For more information, please see full course syllabus of Math Analysis

For more information, please see full course syllabus of Math Analysis

### Systems of Linear Inequalities

- Remember, we solve an inequality in much the same way we solve an equation. However, instead of giving just a single answer, an inequality has an infinite variety of solutions. Also, remember, when doing algebra with an inequality, multiplying or dividing both sides by a
__negative__number will__flip__the direction of the inequality sign. - When we graph the answers to a linear inequality, we shade in the entire region that satisfies the inequality. Furthermore, the boundary line changes depending on if the inequality is
*strict*( < or > ) or not strict ( ≤ or ≥ ). If it is__strict__, the line is__dashed__, while if it is__not strict__, it is a__solid__line. - To help figure out where to shade, begin by just graphing the boundary of the inequality (as if it were an equation). Then, once you've drawn the boundary (with the appropriate solid or dashed line), test some point on one side of the boundary. If the point satisfies the inequality, shade that side. If not, shade the other side.
- When finding the solutions to a system of linear inequalities, it is best to graph each inequality in the system. When we worked with linear equalities (=), we could use substitution or elimination, but that was because we were solving for a single answer. With a system of inequalities, though, we aren't going to get just one answer-we're going to have infinitely many (or none, if the shadings don't overlap). Thus, graph each inequality, shade each appropriately as above, and the solutions to the system are where all the shadings overlap.
- An excellent application of linear inequalities is
*linear programming*. It helps us optimize systems and make the best choice given certain requirements. We start with a linear*objective function*that we are trying to maximize or minimize (such as profit or cost). The objective function is based off some number of variables. The variables in the objective function also have various constraints (given as a system of linear inequalities). This gives a region of*feasible*solutions that are allowed by the constraints. Our objective is to find the maximum (or minimum) of our objective function within those feasible solutions. - The theory of linear programming says that if there is a max/min for the objective function within the constraints, it occurs at a
*vertex*("corner") of the feasible solutions. Thus, to find the max/min of an objective function, we begin by finding the locations of the vertices by solving for intersection points. Once we know all the vertices, we can try each of them in our objective function, and whichever is highest/lowest is our max/min.

### Systems of Linear Inequalities

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro
- Introduction
- Inequality Refresher-Solutions
- Refresher--Negative Multiplication Flips
- Refresher--Negative Flips: Why?
- Refresher--Stick to Basic Operations
- Linear Equations in Two Variables
- Graphing Linear Inequalities
- Why It Includes a Whole Section
- How to Show The Difference Between Strict and Not Strict Inequalities
- Dashed Line--Not Solutions
- Solid Line--Are Solutions
- Test Points for Shading
- Graphing a System
- Solutions are Best Found Through Graphing
- Linear Programming-Idea
- Linear Programming-Method
- Example 1
- Example 2
- Example 3
- Example 4

- Intro 0:00
- Introduction 0:04
- Inequality Refresher-Solutions 0:46
- Equation Solutions vs. Inequality Solutions
- Essentially a Wide Variety of Answers
- Refresher--Negative Multiplication Flips 1:43
- Refresher--Negative Flips: Why? 3:19
- Multiplication by a Negative
- The Relationship Flips
- Refresher--Stick to Basic Operations 4:34
- Linear Equations in Two Variables 6:50
- Graphing Linear Inequalities 8:28
- Why It Includes a Whole Section
- How to Show The Difference Between Strict and Not Strict Inequalities
- Dashed Line--Not Solutions
- Solid Line--Are Solutions
- Test Points for Shading 11:42
- Example of Using a Point
- Drawing Shading from the Point
- Graphing a System 14:53
- Set of Solutions is the Overlap
- Example
- Solutions are Best Found Through Graphing 18:05
- Linear Programming-Idea 19:52
- Use a Linear Objective Function
- Variables in Objective Function have Constraints
- Linear Programming-Method 22:09
- Rearrange Equations
- Graph
- Critical Solution is at the Vertex of the Overlap
- Try Each Vertice
- Example 1 24:58
- Example 2 28:57
- Example 3 33:48
- Example 4 43:10

### Math Analysis Online

### Transcription: Systems of Linear Inequalities

*Hi--welcome back to Educator.com.*0000

*Today, we are going to talk about systems of linear inequalities.*0002

*In the previous lesson, we worked with systems of linear equations.*0005

*In this lesson, we will work with systems of linear inequalities.*0009

*Remember: equations are based on equals signs--left side = right side.*0012

*But inequalities are relationships based on less than (I should perhaps use something different),*0016

*less than or equal, greater than or equal, or greater than.*0023

*Since we haven't really worked with inequalities before in this course, we will first have a brief refresher.*0028

*After that, we will consider systems of linear inequalities and how their solutions are based on all the inequalities at the same time.*0033

*Finally, we will see how systems of linear inequalities can be applied through linear programming, which will allow us to solve optimization problems.*0039

*Let's go: when solving a linear equation in one variable, we always find exactly one solution.*0046

*For example, if we have x + 1 = 0, we subtract by 1 on both sides, and we get x = -1.*0051

*So, on the number line, we would see that the answer to this is when x is -1 on the number line.*0056

*On the other hand, when we solve a linear inequality in one variable, we find an infinite variety of solutions.*0063

*So, if we had x + 1 ≥ 0, we subtract 1 on both sides, and we get x ≥ -1, which means we have x equal to -1 as one of the solutions;*0070

*but we also are allowed to go anywhere above that.*0080

*+3 would be an answer to this; +52 would be an answer to this; -1/2 would be an answer to this.*0083

*Everything from -1 on up is an answer to this--everything there will satisfy our inequality.*0089

*This idea of a wide variety of answers will be important to us as we work with systems of linear inequalities, so we want to keep this in mind.*0096

*Mostly, when you are doing algebra with an inequality, it is just the same as doing it with an equation.*0104

*We can do any arithmetic operation--addition, subtraction, multiplication, and division--on both sides, just like normal algebra.*0109

*However, there is one special thing to be aware of: if you use algebra to multiply or divide by a negative number, the relationship flips.*0117

*For example, if we have -x + 1, we can subtract 1 on both sides, and we get -x < -1.*0126

*At this point, we want to make our x positive, so we multiply both sides by -1.*0131

*That causes our inequality symbol to flip to the other way; so we go from less than to greater than now.*0136

*So, we now have x > 1; this is really important to remember.*0146

*You don't want to forget about this when you are working with inequalities.*0151

*Any time you multiply or divide by a negative number, it will cause*0153

*your less than/greater than/less than or equal to/greater than or equal to/whatever you are dealing with...*0157

*it will flip; your inequality symbol will flip to the other direction.*0163

*It is really easy to miss this flip; it is a really easy mistake to make, so just be careful with this.*0167

*Every time you are doing either multiplication or division, and it is a negative number that you are dealing with--*0172

*negative anything that you are multiplying on both sides--you want to make sure you remember about this flipping thing.*0178

*Notice that we can also get the same thing as we have here if we just added x to both sides.*0184

*If we had had +x and +x, we would have had 1 < x, which is equivalent to saying x > 1.*0188

*They are just the same statement, but two different ways of saying it.*0197

*What causes this to happen--why do we see this negative flipping?*0200

*You probably learned about it in algebra for a very long time, but you might not have a great understanding of it.*0204

*So, let's actually get this: consider if we had some a and b, where a is less than b.*0208

*On our number line, a would come before b; a would be closer to the 0 than b would be.*0213

*So, what would it look like if we multiplied a and b by -1?*0219

*Well, that would cause this mirroring around the 0; they would pop to the side opposite, depending on the distance that they originally were from the 0.*0223

*So, -b and -a appear here; but notice that -b is now farther to the left than -a.*0232

*Just as b was farther to the right, it is now going to be farther to the left.*0238

*So, there is still a relationship, but now the relationship has flipped.*0242

*Because b was more positive than a--b was originally more right than a--we know that -b has to be more negative,*0245

*has to be more left, than -a; thus, we have this -a > -b,*0252

*because originally a is less than b, but now -b is less than -a, so that causes our sign to effectively flip,*0258

*because -b < -a is the same thing as saying -a > -b.*0266

*Great; that is what is going on.*0274

*You want to stick to basic operations; when you are working with inequalities, you want to try to keep using addition, subtraction, multiplication, and division.*0276

*You can use more complicated algebra, like raising both sides to a power or taking a square root or taking a logarithm or taking some trigonometric function.*0284

*But they can cause little problems to come up.*0293

*Consider if you have x ^{2} > 0; now, that does not imply that x is greater than 0.*0296

*We might think, "x ^{2} > 0, so I will take the square root of both sides."*0302

*√x ^{2} > ±0...well, that is just 0, so we have x > 0, right?*0306

*No, that is not true at all: over here, x = -3 is true--it checks out: (-3) ^{2} is greater than 0, because 9 is greater than 0.*0312

*So, that is great; but -3 > 0--that is completely false; that is not a true statement.*0322

*So, we do not have this ability to just toss out other, more complicated algebra things*0330

*without really thinking about all of the implications that are going to happen here.*0335

*So, this idea here doesn't work, because we are dealing with an inequality.*0339

*We can't just take square roots, like we can when we are dealing with equations, which is how we built up all that previous theory.*0344

*So, when you have an inequality, and you want to do something more than just a basic operation on both sides,*0350

*you really have to be careful and think about what you are doing.*0355

*That means that it is best to stick to basic arithmetic;*0359

*so we really just want to stick with addition, subtraction, multiplication, and division,*0363

*not because it is impossible to do this other stuff--not because you wouldn't get it--*0367

*but just because it is easy to make a mistake doing it.*0370

*You can end up making mistakes, because you haven't thought about absolutely all of the ramifications.*0373

*So, you want to stick to the basic operations, because it is easy to see what happens.*0378

*And remember: even with the basic operations, when we have multiplication and division,*0381

*there is still a little bit of danger there, because when you do it with a negative number, it causes flipping of your inequality.*0385

*So, with all of this in mind, it is really just best to stick to basic arithmetic.*0391

*Happily, since we are only going to be working with linear inequalities, that means we can get through everything*0395

*just using these basic operations--addition, subtraction, multiplication, and division.*0400

*They are going to be enough to manipulate our linear inequalities into forms that make it easier for us to understand.*0404

*All right, that finishes our refresher; now we are ready to actually talk about linear inequalities in two variables.*0409

*A linear inequality in two (or more) variables tells us about a relationship between those variables.*0415

*Consider -x + y ≥ 1: we can make it easier to interpret by just having y on one side and other stuff involving x on the other.*0420

*So, we add x to both sides: + x, + x; and we get y ≥ x + 1.*0429

*This gives us a relationship between x and y--how they are connected.*0435

*If x is equal to 1, we know that y has to be greater than or equal to 2.*0439

*If x is 1, then y has to be greater than or equal to 1 + 1, so y must be greater than or equal to 2.*0444

*Similarly, if x were equal to -9, then that would mean that y is greater than or equal to -8.*0451

*So, depending on what the x is, we will change the requirements on our y.*0456

*It is a relationship: it is not this definite single static defined thing.*0459

*It is this relationship of "as x shifts, y has to shift in certain ways to accommodate the shifting x."*0464

*Now, notice: unlike a linear equality, where a single x would give us just a single y, a single x in an inequality gives infinitely many possible y's.*0470

*x = 1 means y is greater than or equal to 2; so x = 1 means that y = 2 is good; y = 5 is good; y = 47 is good; y = 1 billion is good.*0482

*We have this stack as possibilities, as long as they follow that inequality relationship.*0493

*If x equals 1, y still can't be equal to -3; so it is not that anything goes--it just has to follow this inequality relationship.*0497

*But if it does, there is an infinite variety of things that it could end up being.*0505

*The best way to understand inequalities is by graphing them.*0509

*This gives us a visual reference to see all of the possible solutions.*0512

*So, for example, if we had y ≥ x + 1, we could graph it like this.*0515

*Because it is a greater than or equal to, everything on our actual line is allowed as a solution,*0521

*because y = x + 1 is going to be true by "greater than or equal."*0530

*That "or equal" means that we can have equality; the line itself will be true.*0534

*But then, in addition to that, everything above the line is also going to be true.*0538

*Everything in here is going to end up being true; and you should notice that it doesn't just stop at the edge there.*0544

*Once again, it is like how we don't draw arrows to show that it keeps going.*0550

*We have gotten used to assuming that the graph keeps going beyond the edges of our graphing window.*0553

*We are going to also be assuming that our shading goes past the edges; it doesn't stop suddenly--it goes forever, infinitely, up.*0557

*So, everything above the line is a solution, because if y is greater than or equal to x + 1, then consider at x = 0:*0566

*then 1 is true for y, but so would 2 be true for y...*0574

*Let's use a different color, just so we can see this a little bit easier.*0577

*2 would be true for y; 3 would be true for y; 4 would be true for y; 5 would be true for y; 6 would be true for y; and anything above that.*0580

*So, as long as we are above this line with our y-value for a given x-value, we will end up being true; the y > x + 1 is true.*0587

*So, a graph is a really great way to get across all of these possibilities, because it is not just this single line.*0597

*It is certainly not just a single point; it is this huge variety--it is this large area of possibilities.*0602

*This brings up a question: what does the line become if the inequality is strict--less than or greater than--*0609

*as opposed to not strict, which is less than or equal to, or greater than or equal to?*0615

*So, if we have a strict inequality, less than or greater than (it is purely less than, or it is purely greater than),*0619

*we show this by changing how we draw the line on the graph.*0628

*So, if it is a strict graph, like y > x + 1, we show that it is strict by using a dashed line.*0631

*It is a little bit hard to see here; but notice that the line is actually dashed, like this.*0639

*So, that says that we are saying that we are not including the line; we are only including the stuff above the line; the line itself is not allowed.*0644

*On the flip side, if it is a not-strict inequality, like greater than or equal to, we show it by using a solid line,*0653

*saying that the line itself is actually going to be an answer, as well.*0661

*If you are on the line, that is an allowed answer, in addition to all of the stuff that is above it.*0665

*So, we show that the points on the line are not solutions to our inequality with a dashed line, when it is a strict inequality.*0671

*We show that they are solutions with a solid line when it is a not-strict inequality.*0684

*And then, we just shade the appropriate side.*0692

*In both of these cases, we have greater than and greater than or equal to; so they both had the part that was above it.*0694

*Now, we don't necessarily have a very great way to see how to shade which way.*0702

*You will get a sense of if it is up, or if it is down, as you work with these more.*0707

*You can pay attention to what the y is, and if it is less than or greater than.*0710

*But a really great way to be sure of which side to shade is by using a test point.*0713

*You can shade by seeing how this test point gets affected.*0719

*Would this test point satisfy our inequality, or would it fail to satisfy it?*0723

*If the point satisfies the inequality (you just choose some point--it is completely your choice--*0729

*you choose some point, and if it satisfies that inequality), then the whole side does,*0735

*because we have to shade in an entire side.*0739

*So, if one point on one side satisfies, that entire side has to satisfy.*0741

*On the flip, if the point does not satisfy the inequality, then the other side will satisfy the inequality.*0745

*So, if your point fails to satisfy, the opposite side will satisfy it.*0755

*This makes it easy for us to shade: for example, if we have y < -5/2x + 3, we could try the test point (0,0).*0759

*(0,0) is almost always a great test point, because you can plug in 0's really, really easily.*0771

*Let's see: would that end up working out, if we have y < -5/2x + 3?*0777

*We plug in 0 for y; we plug in 0 for our x; plus 3...so we have...is 0 less than 3?*0783

*Sure enough, that checks out; so our test point is good.*0792

*If our test point is good, that means that the entire side must be good; so we just shade this side to say anything below that line is going to be true.*0795

*And notice: how did we get this line in the first place?*0808

*We just take our inequality; we turn it into a line, as if it had been an equation, y = -5/2x + 3;*0810

*and then, we graph that, because we know that that is going to be our line of demarcation.*0817

*It is going to be the place where the shading changes over.*0821

*And then, it gets dashed, because it is a strict inequality; so that is why we have that dashed line.*0824

*And then, we shade off of that line that we drew.*0830

*Alternatively, we could have tested a different point, like, let's say, (4,4).*0834

*Maybe we had wanted to test (4,4); well, if we had tested that, it would be 4 for our y, less than -5/2 times 4, plus 3.*0838

*So, 4 < -5/2(4)...so that gets us -10...+ 3; so is 4 less than -7?*0848

*No, that fails to be true; so this side cannot be the side that get shaded in; that side is not going to end up being true.*0865

*But really, I recommend using the point (0,0).*0873

*You could use any point; you could use (1,0), (-5,3)...whatever point is useful for your specific case.*0875

*But (0,0) is almost always going to be something...you want to choose something that is definitely on one side or the other.*0881

*You don't want to choose something on the line itself.*0886

*But (0,0), as long as it is not on one of your lines, is going to be a really good test point for using.*0888

*All right, if you are graphing a system of inequalities, you do it in the same way.*0893

*You graph each of the lines as if they were equalities; and you use either a dashed line*0898

*(if it is a strict inequality, less than or greater than); and you use a solid line if it is a not-strict inequality*0904

*(less than or equal, or greater than or equal); and then you shade in each inequality appropriately.*0912

*And wherever the shadings overlap is the set of solutions to the system.*0917

*So, let's see how that would work out here.*0921

*For this one, y < 2/3x - 2, that is our red dashed line--dashed because it is a less than.*0922

*So, let's try the test point (0,0) for that: (0,0)...we plug that in; we have 0 < 2/3(0) - 2.*0930

*So, 0 < -2 is not true; 0 is greater than -2; so this fails.*0945

*So, that means that the side opposite our test point is going to have to be what we shade in for the red one.*0955

*Our test point was in the very center, in the origin; so we are going to be opposite that side.*0961

*So, this is going to be what we shade in for the red line: y < 2/3x - 2 is true for everything in the shading, but not on the dashed line itself.*0966

*Then, if we have y ≥ -2x + 1, let's use that same test point; it is not on that line, either.*0979

*So, (0,0): 0 ≥ -2(0) + 1; so 0 ≥ 1; is that true?*0985

*No, 0 is not greater than or equal to 1; so that fails here, as well; 0 is not greater than 1.*0998

*That fails; so by the same logic, we are going to have to be opposite that side, greater than or equal;*1006

*so we are going to be above it; so we shade in the side that is opposite.*1013

*Also, you can think of this as being that y is greater, so it must be above the line;*1019

*and the red one was "y is less than," so it must be below it.*1023

*But the test points are a nice way to be absolutely sure of it.*1026

*And notice how we have overlapping; we have red and blue overlapping in this bottom-right portion.*1029

*So, that means the part that really, truly makes up the answers is everything in here.*1034

*And that is going to keep going out forever, as long as it is in there.*1043

*Everything on the blue line actually would be an answer, but things on the red dashed line...*1047

*The blue solid line here is going to be answers; but the red dashed line here, since it is dashed, fails to be answers,*1052

*because it doesn't satisfy our inequality y < 2/3x - 2, because that is strict.*1064

*So, if it is on the blue line in that shaded portion, it is an answer; if it is on the dashed red line, it is not an answer.*1070

*And there we are--we can see how the system comes together; we can see all of the solutions to it at once.*1077

*All right, now: solutions are best found through graphing.*1086

*Graphing is usually the best way to understand the solutions to a system of linear inequalities, as long as it is in two dimensions.*1089

*For dealing with really high numbers of dimensions, it gets kind of hard to graph.*1095

*But if we are in two dimensions, it is a good way to understand what is going on.*1098

*It gives us a way to visually comprehend what we are seeing.*1101

*It lets us see all of the requirements that the system places on the variables.*1105

*So, this system of inequalities makes certain requirements of our variables.*1110

*And by graphing it, we are able to see all of the requirements at the same time.*1116

*Now, with a linear equality--if we had a linear equality (an equals sign between the left and right side),*1119

*we could use substitution or elimination, because we were solving for a single answer.*1126

*A linear equality is solving for one answer.*1131

*But with a system of inequalities, we aren't going to get just one answer; we are going to have infinitely many answers, because we have shadings.*1135

*As long as we have overlapping shadings for our system, that entire area of overlap,*1145

*where all of our inequalities overlap together--that is going to be all of our answers.*1150

*And because it is an area, there are infinitely many things inside of that area.*1155

*Now, it is possible for us to have no answers whatsoever, if the shadings don't overlap--*1158

*if one side shades up, and the other side shades down, and they never touch each other.*1163

*That is never going to have any answers, because the two can never agree; they are inconsistent.*1168

*But as long as they end up agreeing in some portion--some amount of area, inside of that area we will have infinitely many answers.*1172

*So, when working with a system of inequalities, you will almost always want to graph,*1178

*because you can graph it and then shade in the solutions, and you are able to see what is going on,*1183

*and get an intuitive, visual sense of how this thing is coming together.*1188

*All right, now we are able to talk about linear programming: an excellent application of linear inequalities is linear programming.*1193

*It is not quite like computer programming; but it helps us optimize...*1201

*It is actually quite different than computer programming; but in any case,*1205

*it helps us optimize systems and make the best choice, given certain requirements.*1208

*What does it mean to optimize? Well, we start with a linear objective function.*1214

*We have something that is our objective, something that is sort of looking at the variables that we are dealing with.*1218

*And we are trying to either maximize the objective function, or we are trying to minimize the objective function.*1224

*So, we might want to try to maximize something like profit; or we might want to try to minimize something like cost.*1229

*These are useful things; business is something where linear programming will definitely pop up.*1237

*For example, we could have an objective function like z = 2x + 7y.*1241

*So, if we have z = 2x + 7y, it uses x; it uses y; it is an objective function.*1247

*We get some number out of 2x + 7y; we want to try to maximize or minimize this quantity, z.*1252

*Now, notice: if there are no restrictions on x and y, z will be at its biggest when x is flying out to infinity and y is flying out to infinity.*1263

*It will be minimized when x is flying to negative infinity and y is flying to negative infinity.*1273

*So, we need to have some limitations on what x and y are going to be for us to actually be able to find any maximum or minimum, for that to be meaningful.*1277

*And indeed, when you are working with linear programming, the variables in the objective function,*1285

*these variables x and y, are going to have various constraints on them--things that keep them from being able to go anywhere.*1289

*And those constraints will be given as a system of linear inequalities.*1295

*So, we might have constraints like x + 3y must be less than or equal to 8,*1300

*2x + y must be less than or equal to -4, and -x + 2y must be greater than or equal to -3.*1304

*So, our objective will be to find the maximum or minimum of our objective function*1309

*(remember: our objective function, in this case, was our z = 2x + 7y) while we are obeying these constraints.*1315

*We can't break these constraints; so we have to stick within these constraints,*1321

*but we are trying to get z to be the biggest or the smallest thing we possibly can get out of it.*1325

*So, once we know what our objective function is (z in the previous slide), the first step is to graph the system of linear inequalities.*1331

*To do that, we probably want to put them in a form that we can easily graph.*1337

*So, x + 3y ≤ 8--we can convert that into y ≤ -1/3x + 8/3: we subtract x and divide by 3.*1340

*We can convert 2x + y ≤ -4 into -2x - 4 by subtracting 2x.*1351

*And -x + 2y ≥ -3--we add x to both sides and divide by 2; so we have things that are easy for us to graph.*1358

*We have slope; we have the y-intercept at this point; so we can graph these.*1366

*We graph these, and we get this; and we also shade in, and according to our shading,*1370

*we find out that what is inside of there is going to be what is allowed by these constraints.*1375

*So, the shaded area is the location of all feasible solutions--and by feasible, we mean those that are allowed by the constraints,*1381

*those that are possible things that we can even begin to consider using.*1388

*So, the location of our optimum value, whether it is a maximum or a minimum, must be inside of this portion.*1392

*It has to be inside of this shaded thing or on the lines, because these were all less than or equal or greater than or equal.*1399

*They are all not strict, so we can be on the lines, as well.*1405

*So, since we can be on the lines, it has to be somewhere either on the lines or inside of that shaded portion.*1408

*We know that for sure; otherwise, it won't have followed the constraint, so we can't even look at it for trying to use our objective function.*1413

*Here is the critical idea: the theory of linear programming tells us that, if the solution exists--*1421

*if there is some optimum maximum or minimum--if we can maximize or minimize our objective function--*1427

*if that exists, then it occurs at a vertex of the set of feasible solutions.*1434

*Now, a vertex of the set is one of the corners of the set, where two or more of them intersect.*1441

*That is going to be one of the vertices; a vertex occurs at a corner.*1451

*Now, we can find the locations of the vertices by solving for intersection points.*1455

*We can see where the red line intersects the blue line, where the green line intersects the blue line, or where the green line intersects the red line.*1459

*We are just using what we learned from systems of linear equations to be able to figure out where two lines intersect each other.*1466

*We can figure out where these vertices are located.*1472

*Once we know all of the vertices, we just try each of them in our objective function.*1475

*We can figure out each of these vertices from what we learned in the previous lesson.*1480

*We can understand what is going on from what we have seen in this lesson.*1483

*And then, from there, we just plug them into our objective function;*1486

*and whichever ends up coming out to be highest or lowest ends up being our maximum or our minimum.*1489

*We will see this process done in Example 3 and in Example 4.*1494

*All right, we are ready for some examples.*1498

*The first one: Give the system of linear inequalities that produces the below graph.*1501

*So, the first thing: let's figure out what are the lines that we are seeing drawn here.*1505

*Before we even worry about dashed...is it greater than? Is it less than? Is it less than or equal to? Is it greater than or equal to?...*1509

*before we even worry about that, let's just figure out what equation could draw each of these lines.*1516

*For our red one right here, we see that it is a horizontal line; it is at y height of -2.*1521

*So, that would be made up by y = -2; that equation would draw that line.*1527

*Of course, it is not going to be a less than or equal or a greater than or equal, because it was dashed.*1532

*But we know that that line would be drawn by y = -2; we will come back to figuring out what it is later.*1538

*Our green one is going to be set at a horizontal location of 4; it has no slope whatsoever (it is vertical).*1543

*So, that is going to be x = 4; we have set our horizontal location at 4, and our y is allowed to freely roam up and down.*1551

*Finally, the only one that might even be slightly difficult is this blue dashed line, which goes from (0,4) to (2,0).*1559

*If that is the case, we can figure out what the slope of our blue one is.*1568

*It manages to go down four; it has a vertical change of -4 over a horizontal change of +2.*1572

*It goes from a height of 4 to a height of 0 by going 2 steps to the right.*1580

*That means it has a slope of -2; we also see that its y-intercept is right here at 4.*1583

*So, its y-intercept is 4, so we have y = -2x + 4.*1589

*At this point, we have equations that would produce each of these lines.*1597

*So now, it is a question of what symbol goes in there; we can't use equals, because that would be an actual line.*1601

*We need inequalities; so now it is a question of less than, greater than, less than or equal to, or greater than or equal to, or each of these.*1607

*All right, we are going to swap out that equals sign for something that can actually be used as an inequality.*1616

*So, the first step: the red has to shade above, because we have stuff above it,*1622

*because if we go below it, that fails to be true; so it must be above it.*1628

*That means that y has to be strictly greater than (because it is a dashed line) -2.*1633

*If we want to be absolutely sure about this, we can try the test point (0,0).*1645

*If we plug in 0 > -2, that works out; so we know that that side is correct; that is the side that we want.*1648

*For the green one, we see that it is one the left side of that.*1655

*So, if it is on the left side, we know that the horizontal values that we are allowed to use must be less than 4.*1659

*Is it less than, or less than or equal to? It is less than or equal to, because it is a solid line: less than or equal to 4.*1665

*We can also check out a test point: if we plug in (0,0), then we have 0 ≤ 4; yes, that is true.*1673

*So, we know that that is correct for that shading.*1679

*And then finally, our blue one: for this one, we know that it is shading up.*1681

*So, if it is shading up, then it must be that our y-values are greater than*1688

*(and it is going to be strictly greater, because it was a dashed blue line) -2x + 4.*1693

*If we want to check that with a test point, we could try a test point, just like we did with the others.*1702

*This one is a little bit harder to do in our head, since it involves more things.*1707

*0 is greater than -2(0) + 4; 0 > 4; that test point fails, so indeed, it has to be shaded on this side--we chose the correct one.*1711

*I'm sorry; I didn't mean to make that look like a greater than or equal to.*1723

*We chose the correct symbol: y > -2x + 4, so in total, these are the inequalities that make that system be graphed like that.*1725

*Great; the second example: Graph the solutions to the system below.*1737

*The first thing we want to do--we probably want to convert this into a form where we can easily graph these.*1741

*Divide by -3; since we are dividing by a negative number, the sign flips.*1747

*We have y ≤ -2; next, 3x + y < 3, so y <...subtracting 3x...-3x...+3...we don't have to worry about flipping, because it is addition or subtraction.*1751

*And then finally, 6x - 4y...-4y < -6x - 4; divide by -4: y...it flips because we divided by a negative number, so -6/-4 becomes +3/2x; -4/-4 becomes positive 1.*1766

*At this point, we have enough for us to graph this thing; so let's draw some axes in.*1784

*We will make a tick-mark spacing; it is just a length of 1, so 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5.*1795

*Great; at this point, we can draw each one of these in.*1818

*We will draw them in as lines first; and then, from there, we will figure out how to shade.*1821

*y ≤ -2; we use a solid line, because it is less than or equal, so it is not strict.*1826

*At height -2, we are going to have a line, because we said y ≤ 2; so we graph in the line as if it was y = -2.*1832

*We have just set a vertical height of -2, and our x is allowed to freely roam.*1842

*Next, the red one: y < -3x + 3, so our first point is at the y-intercept of 3.*1847

*If we go over one step, we will have our slope of -3 kick in; we will go down 3, so we can draw this one in.*1853

*All right; oops, I should have had that be dashed, because it was a less than, so it is strict; I will dash that with the eraser.*1865

*And then finally, our green one: I won't forget to dash this one.*1879

*3/2x + 1; so we have a y-intercept at 1; if we go over 2, we go up 3, because our slope is 3/2: over 2, up 3; 1, 2, 3.*1882

*We draw this one in; all right, and now let's figure out our shading.*1894

*y ≤ -2; well, if it is less than or equal to -2, we have to have shading below that; so we shade below that here,*1909

*because we can use a test point, like (0,0), but we also see that y has to be below -2; otherwise that is going to not be true.*1925

*So, we shade in below that.*1933

*For our red one, y has to be less than -3x + 3; once again, the y's will have to be below this value.*1934

*We can use a test point, like (0,0); if we plug that in, we will have 0 < -3(0) + 3, which means 0 < +3, which is, indeed, true.*1940

*So, we shade towards that test point of (0,0).*1950

*And then finally, we have a green one: y > 3/2 + x; 3/2 + x is going to get us this one here.*1956

*So, that is going to cause us to shade up; if we plug in a test point of (0,0), we have 0 > 3/2 times 0, which means 0 > 1.*1975

*Oops, if 0 is greater than 1...that would fail to be true, so we are shading the opposite way; we are shading away from our original test point.*1987

*Our original test point was (0,0), so we are shading in the opposite direction of that.*1997

*Let's extend our red so we can see a little bit better where we are going.*2001

*At this point, we see that the only place that ends up having green, red, and blue together is this section over here.*2005

*Here (where I am shading in with a zigzag purple)--this section over here, and continuing out in that arc, will end up being the answers to our solution.*2013

*We see the solution graphed in that way.*2025

*All right, the third example: Find the maximum and minimum values of the objective function z = 2x + 7, given the below constraints.*2028

*The first thing we do is change them into a form so we can easily graph them and have a better understanding of what is going on.*2037

*Just like we had before, this is the same example that we were working with*2045

*when we were talking about the idea of linear programming, maximizing and minimizing objective functions.*2048

*So, this is the same one that we had before: y ≤ -1/3x + 8/3; 2x + y becomes y ≤ -2x - 4;*2054

*and then, we have y ≥ -1/2x...sorry, not -1/2; it gets canceled out as it adds over, so +1/2x - 3/2; great.*2067

*So, let's just give a rough sketch, so we can see what is going on.*2082

*We don't have to worry about having this perfectly precise.*2085

*The idea is just to be able to see this, because unlike graphing the solutions to an inequality,*2087

*where we never really figure out the thing, so we want to have a solid graph; in this case,*2093

*the graph is just a reference, so we can have a better understanding of how the solving is working.*2096

*So, -1/3x + 8/3; I will start at around some height of about 3.*2101

*And so, let's actually put in marks, so we have some rough sense of where we are going here.*2109

*1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5; great.*2116

*OK, the red one: positive 8/3, so we are almost at the 3; and then 1/3...it will slowly slope down to...*2126

*by the time it gets here, it will be about here; you could drop down...and it is going to be a solid line,*2134

*because it was less than or equal to; so it is like that.*2140

*And then, y ≤ -2x - 4; at -4...and then, for every one it goes to the left, it will go up 2, 2, 2...*2143

*Also, it is solid once again, because it was a less than or equal.*2158

*And then, the final one is also going to be solid; it is less than or equal to 1/2x - 3/2, so halfway between the -1 and the -2.*2162

*At 1/2x...so there...OK; it is not a perfect graph, but it gives us a rough idea of what we are looking for.*2170

*We are looking for something that is going to be in this part, including the lines.*2181

*Now, the theory of linear programming--what we learned about that--the critical idea--was that we know*2187

*that the answer is going to be (if there is an answer at all) on one of these vertices.*2192

*For the maximum and the minimum, each will show up as a vertex on this triangle (or whatever kind of picture we have).*2198

*It is going to show up as a vertex on this image--one of the corners to our set of allowable points/feasible solutions.*2205

*So, let's figure out what it is: let's do the red and the blue together first.*2213

*If that is the case, we can treat this as y = -1/3x + 8/3, and y = -2x - 4,*2217

*because for this, we are trying to figure out where the lines intersect, because we are looking for vertices.*2224

*-1/3x + 8/3 = -2x - 4; where do our red line and our blue line intersect?*2228

*Multiply everything by 3; at this point, I am just going to use blue.*2240

*Multiply everything by 3 for ease: -x + 8 cancels out those fractions; -6x - 12...*2242

*We add 12 to both sides; we get 20 over here; add x to both sides; we get -5x; divide by -5 on both sides; we get -4 = x.*2250

*We can plug this into one of the two functions; let's go with the -2x - 4, because that will be easier.*2262

*Once again, we can use it as an equality, because we are looking for a line intersection.*2267

*We aren't worried about the inequality effect; we are worried about "as if it were a line," because we only care about the vertex right now.*2271

*So, we can plug that in, and we will have that y = -2...plug in our x, -4; minus 4; y =...-2(-4) gives us + 8, minus 4; so y = 4.*2277

*Great; so one of our vertices is (-4,4).*2290

*Next, what if we had our red intersect our green?*2295

*-1/3x + 8/3--if that was the line--would be equal to (because we are looking for the intersection) 1/2x - 3/2.*2299

*So, where do those intersect? Let's multiply everything by 6, just so we can get rid of these fractions; I think that makes it easier to look at.*2309

*So, multiply everything by 6; I will just choose red for the color I will use here.*2316

*That gets us -2x plus...8/3 cancels out the 3 portion of the 6, which means it is going to be 2(8), 16.*2322

*Now, we are multiplying by 6 still, so it is going to give us 3x, multiplying by 6, minus 9, because 3/2 times 6 becomes 9.*2330

*At this point, we can solve this out: add 2x to both sides; add 9 to both sides.*2338

*Adding 9 to both sides, we get 25 =...add 2x to both sides...5x/5; 5 = x.*2342

*We can now plug this into either one of them; to me, the y = 1/2x - 3/2 looks ever so slightly friendlier, so I will plug into that one.*2348

*So, y = 1/2x, this one right here, so y = 1/2(5) - 3/2; so 5/2 - 3/2 = 2/2, which equals 1.*2358

*At this point, we have another vertex; 5 = x and y = 1; so that is going to occur where the red and green intersect over here, (5,1).*2375

*And then finally, let's look at where the blue and the green would intersect, the one that we haven't looked at yet.*2390

*-2x - 4 = 1/2x - 3/2; multiply everything...*2396

*I'm sorry; let's change that into colors, just so we can keep our color scheme going.*2404

*1/2x - 3/2...we will multiply everything by 2 to get rid of that fraction; so -4x - 8 = x - 3.*2409

*Put our x's together; put our constants together; add 3 to both sides; we get -5.*2420

*Add positive 4x to both sides; we get 5x; divide by 5 on both sides; we get -1 = x.*2424

*We can now plug that into either one; this one, to me, looks friendlier, so y = -2(-1) - 4; y = +2 (because the negatives cancel out) - 4; y = -2.*2431

*At this point, we have y = -2 and -1 = x; so a point here will be (-1,-2).*2447

*We have found the three vertices to this; now, we need to go use that information.*2458

*They are at (-4,4), (5,1), (-1,-2); so now, we want to evaluate them in our objective function, z = 2x + 7y,*2464

*and figure out which one makes z biggest, which one makes z smallest, and which one is just somewhere in the middle.*2473

*All right, so where are our maximum and minimum?*2479

*Our vertices are (-4,4), (5,1), (-1,2); we have (-4,4) here, (5,1), and (-1,-2) here.*2481

*All right, so let's start with (-4,4); we plug that in for z = 2x + 7y.*2494

*So, what will our z end up coming out to be?*2503

*2 times our x component is -4; plus 7 times our y component is positive 4, so that gets us -8, plus 7 times 4 (28), which equals 20.*2506

*So, we get 20 out of (-4,4).*2517

*Let's try our next one, (5,1): plug that in--we get z = 2 times our x component of 5,*2519

*plus 7 times our y component of 1 which equals 10 + 7, so we get 17.*2527

*And then, finally, we plug in our (-1,-2), and that gets us z = 2 times x component, -1, plus y component, 7(-2).*2535

*So, we get -2 - 14, which is equal to -16.*2552

*All right; at this point, we can see who is our winner for maximum; that is going to be z = 20, which occurred at (-4,4).*2558

*So, (-4,4) is the maximum for our objective function.*2566

*And (-1,-2) put out -16, so that was the minimum value that we ended up seeing.*2571

*And (5,1) gave us +17, but that is neither a maximum nor a minimum, so we just forget about it.*2581

*All right, the final example: A car lot needs to choose what cars to order for selling on their lot.*2589

*They have two options: the Nice, which they purchase at 20000 and sell at 30000, and the Superfine, which they purchase at 50000 and sell at 65000.*2596

*If they have a total budget of 2.9 million dollars for purchasing cars, and 100 spots on the lot for new cars,*2606

*how many of each car should they buy to maximize profit?*2613

*Assume that they manage to sell whatever they purchase.*2616

*The idea here is that they purchase a car, and then they mark it up and sell it.*2619

*And so, the difference between those two is how much money they make.*2623

*But they have limits on how much money they start with for purchasing cars, and how many spots they have to actually put the cars on the lot.*2626

*So, they have to figure out what is the best combination of cars to purchase, to be able to maximize how much money they make out of it.*2633

*So, with all of this in mind, let's start setting some things up.*2640

*First, we are going to have to figure out how many numbers of Nice they buy.*2643

*Let's say N is going to be the symbol that we use for saying the number of Nice cars that they buy.*2647

*The Nice cars that they buy will be N, however many that is.*2655

*And then, the Superfines that they buy--S will symbolize the number of Superfines that they buy.*2659

*OK, now another thing that we have here is a lot of really big numbers: 30000, 20000, 50000, 65000, 2.9 million dollars.*2666

*Now, we could work with the numbers as they actually are, and everything would end up working out.*2678

*But we would have this extra factor of 100 showing up; all of these things are divisible by 1000.*2681

*So, I am going to say, "Why don't we make things a little bit easier on ourselves?"*2686

*And since we are dealing with lots of money--we are dealing with all of these big things,*2690

*measured in the thousands and tens of thousands, let's convert everything.*2694

*For ease, let's work in terms of thousands of dollars; let's talk about everything in terms of how many K it is--*2699

*how many thousand (K, kilo, thousand)--how many thousand dollars everything is.*2708

*OK, that will make things a little bit easier, just in how many 0's we have to write down.*2713

*So, next, let's figure out how much profit they make off of a single Nice,*2718

*because what we want to do is maximize the profit; so we need to create some function p*2724

*that is going to be p = the profit, so we need to come up with some equation that describes p*2729

*in terms of how many Nice and how many Superfines they buy--how much N and S we have to go around--*2735

*because we know for sure that they will sell all of them.*2740

*So, how much profit do they make if they buy such a number of Nice and such a number of Superfines?*2742

*First, we have to figure out how much profit they make off of selling a Nice and how much profit they make off of selling a Superfine.*2747

*A Nice: if they buy a Nice, it costs them 20K, 20000; it costs them 20000, which is 20K in our new form.*2753

*They can sell that for 30K, which would net them a profit of the amount that they sold it for, minus the amount that they had to buy it for.*2766

*That would net them a profit of 10K.*2778

*For the Superfines, they cost them 50K to buy; they sell them at 65K; so each one they sell is going to net them a profit of 15K.*2782

*OK, with this in mind, let's write p in green for American money, since we are dealing with U.S. dollars here.*2800

*p = 10 times N, the number of Nices that we sell, plus 15 times S, the number of Superfines that we sell,*2810

*because remember: for each Nice we sell, we manage to make 10K; for each Superfine we sell, we manage to make 15K.*2822

*So, the total profit is going to be 10N + 15S...K, but we don't have to worry about the K's now.*2829

*We won't worry about it when we are actually dealing with the numbers--the fact that they have the unit of thousands of dollars.*2836

*All right; but there are some restrictions here.*2842

*We were told that there is a restriction of 2.9 million dollars for purchasing cars.*2845

*They don't just have unlimited amounts of money to buy cars.*2852

*If they had unlimited amounts of money, they would want to stock their entire car lot with nothing but Superfines.*2855

*They would just want to fill the thing all up with Superfines; but they have a cost to these things.*2860

*So, we have to deal with that; so if it is 2.9 million for purchasing cars, then...*2865

*well, each Nice that we buy is going to be 20N; and each Superfine that we buy is going to cause us to spend 50S.*2873

*So, 20N + 50S is how much we have spent on cars.*2884

*Now, they can spend up to 2.9 million; so 20N + 50S has to be less than or equal to their total budget.*2887

*And their total budget is 2.9 million, which we can write out as 2900, because 2.9 million is the same thing as 2900 thousand.*2895

*So, since we divided everything by 1000, we are now dealing with 2900.*2904

*All right, another piece of information that we had was that there are only 100 spots on the lot.*2908

*There is a maximum number of spots that the can fit cars into.*2914

*So, if that is the case--if they only have 100 spots to fit cars into--well then, the number of Nices that we are buying,*2918

*plus the number of Superfines that we are buying, has to be less than or equal to 100,*2924

*because they can only fit up to 100 of these cars on the lot.*2929

*So, that is two requirements so far.*2932

*Now, there are two hidden requirements that we might not see, but they will make it easier to actually graph the thing.*2934

*Think about this: is it possible to have a negative number of Nices--can they buy negative cars?*2942

*No, the lowest number of Nices that they can buy, or the lowest number of any car that they can buy, is 0.*2947

*So, we know that however many Nices that they buy, it has to be greater than or equal to 0.*2952

*And similarly, S must be greater than or equal to 0.*2958

*We have four requirements on this: the amount of money that have to spend (20N + 50S has to be less than or equal to 2900);*2961

*N + S has to be less than or equal to 100 (the number of spots on the lot); and the fact that they can't buy negative cars--*2968

*N has to be greater than or equal to 0, and S has to be greater than or equal to 0.*2974

*And our objective function that we are trying to maximize, because we are trying to maximize our profit, is this p = 10N + 15S.*2977

*We want to maximize p; at this point, we bring to bear the power of linear programming.*2987

*First, let's convert some of this stuff into things that we can easily graph.*2993

*So, if we are going to graph this, so we can see what is going on, we have to choose something to be horizontal and something to be vertical.*2998

*So, just arbitrarily, let's make N the horizontal and S the vertical; N gets to be horizontal; S is vertical.*3009

*With this idea in mind, we can now talk about ordered pairs that come in the form (N,S), just as (x,y) is (horizontal,vertical).*3022

*Since we made N horizontal and we made S vertical, it is going to come in (N,S).*3031

*Now, there is no reason it couldn't be flipped; but we just chose one, and we stick to it.*3036

*And as long as we stick to it, it will work out.*3039

*OK, at this point, if that is the case, then since S is the vertical, we normally solve for y (the vertical) in terms of other stuff;*3041

*so we want to solve for S (the vertical) in terms of other stuff, because that is what we are used to.*3047

*N + S ≤ 100 means that S has to be less than or equal to -N + 100.*3051

*20N + 50S ≤ 2900; 50S ≤ -20N + 2900.*3058

*We divide both sides by 50; we have S ≤ -2/5N + 2900/50 (becomes...give me just a second...58; I had to use a calculator for that one).*3068

*So, 2900/50 becomes 58; we have S ≤ -2/5N + 58.*3084

*So, with this in mind, we have enough to be able to see what we are allowed to go for here.*3091

*If N has to be greater than or equal to 0, then we have this graph...*3099

*If N has to be greater than or equal to 0, this is actually going to be the green one; so for a second, forget about that mistake.*3104

*N has to be greater than or equal to 0, so we set N equal to 0 and draw a line.*3114

*S is allowed to go wherever it wants; everything on this side is fine by our N > 0.*3119

*S has to be greater than or equal to 0: we set a line S = 0; everything greater than that is fine by S ≥ 0.*3126

*Now, the interesting parts are actually going to end up being the S ≤ -N + 100...*3136

*If that is the case, we can get a good sense of this; let's say 100 is up here; at -1, it marches down like this at a 45-degree angle.*3141

*And we know that S has to be less than or equal to -N + 100, so it will be below this.*3152

*And S ≤ -2/5N + 58; it is going to start at a y-intercept height of 58, but it goes much less steeply down at -2/5.*3159

*So, it goes like that; and it also will allow for the part below it.*3172

*So, the total set of things that is allowed is this highlighted part in yellow.*3177

*However, the only things that are really going to be interesting are going to be our vertices--the vertex here, here, here, and here.*3184

*We will circle those in yellow, just to make them a little bit more obvious.*3197

*But it is probably becoming pretty hard to see through the loudness of those colors.*3200

*We have an idea of those corners.*3203

*Now, notice: (0,0) is one of the points--we have an origin corner.*3205

*Is it possible to make any money if we plug that into p =...0 for N and 0 for S, if we buy no cars to sell?*3209

*No, we can't make any money; so we only have three things to care about:*3217

*where red intersects green, where red intersects blue, and where blue intersects purple.*3220

*So, if that is the case, let's look at where red intersects green first.*3227

*That is -2/5N + 58 gives us N ≥ 0.*3231

*If that is the case, if it is N ≥ 0, that is a little bit difficult, because we are set up for solving for S.*3241

*So, where is that going to intersect? Well, at N ≥ 0, we want to figure out what that comes out to be.*3248

*If S is equal to -2/5 times 0 plus 58, if we know that this is going to be equal to S, and N is greater than or equal to 0,*3259

*we can treat it as N = 0, because we are only dealing with lines here.*3269

*So, we plug that in here; we have -2/5 times 0, plus 58, equals S; so 58 is S, and we have an N of 0.*3272

*So, the point that we get out of this is going to be N = 0, 58 = S, so (0,58) is one of our points so far.*3285

*Next, let's look at where the blue intersects the purple.*3294

*-N + 100 =...the purple had a height of S = 0, no height at all, so that is just going to be 0, so 100 = N.*3299

*That means we have (100,0), because if we were to plug that into N + S = 100,*3311

*since we are dealing with the lines now--if we plugged in that, we would have 100 + S = 100; therefore, S = 0.*3321

*So, we have (100,0); notice that these are the two extremes possible for the car lot to sell.*3328

*It could sell all Superfines--it could sell nothing but Superfines--and it has a maximum amount of money that it can spend, 2.9 million.*3334

*So, it can't buy its lot entirely full of Superfines; it could only have 58 Superfines on the lot before it would run out of money to purchase cars with.*3342

*On the other hand, it could buy 100 of the Nices; and it would still have money left over, but it has run out of spots on the lot.*3351

*So, it could buy 100 Nices and no Superfines, and have extra money left over;*3358

*or the lot could buy all Superfines and run out of money before it runs out of spots.*3363

*It would have 58 Superfines, but still, that gives us another 42 spots left for Nices.*3370

*So, those are the two extremes possible, so far.*3375

*Finally, we could also have the possibility of if we have the blue and the red intersect.*3378

*If we had -N + 100 equal to -2/5N + 58, we can multiply everything by 5, just to make it a little bit easier on us.*3385

*-5N + 500 = -2N + 58...we turn to a calculator to figure out what 58 times 5 is.*3402

*We get 290 out of that; we add 5n to both sides; we subtract 290 on both sides.*3411

*500 - 290 gets us 210; we add 5n to both sides; that gets us 3n.*3423

*That gets us...dividing by 3...I'm sorry; not 80, but 70, equals N.*3429

*We use that to figure out how many Superfines that would bring us.*3436

*By our lot equation of S ≤ -N + 100, we want to be on the line (equals), so S = -N + 100; so S = -70 + 100, or S = 30.*3440

*The three possibilities are (0,58), (100,0), and (70,30).*3457

*At this point, we just want to actually plug each one of them in and figure out which of these works best.*3466

*Let's start with the point that we have arbitrarily made the red point.*3473

*If we sell nothing but Superfines--we spend all of our money on Superfines--then that is 10 times 0, plus 15 times 58, just so we keep up the same pattern.*3477

*15 times 58 becomes 870; so we would make a total of 870 thousand dollars at the lot if we bought nothing but Superfines.*3490

*Next, we will do the one that we arbitrarily made the blue point: the profit out of that one would be 10(100) + 15(0), the number of Superfines.*3501

*So, the profit that we would make off the lot would be 1000 + 0.*3512

*So, if we sold nothing but Nices, we would be able to get 1000 thousand, or 1 million, off of our sales.*3516

*But what if we sold a combination of Nice and Superfine?*3526

*If we sold a combination, the one we arbitrarily made the green point, then our profit would be 10(70) + 15(30).*3529

*10 is the profit from the Nices; 15 is the profit from the Superfines; we are selling 70 Nices and 30 Superfines.*3541

*We know that it is possible, because it is on our vertex.*3547

*That gets us 700 + 15(30) is 300 + 150; 450; so this is a total profit of 1150 thousand from 70 and 30.*3550

*So, the best, the optimum solution, for maximum profit, is to sell 70 Nices and 30 Superfines.*3567

*And if you are curious, that came out to be a profit of 1150 thousand, so the total profit,*3582

*the maximum possible profit for the lot, will end up being 1150 million dollars.*3588

*All right, cool; that finishes for linear inequalities.*3596

*There is this idea of linear programming, and there is also just being able to figure out what the things that our linear inequality allows for are.*3600

*Linear programming is a great thing that we can use all of our information about linear inequalities for.*3605

*But you can also just work with linear inequalities.*3609

*All right, we will see you at Educator.com later--goodbye!*3611

3 answers

Last reply by: Professor Selhorst-Jones

Sun Jul 24, 2016 12:59 PM

Post by Joshua Jacob on July 23, 2014

Professor Vincent,

Could you explain the second equation in blue for example 2?

I graphed it out and even graphed it in my calculator and the shaded area is under the line, not above it.

Thanks for the amazing lectures!

Josh