Hello everybody, HoodedCoverCommander788 here, and it's time for another vintage GI Joe toy review, and based on my last two videos, the viewers don't seem to be quite as interested in looking at these 1982 GI Joes, so naturally I'm going to do another one. We've already looked at Scarlet and Rock and Roll, so this time we're going to look at the 1982 and 1983 Breaker. Now I have already reviewed this figure, but that video was the first video I ever did on this channel, and that video wasn't very good.  This figure is kind of special to me personally, so rather than sitting around talking about it, let's look at the toy. This is Breaker, GI Joe's first communications officer from 1982. He was first available in 1982 in this so-called straight arm version. He was called straight arm because he had a hinge here at the elbow, but he could not swivel his arms. The following year, in 1983, he was reissued with swivel arm battle grip, which added a swivel here just above the elbow. This 1982 straight arm version was only available for that year. The swivel arm version was available in 1983 and 1984. It was discontinued for the year 1985, and he didn't really have a replacement in 1985. However, in 1986 there was a new GI Joe communications specialist, Dial Tone. On a personal note, Breaker was the first GI Joe figure I ever got as a child. I got the straight arm version back in 1982. Let's take a look at Breaker's accessories, starting with his communications headset, which pegs into the holes on the sides of his helmet, like that, and it has this long wire, which pegs into the hole in his backpack. This communications headset does look like big 80's headphones. It has a microphone that comes down, it has some detail, has some buttons here on the top, and of course this long thin wire is just made of the same kind of hard plastic, so that can break off. Those early GI Joe's did tend to use these long wires coming off the accessories to connect to the backpacks. As did the 1982 flash, he had this long wire from his laser rifle that connected to his backpack. Later GI Joe's replaced these long thin wires, these breakable wires, with removable black rubber tubing that would serve that purpose, that worked a lot better and was much less likely to break. One unfortunate thing about Breaker's headset is when it's plugged into the backpack, it does restrict the head movement a little bit, it naturally wants to pull the head to the left. In 1983, Hasbro came out with battle gear accessory packs, which were replacement accessories usually made in a different color plastic than the original, and there was an accessory pack communication set, and as you can see, it's just a slightly lighter gray than the original. The original is a dark gray, almost black, and the accessory pack version is a lighter gray. So that is something you would want to watch out for if you want an original Breaker communication headset. Breaker's next accessory is his helmet, and this is a standard helmet. This is the same helmet that was issued with most of the 1982 GI Joe action figures. Has holes in the side there, that's where his communications headset pegs in. A lot of other GI Joe figures from that year had this helmet, light grunt, pretty much identical, and as you can see, the color of the helmet should pretty closely match the color of Breaker's uniform. The last or perhaps the most important accessory is what the contents of the card on which Breaker was packaged call, a TV radio backpack. And as you can see, it is kind of lacking in detail. This is supposed to be a very modern, high tech device, but does not have a lot of detail there. I guess this is supposed to be the TV screen on the TV radio backpack. There's the hole in which the communications headset plugs in. And that's about it, really kind of minimal. Comparing this communications backpack to the one that came with the 1986 dial tone, you can see dial tone's backpack is loaded with details, lots of high tech wiring and things like that. Looks very futuristic, very high tech, and by comparison Breaker's communications backpack looks kind of primitive. There's a difference in the pegs on the backpacks that came with the 1982 figures and the ones that came with the 1983 swivel arm figures. The 1982 backpacks had kind of a shorter stubbier peg here and the 1983 backpacks had a slightly longer peg with a more rounded end. And the holes on the back of the action figures are also slightly different, such that the 1982 backpack fits very loosely on the 1983 swivel arm figure. After looking at all these accessories that come with Breaker, do you notice anything missing? That's right. Breaker comes with no weapon whatsoever, and that I think is one reason why Breaker was never a fan favorite. He only comes with his helmet and communications set, no weapons, no guns at all, and I think most kids would want their new army action figure to come with some kind of firearm. But Breaker didn't, and so it kind of looks like he is intended for more of a support role rather than a combat role. If you have an extra M16, perhaps from an accessory pack, or you could take it from Grunt since Grunt gets no respect, give that to Breaker and that looks pretty good. I think that looks appropriate. There is another alternative for the weaponless Breaker that I'll discuss a little bit later in this video. Let's take a look at the articulation on Breaker. The 1982 Breaker had the typical articulation for GI Joe figures of that year, meaning he could turn his head from left to right like that. He could lift his arm up at the shoulder about so far, and he could swivel his arm at the shoulder all the way around. He had that hinge at the elbow so he could move at the elbow a little bit. The figure was held together with a rubber O-ring that looped around the inside. That allowed the figure to move at the torso a little bit. He could move his legs apart about so far. He could move his leg at the hip about 90 degrees, and he could bend at the knee about 90 degrees. The 1983 Breaker had the same articulation except it added a new point of articulation at the bicep. He could now swivel his arm at the bicep all the way around on both arms. That was supposed to allow the figures to hold their weapons with a two-handed grip, but Breaker did not come with any weapons. Let's take a look at the sculpt design and color of Breaker starting with his head. His head as you can see has brown hair and he has a brown beard. Looks a little bit like Jesus. The Breaker action figure shared a head with clutch and rock and roll from that same year. This is pretty typical. A lot of those 1982 G.I. Joe action figures have shared a lot of parts among them. As you can see they have different hair colors and that's really the only way to distinguish the heads of these different figures. Allegedly, the Breaker action figure was not intended to have a beard. Instead of sharing a head with other action figures, he was supposed to have a unique head sculpt and supposedly collectors have found a pre-production packaged version of Breaker with his unique head sculpt. Breaker's chest features black straps with an unpainted knife here and what is supposed to be an unpainted hand grenade on this side, but it looks like a previous owner has taken some silver paint and kind of washed over that hand grenade, but that is supposed to be unpainted like the hand grenade on this swivel arm version. Those straps continue to the back in this Y pattern. The chest and back piece that have the straps that come all the way down were shared by other figures from 1982, including grunt, hawk and snake eyes. There's a difference between the 1982 waist piece on Breaker and the one on the 1983 Breaker. The 1982 waist piece was thicker like this and the 1983 waist piece was slimmer. Also the belts were different and these belt buckles look like they're both intended to be brand stamps for Hasbro. Breaker's arms feature rolled up sleeves and those are parts that he also shared with clutch and rock and roll. You can see the difference in the 1982 and 1983 arms. The swivel on the 1983 arms was added right at that cut where the sleeve ends. These legs are pretty standard. They are shared by many of the 1982 GI Joe figures. It has some pouches on the sides and these are painted in silver and then of course it has some pretty standard black boots. Let's take a look at the file card. This file card was printed on the back of the card on which the action figure was packaged. You can see some of the artwork from the front of the card there. You were encouraged to cut these out and keep them. It shows his faction as GI Joe and it has a portrait of Breaker there and do note that the beard on the portrait of Breaker is very faint there. This looks like a shadow and it may have been that this was added later. Breaker may not have originally been intended to have a beard. This says he's the communications officer and his code name is Breaker and Breaker is a slang used by CB radio operators and it's just a signal to indicate that the operator is ready to send a message. His file name is Alvin R. Kibbe. His primary military specialty is infantry. He has no secondary military specialty which is kind of odd because he is a communications expert. His birthplace is Gatlingburg, Tennessee and his grade is E4. This section says Breaker is familiar with all NATO and Warsaw packed communications gear as well as most world export devices, specialized education, signal school, covert electronics and project gamma. The signal core handles all communications and information systems of the combined armed forces so that would be appropriate for Breaker as a communications officer. Project Gamma is a special forces detachment from 1966 to 1970 that ran covert intelligence operations in Cambodia. This means by extension that Breaker served in the Vietnam conflict and if he served in Project Gamma no later than 1970 then Breaker must be older than he looks. Grade expert M16, M1911A1, MAC-10 Ingram, none of which the action figure comes with. In parentheses classified, don't tell anyone, but he speaks seven languages, nine if you include Pig Latin and Klingon. This bottom section has a quote. It says he's efficient and self-assured and has an uncanny ability to turn adverse situations to his favor. What does this file card tell you about the character of Breaker other than just his job? Nothing really. This bottom quote down here reads like one of those motivational posters that hangs in your supervisor's office. Taking a look at Breaker overall, the first complaint is pretty obvious. This is a very generic figure. He's really made up of parts from other action figures. He has no unique parts all on his own and he doesn't even come with a weapon even though I've let him borrow one for the purposes of this video. He did not come with any firearms. So he really didn't work very well as an action soldier in that first line of GI Joe figures where other figures had big machine guns and submachine guns and vehicles. Breaker had none of that. Comparing Breaker with his replacement dial tone, the dial tone figure really just goes all out with paint color applications and detail and accessories and the Breaker action figure just doesn't. That's why it's really hard for fans to be very excited about this Breaker action figure. The Breaker action figure is generic. That cannot be denied. But some collectors have complained that these 1982 GI Joe figures are boring. But I disagree. These 82 figures are wearing essentially modified uniforms of American soldiers. And people wearing uniforms very much like those were present at some of the most important events in all of human history. So generic? Yes. But boring? I don't think so. They may lack the flashy colors and details of GI Joe figures that we would get later. But as soldiers they represent the core of what GI Joe is supposed to be. And the very reason that the line was called GI Joe in the first place. Breaker appeared in both the GI Joe comic book and the cartoon and in both of those media he was more than just a radio man. He was also a technology nerd and a computer whiz. That gave him a little bit more to do. The Breaker action figure as you can see has a beard. But in the GI Joe comic book Breaker did not have a beard. He was clean shaven and he was known for chewing bubble gum all the time and blowing bubbles. In the cartoon Breaker did have a beard. And I think I actually prefer Breaker with the beard because in the comic book depending on who the artist was in any given issue if Breaker wasn't blowing his bubble gum then he was almost indistinguishable from grunt. So his beard makes him a little bit more distinguishable from the other very generic characters from that year. Breaker had a lot of appearances in the first year of GI Joe comic books but in the later years he kind of took a backseat as new GI Joe characters were introduced so you definitely didn't see him as often. However he did meet a very tragic end as in issue number 109 he was killed along with some other GI Joe's. Breaker appeared in the live action movie GI Joe Rise of Cobra but his nationality was changed to Moroccan. In the GI Joe animated series Breaker's role was partially taken over by a character called Sparks. Sparks was a character invented specifically for the cartoon series. He did not have an action figure in the vintage line. I much would have preferred to see Breaker more in the cartoon rather than have a character invented that sort of pushed him aside. Breaker was the first GI Joe figure I got as a child way back in 1982 and I got the straight arm version. Now why would Breaker be the first GI Joe that I would select to buy and own for myself? I've been thinking about that and I think there are a couple reasons. First even though there was one GI Joe commercial that came before it the first GI Joe TV commercial that I remember seeing was the one with Breaker on the Ram motorcycle so that was my first exposure to GI Joe. Secondly I think I got this action figure with my dad and my dad not knowing anything about 80s GI Joe may have associated the bearded Breaker with the 1970s flocked bearded action team GI Joe and so Breaker may have seemed like the main character and naturally the first one to get. Once I got the figure though my mind was blown. Before that like a lot of other kids I collected Star Wars and the Star Wars action figures they were fine I didn't really have a problem with them but then GI Joe comes out with movable elbows and movable knees and it was just a whole new world and after that I'd still get Star Wars action figures every once in a while but I was a GI Joe kid. As a reviewer I have to be honest with you about this action figure's shortcomings and it has many but on a personal level this figure is very special to me it has a lot of sentimental value this was my entry point into GI Joe if not for this Breaker action figure there might never have been a hooded Cobra Commander 7AA channel you probably wouldn't be watching this video the course of human events may have been altered irreparably maybe the communist would have won the Cold War ever think about that and I think it's saying something when the most generic GI Joe figure that was released that year still blew away any Star Wars figure that had been released before then it just took one GI Joe to displace Star Wars the other solution to the weaponless Breaker that I mentioned earlier is to put Breaker on the RAM motorcycle that came out that same year even though the GI Joe comic book often features rock and roll writing the motorcycle I think it's much more appropriate with Breaker he can still wear all of his communications gear and while he's on the motorcycle he has this huge Gatling gun that he can use to fight Cobra and if he's not on the RAM I have to admit I just put him at the computer console in the 1983 GI Joe headquarters command center because I don't know what else to do with them that was my review of the 1982 and 1983 releases of Breaker I hope you enjoyed it and if you're thinking of getting a Breaker action figure I hope you found it informative if you like it don't forget to give it a big ol thumbs up on YouTube and don't forget to subscribe I've got a lot of great new GI Joe toy reviews coming your way you don't want to miss them and don't forget to like me on Facebook and follow me on Twitter you get a lot of updates there you don't get anywhere else thanks for watching I'll see you next week with another vintage GI Joe toy review I'm picking up a Cobra plane on radar that we've captured the GI Joe anti-aircraft gun Cobra bombers can take out their command center flash to Breaker flash to Breaker Cobra has taken the flack we need some help over here


TLO what's We are on twitch. We are not live, but you can leave a like comment subscribe turn on your post notification bells Now let's continue to grow the family from Chicago to the UK. I honestly don't know what happened to my audio and that other one I'm gonna redo the police academy one, but just not today I'm gonna let y'all know that I've done it already the police academy one this one though I haven't done so I'm just gonna do this real quick. Yo diggity check it yourself Yo diggity check yourself before you wreck yourself Drug is ruining people lives. That's why I see with none other than will glass be Be a special Asian from the drug enforcement agency Say Sorry You too Y'all don't see the this is y'all see this on the table. It's actually set up in line That's great Say thanks for having me on so let's go through these This is Moroccan black in it. We've got to be a geek not to know that looks like looks like some hash very very very similar to marijuana And it's used in quite the same way. I never understood the point of hash Like why was hash created when there's marijuana, I'm just curious, you know You got to remember weed is legal in the state that I am in and the county that I am in Cook County, Illinois Chicago Cook County, Chicago, Illinois What is its effects it slows down your reaction it slows your ability to learn as well Your brain just can't really slows down and doesn't stop functioning, but it slows down the learning curve And is there any scientifically proven the slowing down the learning Just some say that on on on the on the greenery you might be more enlightened negative effects that Being And how much would something like this cost it really depends on the quality of the weed. I mean some weed now I can go for over a thousand dollars a pound. That's really being ripped off. I can get you What year is this? I'm sorry. This is I think it's in like Well I think people move on from the herb to other drugs the problem with marijuana is it is So does you recommend this as a good first smoke? Absolutely not. So just to make clear All this stuff so far is legal What about the other stuff all this is it just make clear all this is illegal what okay? So it ain't legal, but if it's just for personal use, let's say you was just a lot off a kilo or more ever Half a key at the crib a personal It's very legal now in the United States and most most states is legal in the United States in the United States Still illegal. What's this little baby in? those right there is Ndma known on the street is ecstasy. Is there chemical indies that Party people in the house to make you dance like a prick I've seen some videos some people that they're using that stuff in dancing and I guess that's the way to describe it if kids out They're absolutely asked to take something. How many is it safe for them to take for somebody to put even one of those in Their body would have you know, it's absolutely crazy because me mate Dave took 24 in one night Hey, look how he looking at him look how the agent look at him up and down low 24 in one night He was really buzzing and loads of energy That's false because the legs they offer X feel you have no energy It takes all the serotonin out of your body and pushes it out at one time When you take it so the next day you you're actually down You have mood swings. You don't really want to converse with nobody and this is all and I know off off the internet that I Watch YouTube videos to learn this type of stuff. So and everyone on the egg and Langley village boss Said that you drove it better than ever before Well again, I've known Heard stories and talked to parents Whose kids took just one tablet As you think it is ever Ryan's cut to give an animal an ecstasy Now if you like your animal, what about if the animal looks really sad first of all in the United States It's illegal even to possess easy if you're gonna possess them and Provide them to your animal to an animal. You probably have some mental problems yourself anyway Now MDMA still is Illegal, yes, is there any other things that is addictive a Lot of different drugs are addictive.  Why don't you give them to charity? Could take illegal drugs like this It'd be harmful to anybody who Who actually got got to them You out there Listen up Learn keep away Stick clear because once you start taking this then he's gonna take this then that mix those together Try a bit of that. You'd be really really mashed if you took all that you want to go, wouldn't you?  Look at how he's still looking at how He looked at him like man. I'm watching my hands man. You're not funny Say Did he just double fist bump? Did he just double he had to do some of that stuff on that table if he just double fist bump out of you Oh, yo, dig it.  I'm going If you're of legal age


Where do you guys see, you know, how do you think about evolving? How do you think about growing? Are you in a good spot? Are you a great acquisition target?  We're 100% trying to become a lifestyle brand, you know, like we are a product that fits in your pocket and you go out and you accomplish what you want to do because you're in the right state of mind to accomplish what you want to do and That means, you know, only putting the content out there for fitness videos like working with the right people if we are ever gonna try to upsell and release things Yes, we're gonna come out with more products, but beyond just more products We want you to rethink your life and how neuro fits into it and can enhance it So I think I mean what they're gonna steering is extremely smart, you know It's like a way luggage say what you want about their management, but they want to take over travel Not just luggage, but that is such a great big picture way of thinking about lifestyle That's so true. When you say lifestyle brand, are you thinking just thinking out loud is it clothing? Is it moving into something else? Is it more community driven besides just things like gum and mints or what is that? Can you elaborate a little bit on what a lifestyle brand looks like to you? For us, it's our bread and butter is having functional gum and mints. It's inherently shareable And I think to your point, it's creating a community around it So what what can we support? It's fitness instructors during this difficult time, you know We've donated 12,000 units of product for free for any healthcare or frontline workers And we're kind of we're pushing another 15,000 like more units to donate And it's just something where it's like it aligns with our strategy like our ethos or culture and yeah So anybody listen to this podcast for a healthcare worker hit us up or we're still giving out free product to try and Hopefully helps you in that lifestyle like somebody on the go and just really trying to create that sense of community around that It's it's like how Red Bull is in many ways like Red Bull differentiates themselves from other energy-driven companies by building that brand. So you guys have any cliff diving events in your near future? I mean, I'm always down to go cliff diving Right now tag coming up in September Over a blast like Lulu lemon is a Not to keep rambling about other companies But Lulu lemon does an amazing job with building lifestyle and the way they built their company was tagging on to the yoga influencers and the yoga instructors and Them teaching the classes and them having big community sessions with yoga instructors and supporting that it became this like Not just a fitness brand, but a like an appreciate yourself find balance brand and that's how they kind of got their first First group of customers as well. So for that is the natural progression for you guys to have People in the MMA world do the same sort of thing as the Lulu lemon yoga instructors So I think like we're saying it's less so the fighters that we're attracted to but it's more So the people that use their mind and do the analysis, you know, like the Joe Rogan's or the arrow Hawaanis At the same time, there's like the CrossFit community is amazing because They're trying to balance their body with every single data point that they have in the same way new tropics does It's less so about the new tropics themselves It's more so about the research the data points and how you improve yourself using the tools that you have So that's why we always talk about the mind refreshing your state of mind more So than being like this gives you energy because one is a singular thing The other one is a comprehensive thing that affects every single person Hey, everyone, thanks for checking out that clip if you enjoyed it be sure to hit the like button down below And if you're interested in hearing the full episode, it's out right now on our YouTube channel We've had a lot of great guests come on this show before and we've got a lot of great guests coming up in the future So hit subscribe so that you don't miss a single episode and one final note We're always looking for new ideas and new companies feature on the show So if you know of someone or know of a company write us a comment down below letting us know who they are and what they do We'd be happy to have them on the show till then I'll just be here waiting for your comments. So See you later


Here's the rewrite of system talking about the plus and minus. The solution to x squared is equal to 4 is plus and minus 2. You need both solutions to be correct, so you're saying, yes. I'm just saying we take square root of x to represent only the positive, because if you don't, it's not a function. Yeah, for sure. If you're talking about functions only, 100%, right? Function. Square roots having two positive outputs would make there be four numbers the quadratic formula would give us. Four numbers, it would give you two numbers, like square root of da-da-da-da-da, would be two values, wouldn't it? Yeah, check this out. Here. Now, what what system is saying is 100% true when we're talking about functions, right? So this comes into play here. Watch this. I'm gonna kick it down for you. One of the first places we encountered this, right? Let me see if I got a pen that's gonna be a little darker. Better coming out. Is that better? Yeah, maybe. I need a whole new one. Let's find the pen that's gonna be nice and dark. That's not it. Let's dump that one. Oh, it's not bad. Brown? We'll use brown. There, we'll keep this one here. Let me see the red one too. Oh, that's nice. That's darker. Let's use the red one. Okay, now take a look at this thing. Talking about why a function would not be a function or a relation would not be a function if you had a positive and negative outcome when you took the square root of a number of square root of function, right? Now, one of the things that happens in mathematics is we try to manipulate functions, change things around to see what happens to them, right? So just imagine, just imagine you had this. y is equal to x squared, right? Let's assume we wanted to graph this function. Now, graph of this function is a quadratic. We've done a lot of quadratics before, or you could just create a table, right? So let's just create a table. Just so for those that want to follow what's going on here, you don't have to know about quadratics and the quadratic formula and completing the square and stuff like this to be able to graph this. So this is just a function, right? A function is just sort of a relationship but a special type of relationship where for given x, you can only have one y, right? For a given input, you can only have one output and right now we have a y is equal to x squared and the x we consider to be our independent variable. It's our input and y is our output, right? So if we want to graph this function to see what it looks like, all we got to do is put in an input and get an output, right? Put it in, get it out. So let's put something in for x. Let's put in zero for x and all you do you say, okay, y is equal to zero squared. What's zero squared is zero. Okay, so that's a point on the graph, zero zero. Let's put another input, one. When x is one, we get one squared, we get one. So when x is one, y is one, x is one, y is one, right? Let's put another point, x is two, right? Well, if you put two for x, you get two squared, that's four. So when x is two, two, three, four, right? Oops, let's graph this better. So we're here, right? Let's put in three. When x is three, you get three squared, you get nine. That's four, five, six, seven, eight, nine. We're up here. We're off the board. So let's just graph this, right? Here's what it looks like on this side. Well, it goes up like that, right? Well, okay, we've got a feeling for what the graph looks like on this side of the y axis, right? Because this is, this is your y axis, this is your x, right? That's where your x, x this way, y this way. Well, let's see what it'll be if x is negative one. What's y when x is negative one? So you plug in negative one for x, right? So you're going to get negative one squared, which is one. Oh, one again, so one. And if you plug in negative two, you're going to get negative two squared, you're going to get four. Oh, so that's a mirror of that, right? Cool. Hopefully that's symmetrical enough, symmetrical enough, right? So that's a graph of your base quadratic function, which is the parabola, which opens up like this. Cool enough.  They go, hey, what'll happen if we switch the x and a y here? Switch them up, switch them up. What do you mean switch them up? Well, make the y and x and the x or y. Y for the hell of it. Let's see what happens. Okay, let's do it. So what happens if we do this? Fine. Yes, I want to do a different color, so we know it's the different stuff. Green, brown, fine. Take, take, I don't want to write all capitals, take inverse of y. Here, take the inverse of it, which means flip the x and the y around. So what you get is x is equal to y squared. Flip the x and the y. That's what an inverse means. What it also means is a reflection about the line y equals x. So you're reflecting a line. I should make this one. Give it a little curve, right? Because there is a parabola after all. Give it a little curve. Give it a little curve. So when you take the inverse of a function, what you're really doing is you're doing this. You're taking a function and flipping it about the line y equals x. Y is equal to x. You're flipping it about that line. So, okay, let's do a little algebra on this. Well, if you're going to do a little algebra, you're going to get y by itself. Y is your function. Y is your independent variable. x is your dependent variable. Oh, sorry. Y is your dependent variable. It's dependent on x, so you want to get y by itself, right? So you take the square root of both sides. So y is equal to square root of x, right? But what we talked about was square root has to be plus and minus, right? So, square root of anything is plus and minus. Now, remember, we don't have a number in there yet, right? So do we have to write in plus and minus here? Not really, because we haven't taken the square root of x yet, right? It's just a variable, right? By definition, you take the square root, it's plus and minus, right? So let's leave it like that. Let's not put plus and minus there, okay? Now, if this is what's going on and you're taking the inverse of this function, which means you're reflecting about the line y equals x, which means all you're going to do is flip the x and the y. You're switching the numbers, right? If you're going to switch the numbers, let's graph it. Let's create a table. Well, if we're going to create a table, all we're going to do is flip these guys. So the x becomes a y, so 0, 1, 2, 3, negative 1, negative 2, and the y becomes an x. 0, 1, 4, 9, 1, 4, okay? We can test this if you want. Right? What's the square root of 0? Put 0 in here. Well, square root of 0 is 0. What's the square root of 1? Square root of 1 is 1. What's the square root of 2? Well, square root of 2 is just square root of 2, right? What's the square root of... I'm putting this in the wrong way. The square root has to go here. So over here, y is equal to... No, no, that was correct. Y is equal to square root of x. So square root of x is equal to 1. Square root of 2 is equal to square root of 2. Square root of 3 is equal to square root of 3. Square root of negative 1... Oh, no, no, I'm already putting the... Here. Square root of negative 1. Square root of 1 could be negative 1, and square root of 2, square root of 4. Look at this. Look at this. I'm messing this up. 4 is 2. Square root of 9. Did I confuse guys enough? Square root of 1 is negative 1. Square root of 4 can also be negative 2, right? I don't usually do it this way. I'm trying to push it, right? So square root of 4... You put 4 in for x. Square root of 4 is 2, right? But it's not just 2. It's plus and minus 2. So instead of putting minus here, I'm going to put it here. Plus and minus, right? Square root of 9 is not just 3. It's plus and minus 3. Square root of 1... Well, it was 1 and negative 1, so it's plus and minus 1, right? So we don't need these bottom guys. We can just take them out, right? I hope that's clear. I sort of mucked it up in the process, right? Crafter, how you doing? Hope you're doing well. So check this out. What does this mean? That means if we're going to graph this, when x is 0, y is 0. So we're here. When x is 1, y is plus and minus 1. When x is 1, y can be here or it can be here. When x is 4, 3, 4, y can be plus and minus 2. Here and here, right? And then 9 is plus and minus 3. We're off the board again, so let's graph this. Are we snagging anything today? I got some grapes. They're pretty good grapes. I got some that are loose here. They're really nice grapes. Just a little. I had a good breakfast. Let's see our grapes. Who's going to focus on my grapes? There you go. Really yummy grapes.  Oh no, it's just the end of it. So it's pretty yummy. As well as learning how to calculate my hand. No, I would not recommend high definition grapes. High definition grapes. So take a look at this thing. This is what the inverse means of this function. Now remember, I keep on calling the function, but you can think of it as a relation if you want. So if you take this function, this relation, and take its inverse, that means switch the x and y around, switch the x and y around, means you're taking this function and flipping it along this, you get this. Now here's where the problem comes in, a system saying the plus and minus. If we say this is a function, then its inverse for it to stay a function means that it has to pass the vertical line test, which means for a given x value, it can't have two different y's. For a given x value, x is equal to one. It can't have plus or minus one as an answer. For x is equal to four, you can't have plus and negative two. You can't have an x pointing to two different y's. That's what it means for it to be the definition of function. If the question was, this is a function, then you have to decide if you're going to kill the top or kill the bottom of this, depending on your system.  Which means the inverse of this function is going to be this function, and this function looks like that. So you don't have the negative results when you take the square root of a number. However, if I said find the inverse of this relation, then if I define it as a relation, that means this can be a relation, that means the negative numbers can remain, which also means that these numbers would still be there. So it's all about definition. It's all about definition. Usually in high school mathematics, you're dealing with functions. So you end up eliminating the bottom. But let's assume you have this. Let's see if I was green pens doing. What if you had this? y is equal to negative square root of x. If y is equal to negative square root of x, then your graph would no longer be the top guy, it would be the bottom guy. Because the square root of x, you would define to be positive, and you're taking the negative of it. So that part would be the legit answer. Okay. Does that clear things up? Is that okay, system? Or anybody else that's wanted to know what this is about? It's interesting. It really digs down deep into the essence of what it is that we're doing. Most people don't appreciate this. And system, thanks for bringing it up, by the way, it's important to have a visual of what it is really that we're doing and why it is that we're doing it. How does the syntax work? And it's all about the word. Function. Function. Oops. Function. y is a function. That would mean take inverse of y, which is a function. Right? If I say this is not a function, or if you don't specify, right? I think it's legit to put it that way. It really depends on the teacher's as well and the correction. Like, that's the kicker with this, right? It's really dependent on how you're learning it. Right? But the syntax, the math, it's just there. It's just there. That's the thing with mathematics. A lot of, unfortunately, they make special rules in math to apply to a certain system, and people think those special rules are universal and you can do that in the language of mathematics whenever you come across it, but that's not true because that applies only to that system, right? So, for example, when you're doing calculations, if you're finding maximum area, maximum area in general, right? Your graphs for maximum area when it's quadratic, you go from here to here, right? You end it there. Your domain and range, right? Why?  You can't have a negative area, so you eliminate anything below the x-axis, okay? Because you can't have a negative value for that, right? That's the definition of the system. However, for quadratics, you can have negative numbers, right? That's their infinite, okay? I hope that helps out. I love talking about this stuff because it gets into the nitty-gritty of what it is that's going on.


In this video, I wanted to explain to you a bit about the concept of attention and how attention creates energy and why managing your attention might just be the most important thing you can do with yourself. So think of two people. Think of your average neighbor, somebody who lives next to you, let's say the average person that makes $100,000 a year, has a decent marriage, has a decent life, nothing above average. And now think about somebody like Bill Gates, who has I think around $130 billion, so this is Bill Gates, and this is Dufus, your neighbor Dufus. So as you can see, they're both regular people. They both have arms, legs, heads, everything is just the same, IQ, maybe somewhat of a difference, maybe IQ is in here, it's let's say the average person 100, let's say Bill Gates 140. This is 40% more. This is not, I don't even know the difference, but this is not a thousand times more. This is 40% more. So obviously the reason Bill Gates is worth $130 billion might be in part because he's very smart, but it's not the main reason. What is the biggest difference between these two people? In my mind, in my estimation, the biggest difference between these two people is one word, focus, you know, two words, focus and attention. So what do these two very important words mean? Focus and attention.  So somebody who puts their attention mostly on negative things would mostly attract negative things to their life. This is not the law of attraction, this is about you being right, this is about you constantly validating what you see. So if all you see around you is negative things, you're going to validate that by creating that reality for yourself. Again, not law of attraction, actual practical reality. Now the next level that people don't really talk about in self-development is that most people in self-development are focused on maybe a bit on small, tiny goals, maybe they're focused on maybe improving their life a bit, but what they're not focused on is this. They're not focused on the big results. What Bill Gates puts most of his attention on is how do I get this Microsoft company to get massive? How do I help billions of people in the world? Trading is always very big and because he constantly puts his focus on the big results that he's trying to create, he creates those big results. On the other hand, this guy, Mr. Dufus, who's the average guy, he puts his attention, sorry, on average. So he puts his attention on let's say the mortgage I have to pay on maybe I can get another $1,000 a month on the loans he has to cover, on his wife nagging him about watching TV together, on his weight, which is not very good. So you have the guy here who's just the average guy who's focusing on average stuff.  He's focusing on, oh, let's improve my life by 10%. Let's do some meditation. Let's work out, just average stuff. I'm improving myself. I'm getting a bit better every day. But there's no serious improvement. There's no 100% increase every month. You don't really see those big jumps. And the way it looks, if you actually make it into a chart, you see that the people who are average, their life basically looks something like this. So just up and down and up and down and you have some sort of an average and they don't really ever cross that average. Now you have people in self-development whose life looks something like this. They are improving. They do see improvements in their life, but they don't see anything big. So these are the self-development. You might not see that. It might be outside the board. So people in self-development, they see some improvements. But then you have the really successful people, the people who take massive action, who think really big, who most of their attention is on getting here. And what happens is they actually go like this. They move much faster and they move much more intensely. Because again, the people who are average, all they're thinking about is what's currently available for me. Oh, this issue I need to get fixed. Oh, this. It's just the present. Just I'm scared about this. I need to handle this. The people in self-development, they're like, oh, I want to get here. You know, I want to get someday, maybe in 20, 50 years, I want to get to some big destination. The really successful people, they're like, I want to get here now. I want to get to 10 billion now. How can I compress time? How can I get here really, really fast? Now why is this so important? Because when you want to get to the moon, for example, you want to get from here all the way to Mr. Moon, you can't get here with a car. You can't get here with a boat.  You need a serious jet. You need some penis-looking jet to fly you all the way to the moon. And so the vehicle is different. Now if you're Bill Gates, you want to get to the moon. You want to get to beyond the moon. So ahead of time, he's thinking big. He's thinking, I need a company. I need thousands of people. I need the best support. I need the best product from the beginning. But when you're Mr. Average, when you're Mr. Just Self-Development, just going like this, you want to get to maybe an extra $1,000, maybe an extra $10,000. Well, you can do that on your own. So in the first place, the vehicle you're taking is a very slow, a very small vehicle that can take you there. So again, the higher you aim, the bigger the vehicle you're going to have to get, the more stronger, the more horsepower, the more pressure it can take, which again, the difference between walking, bicycling, car, a plane, or a spaceship. Now you want the spaceship. Now they don't have teleportation yet, but until they have that, you want to create a spaceship. You want to make sure your ship is so tight that you can fly all the way to the moon. And again, if you don't aim there in the first place, you're never going to get there. I hope this message helped you. Let me know if you have any questions and I'll see you in the next video.


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.  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.  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.  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.


Good morning everyone. It's so excited to be here. I'm Chen Wang from IBM Research. I'm a research scientist working on cloud native AI platform and I have been working on Kubernetes for five more years. This is Abhishek. Hello folks. I'm a senior software engineer here at IBM and excited to be here. So today we are going to introduce some of our experienced best practices on ARM serving. So before that we all know has been attracting a lot of attention and then serving ARM will help a lot in modernize the existing business use cases and applications. So however serving large language models is very expensive. So first it needs to run on high-end GPU accelerators such as A100 or even H100 and then the sequential nature of large language models makes the inference processing time very long. So for one A100 accelerator if you want to precise and request it's less than one request per second. So if you are thinking about production use cases or our business use cases you may need tons of those requests inference requests going through and then it will need tons of GPUs which is apparently very costly and then the diagram just shows a simple toy example of how is making the inference iteratively and so it is generating token per token and every time it generates a new token which is a word it needs to cache all the previous tokens in your cache. So this sequential nature mix is very like takes a lot of resources and makes the inference very slow. So that's why in production use cases we really want to have some techniques to improve the support and performance of inference and then there are two popular techniques in the open source of wider academia community to improve the serving. One is called continuous batching and the other is patient attention. So for the batching part continuous batching is basically derived from the static batching which is utilized more memory to batch more requests to use GPU more efficiently and then continuous batching is just continuously understanding how many requests are coming in and then utilize the memory better to continuous put requests, candidate requests to the previous one so you maximize your memory utilization as well as your GPU utilization. So the patient attention kernel technique is similar it's tried to might be the logical blocks of KV cache which is necessary to generate to the next token to the physical KV cache blocks so you can have more efficient utilization of memory and reduce the resource fermentations in the memory space allocation. So in our case we have research clusters and we want to serve a lot of GPU models for the wider users in our research lab and then we find out if we will serve a wider range of models and then some models are very popular and some models may be idling for a long time but researchers still want to use those and in this case for example we if we want to serve 50 models and find out 30 of those unpopular models but necessary to serve and then if we use one GPU to serve those unpopular models and then we will find out we still have a very long long tail GPU underutilized so how can we solve this problem and we are thinking about like packing more unpopular models in fewer number of GPUs and then the available techniques of GPU sharing available nowadays can be time sharing, MPS and make and then due to the nature of the batching and the memory page attention kernel or flash attention kernels memory optimizations in those servers so we really think if you limit the memory allocation you are not using the compute efficiently enough so because MPS and time sharing really targeting the dynamic sharing of HBM space and those unpredictable memory allocation may lead to exceptions easily when there's there are birthday requests are coming in so then we want to try the make partition initially which is a static partitioning the memory space so we did some simple experiments on benchmarking the MPS model on different varying size of make petitions and then if we set the per token generation latency to 50 millisecond per token and then we found out like if the load is low enough like there's less fewer than 32 concurrent users of sending requests then we can guarantee the latency pretty well using smaller make sizes like 4g 4tgb however in our practice we find out if we use the NVIDIA default GPU operator to enable make every time we want to reconfigure the make petitions we need to evict all the workloads on our GPUs on our server so this but from because the optimal make partition we need for serving the model really change over time over very low so we want a dynamic way to create make partitions so that's why Abhishek will talk more about how we use DRA. Thank you Chen for working us through the importance of using make slices for inference workload let's quickly dive into DRA so DRA stands for dynamic resource allocation and it provides two new APIs basically resource claim and resource class to request GPU resources and while DRA is a blanket statement but it solves a very important use case for us which is the ability to have incremental mix slices on some of the other end of GPUs so as we see that in the DRA world there is a quite a lot of setup that is needed to enable GPU sharing so on the right hand side of the screen in the middle we see that there are different resource claims like GPU claim parameters and make device claim parameters that are needed to be set up and those claim parameters are then called on a reference into this OPD model workload that we see over here let's quickly dive into the demo here so what we do here is we submit the same workload that we just saw on the previous slide as we submit this workload we see few resources that are created but notable resources here are the resource claims that are created once the desired resources are created then the container comes up and inside the container what we have is the VLLM server we wait for some time for the VLLM server to come up and enable port forwarding to interact with it now we send a prompt or a sentence completion request to the model saying San Francisco is and and finally we do get a response that it's a great place to to live in by the OPD model workload here thank you for watching this demo if you want to learn a bit more about DRA we do have another full talk here on Thursday questions thank you thank you okay we do have the the barcode of the talk and also the other tutorial on how we deploy VLM server using DRA in the previous slides so and the demo link is also available