dotted circle. If you'll have to take my word for it, that's a circle. That's the square root of five there and the square root of five here. And would I wanna pick the points on the inside or the outside of the circle to be the solutions of that inequality? Whoops, I better move that over a little bit. Hadn't I? Inside the circle. It'll be the points inside. How did you know that, Steven? I looked at the origin and zero squared plus zero squared is less than five. Yeah, you know what? As main goes, so goes the nation, you bet. So what he's saying is choose the points on the inside of the circle, okay. Now I'm not gonna shade it in because I'm gonna have several graphs here and if I shade them all in, it's gonna be hard to read them. So I'll just put, maybe I should just put a couple arrows in there to remind us we want the inside. Okay, the next inequality is X minus Y less than one. So of course what I'm gonna be graphing is X minus Y equals one. How would you graph this now? What would you do to graph it? I'm thinking, why don't we find the intercepts? That seems really quickly, quick to me. What's the X-intercept here? A negative one. The X-intercept, let's see, would let Y be zero. So we'd get a one. Let's see, now the square root of five is a little over two. So that's probably two. This is probably one. So here's my X-intercept of one. What's the Y-intercept? That's negative one. That's negative one, yes. So that's probably negative two. This is about negative one. So we said negative one right there. And when I draw that line, will I make it solid or dotted? Dotted. Dotted, of course. So dotted goes by here. Okay, now one of the things you have to be aware of when you're graphing multiple inequality solutions, you need to find these points of intersection so that you know where the curves end and you switch to the other curve. So I'd need to find out where does the line cross that circle. And to do that, let's just come over here and quickly solve the system of equations. That's where X squared plus Y squared equals five meets X minus Y equals one. And if I solve for Y, I think Y would be X minus one there. So if I substitute that in, X squared plus X minus one squared would equal five. So what I'm doing is solving the system of equations. And this would say X squared plus another X squared minus two X plus one is equal to five. So two X squared minus two X minus four is zero. X squared minus X minus two is zero. X minus two X plus one plus one is zero. I'm kind of hurrying along here because this is, I don't want us to get slowed down too much by this. So either X is two or X is negative one. So these intersect at two and here they intersect at negative one. Now by the way, when X is two, when X is two, what is Y in this equation right here? When X is two, what is Y? One. One. So that says along this line, this must be the point two, one. And at the other end, when X is negative one, what is Y? When X is negative one. Negative two. Negative two. Okay, so I found the intersection for the region. I don't know if it's this side or this side. We'll have to decide. In fact, let's do that now. The inequality said X minus Y was less than one. Does the origin satisfy that? Yes. Yes, so that means I want to choose points above. So we're restricted now to this portion of the circle. The other inequality, oh let's see, the other inequality, sorry I wrote on top of that, is X greater than or equal to zero. That says over here and Y greater than or equal to zero. That says we're, let me draw it over here. We're above the X axis. So I think what that limits us to is this region. Now I have two end point or two vertices located. There's a vertex here, of course that's zero, zero. And there's a vertex here and that's zero, square root of five. So I have all the vertices located and I have the curve. This would be solid. This would be solid. This is dotted. And this is dotted. These two are solid because it did allow for a quality right there. Okay, what I'd like to do now is to move on to the linear programming problems. And let's bring up the next graphic and I think I can use this to explain how linear programming works. This is a not uncommon business type problem and it allows me to show you what linear programming is about. The problem says a company manufactures color and monochromatic computer monitors. Now let's suppose that the profit for each of these is $75 and $50 respectively. That doesn't sound like much of a profit but we'll say that's the number for the purpose of the argument. And suppose that only 12 monitors can be produced per day. This must be a pretty small company. They can only do 12 of them a day. That's what, one and a half an hour or something like that. And at least twice as many monochromatic monitors are sold as color monitors. Now under these conditions, what is the most profitable way to manufacture these monitors? Okay, well let's come to the green screen and let me write down the information that we have. We have color monitors and we have the monochromatic monitors. The profit on a color monitor is $75. The profit of a monochromatic monitor is $50. We said that we could manufacture at most 12. So no more than 12, that would be per day. And we said that we could sell, let's see, at least twice as many monochromatic monitors as color monitors. Okay, and the question was, we wanna maximize profit. I'm gonna call that Pmax for profit. So we wanna know what would be the most profitable way to manufacture these monitors? Like should we manufacture all monochromatic? Well, not necessarily. We do sell more monochromatic than color, but you see color makes more profit for us than monochromatic does. So let's see if we can assemble this information into a mathematical form. You see, the problem was written in English. I've now abbreviated the English and now we're gonna go to a mathematical form. First of all, let's focus on the profit function because we're trying to maximize that. I'm gonna suppose that we manufacture X color monitors and Y monochromatic monitors. If I make $75 in every color monitor, then I'll make $75X when I sell or when I make X of those monitors. And then I'll get $50 for each Y. This is referred to as the objective function. It's the one that you wanna maximize or minimize. This is the objective function right here. Now we have certain constraints and the constraints I'll write below here. First of all, we can manufacture no more than 12 monitors altogether. In other words, X plus Y must be less than or equal to 12. On the other hand, we know that we can sell at least twice as many monochromatic as color. So that says that the Y is greater than or equal to two times the X because the monochromatic is at least two times as big as the X. You know, there are two other conditions. They are that X is greater than or equal to zero because you can't manufacture a negative number of monitors and Y is greater than or equal to zero. You know, there is yet another condition and that is that X and Y, I suppose, would be integers. We're thinking about whole monitors here. I suppose you could manufacture half of a monitor but there might be some argument you could make that X and Y would have to be integers as well. Okay, now let's put this information together in a graph. So let's see. I'm not gonna graph the profit function. I'm gonna graph only the constraints. First of all, I'm gonna graph X plus Y equals 12. Now this equation crosses the X and the Y axis both at 12, so I'm gonna put 12 here and I'm gonna put 12 here and I'm gonna put a diagonal line connecting them. I'll make it solid because this said that it could equal 12. The next equation is that Y is greater than or equal to two X. So if I graph Y equals two X, that's a line coming out of the origin with slope two. Okay, this line is supposed to have slope two right here. This is Y equals two X that I'm graphing and by the way, this was the line X plus Y equals 12. And if I graph the line X equals zero, that's this line, X equals zero. And if I graph Y equals zero, that's this line coming across here, Y equals zero. Okay, now where's my solution region? Well, in the first case, X plus Y less than 12, I think that has me go underneath because the origin satisfies that. Y is greater than or equal to two X. Now the origin wouldn't be a good test point there because that's right on the line. So what if I were to come over one to one zero? If I plug in one for X and zero for Y, does that satisfy this inequality? No, it doesn't. Zero is not greater than or equal to one. So that says I have to go, I have to go up. Okay, X greater than or equal to zero, that tells me X is over here. Y greater than or equal to zero says I'm above this line. So if you put that all together, I think what we've narrowed it down to is this region right here. I think that's my solution region. So the question is, which point in that region will make my profit function maximum? Well, you know, there's infinitely many points in that region if you allow fractions and decimals, but if you consider only the integers, there are not so many to try, but I'd like an efficient way to decide which point in there will maximize this function. Well, here's how I do this. I'm going to rewrite my profit function this way. I'm gonna solve for Y. 50 Y is equal to negative 75X plus P. I'm going up here and solving for Y. Y is equal to negative 75X plus P over 50. Now, if I graph this line somewhere in that picture, the Y intercept is P over 50. So let's say I just pick an arbitrary place. I'm gonna call that P over 50. Steven. Wouldn't you also have to divide negative 75X by 50? Oh yeah, you're right. You're right, I should have divided 50 there. Thank you, that should be three halves. Thank you. And so this is a line that sort of tapers down like that. Now, if I want to maximize the profit, that means I want to maximize this fraction. So if you imagine moving this line up, as I move it up, I get more lines that are parallel to it. And at the moment that I exit the region, that's when I have my maximum profits. That's my biggest Y intercept. Now that will occur at an endpoint, either here or here. I don't think it's gonna happen here this time. So here's the summary of what I'm trying to get across. And that is the profit functions maximized at one of the vertices of this region. Because as this line goes up, at the moment it leaves the region, it'll be leaving at a vertex. So I need to figure out the profit at the three vertices. One of them obviously won't be the answer, zero, zero, but I'll include it to make sure I mention all of the vertices. The vertex there is zero, 12. And the vertex here, I'd have to solve simultaneous equations. If I remember correctly, this is four, eight. Four, eight. That's because Y is two times X, and X plus Y is 12. Just to save time, I'll go ahead and write that one in. Four and eight. Now what's the profit in each case? Well the profit at zero, zero is zero. The profit at zero, 12 is zero plus 50 times 12. That's $600. And the profit for four, eight is 75 times four, plus 50 times eight. That'll be 400 plus 300 is 700. And so what this tells me is the maximum profit is $700. And I should manufacture four color and eight monochromatic monitors to get that maximum profit. This is called linear programming. You will see more examples in the textbook on linear programming, but in every case, you need to find the vertices of the region and check your objective function to see which one maximizes it or minimizes it, whatever it is you're looking for. Okay, we'll stop there and I'll see you next time for episode 27.