For more information, please see full course syllabus of Pre Calculus
For more information, please see full course syllabus of Pre Calculus
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 answerwe'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

 It helps to begin by putting the inequalities into slopeintercept form (y=mx+b, where m is the slope and b is the yintercept). Of course, since we're dealing with inequalities, we won't be able to precisely fit the form (since it has an `=' sign), but we can use the same structure. In this problem, both inequalities are already in that structure, so we're ready to start plotting them.
 Plot out a line based on each inequality. While plotting the line on the graph, treat it as if it were a normal equation that you were graphing. The only thing to keep in mind is that ≤ and ≥ get solid lines, while the strict inequalities of < and > get dashed lines.
 Once the lines are plotted on the graph (and solid or dashed, according to the inequality type), shade each one according to its inequality. This can be done in one of two ways. First, if it's in the slopeintercept structure (like ours are), you can look at the "direction" of the inequality sign. If the y is greater (y ≥ or y > ), then we shade above the line. If the y is lesser (y ≤ or y < ), then we shade below the line. Alternatively, if you're not sure of the above or the inequality is not in slopeintercept, choose a "test point" on the graph. Plug it in to the inequality. If it works, shade the side that the test point is on. If it does not satisfy the inequality (it "breaks"), shade the opposite side.
 Once you have shaded each of the inequality lines appropriately, look for where the shadings overlap. The overlap is the set of solutions.

 It helps to begin by putting the inequalities into slopeintercept form (y=mx+b, where m is the slope and b is the yintercept). Of course, since we're dealing with inequalities, we won't be able to precisely fit the form (since it has an `=' sign), but we can use the same structure.
3x+y < 5 ⇒ y < −3x + 5 2x+2y < −4 ⇒ 2y < −2x−4 ⇒ y < −x−2  Plot out a line based on each inequality. While plotting the line on the graph, treat it as if it were a normal equation that you were graphing. The only thing to keep in mind is that ≤ and ≥ get solid lines, while the strict inequalities of < and > get dashed lines.
 Once the lines are plotted on the graph (and solid or dashed, according to the inequality type), shade each one according to its inequality. This can be done in one of two ways. First, if it's in the slopeintercept structure (like ours are), you can look at the "direction" of the inequality sign. If the y is greater (y ≥ or y > ), then we shade above the line. If the y is lesser (y ≤ or y < ), then we shade below the line. Alternatively, if you're not sure of the above or the inequality is not in slopeintercept, choose a "test point" on the graph. Plug it in to the inequality. If it works, shade the side that the test point is on. If it does not satisfy the inequality (it "breaks"), shade the opposite side.
 Once you have shaded each of the inequality lines appropriately, look for where the shadings overlap. The overlap is the set of solutions.

 One way to approach this problem would be to graph all the solutions to the system of inequalities, then plot the point, and look to see if it is within the solution set on the graph. If it is, then it's a solution. If not, then it is not a solution. However, a much easier way to do it is to just check each of the two inequalities with the point. If it satisfies each inequality, then it's a solution. If it fails to satisfy one or both, then it is not a solution.
 Check to see if it satisfies each of the inequalities. [Remember, (4, −7) means the combination of x=4 and y=−7.]
−3x−y < 6 ⇒ −3(4) − (−7) < 6 ⇒ −12 + 7 < 6 ⇒ ⇒ − 5 < 6 4x+2y ≥ 2 ⇒ 4(4) + 2(−7) ≥ 2 ⇒ 16 − 14 ≥ 2 ⇒ 2 ≥ 2  Thus, since the point satisfies ("is true in") each of the inequalities, it is a solution to the system. [Notice that just because it is a solution to the system does not mean it is the only solution to the system. There are infinitely many solutions, this point is just one of them.]
 Begin by identifying the equation for each line first. Just treat it as a line equation to begin with, even though they will later become inequalities.
 Let's start with the red, upper line. We see it has a yintercept of b=7. From there, it crosses the xaxis by dropping down −7 and moving right 7. Since slope is the change in the vertical divided by the change in the horizontal, we get
Putting these two things together, we have y=−x + 7 for the red line equation.m = −7 7= −1  Now we will work on the blue, lower line. We see it has a yintercept of b=−6. From there, it crosses the xaxis by rising up 6 and moving right 2. Since slope is the change in the vertical divided by the change in the horizontal, we get
Putting these two things together, we have y=3x −6 for the blue line equation.m = 6 2= 3  Now that we know equations to describe the lines, we need to figure out the appropriate inequality. For the red, upper line, we see that it is solid, so we will use either ≤ or ≥ . Notice that the solution set is always below the red line, so the y value must be less, giving us
For the blue, lower line, we see that it is dashed, so we will use either < or > . Notice that the solution set is always above the blue line, so the y value must be greater, giving usy ≤ −x + 7 y > 3x −6

 It helps to begin by putting the inequalities into slopeintercept form (y=mx+b, where m is the slope and b is the yintercept). Of course, since we're dealing with inequalities, we won't be able to precisely fit the form (since it has an `=' sign), but we can use the same structure. For this problem, everything is already in the slopeintercept structure, so we can move on to the next step. [If you're confused by how to graph y ≥ 0, remember, y=0 would just graph as a horizontal line at height 0.]
 Plot out a line based on each inequality. While plotting the line on the graph, treat it as if it were a normal equation that you were graphing. The only thing to keep in mind is that ≤ and ≥ get solid lines, while the strict inequalities of < and > get dashed lines.
 Once the lines are plotted on the graph (and solid or dashed, according to the inequality type), shade each one according to its inequality. This can be done in one of two ways. First, if it's in the slopeintercept structure (like ours are), you can look at the "direction" of the inequality sign. If the y is greater (y ≥ or y > ), then we shade above the line. If the y is lesser (y ≤ or y < ), then we shade below the line. Alternatively, if you're not sure of the above or the inequality is not in slopeintercept, choose a "test point" on the graph. Plug it in to the inequality. If it works, shade the side that the test point is on. If it does not satisfy the inequality (it "breaks"), shade the opposite side.
 Once you have shaded each of the inequality lines appropriately, look for where the shadings overlap. The overlap is the set of solutions.

 It helps to begin by putting the inequalities into slopeintercept form (y=mx+b, where m is the slope and b is the yintercept). Of course, since we're dealing with inequalities, we won't be able to precisely fit the form (since it has an `=' sign), but we can use the same structure.
 Converting them to slopeintercept structure, we get the following:
2x+3y > −12 ⇒ 3y > −2x − 12 ⇒ y > − 2 3x −4 3x−y ≥ − 8 ⇒ −y ≥ −3x − 8 ⇒ y ≤ 3x + 8 −3x+6y < 30 ⇒ 6y < 3x + 30 ⇒ y < 1 2x + 5  Plot out a line based on each inequality. While plotting the line on the graph, treat it as if it were a normal equation that you were graphing. The only thing to keep in mind is that ≤ and ≥ get solid lines, while the strict inequalities of < and > get dashed lines.
 Once the lines are plotted on the graph (and solid or dashed, according to the inequality type), shade each one according to its inequality. This can be done in one of two ways. First, if it's in the slopeintercept structure (like ours are), you can look at the "direction" of the inequality sign. If the y is greater (y ≥ or y > ), then we shade above the line. If the y is lesser (y ≤ or y < ), then we shade below the line. Alternatively, if you're not sure of the above or the inequality is not in slopeintercept, choose a "test point" on the graph. Plug it in to the inequality. If it works, shade the side that the test point is on. If it does not satisfy the inequality (it "breaks"), shade the opposite side.
 Once you have shaded each of the inequality lines appropriately, look for where the shadings overlap. The overlap is the set of solutions.
 Begin by identifying the equation for each line first. Just treat it as a line equation to begin with, even though they will later become inequalities.
 Let's start with the red, vertical line. This is possibly the most difficult because we can't give an equation in slopeintercept form. We have to realize that it has a fixed value of x=−5, while the y value is allowed to "roam freely". Thus, it can be described with the equation x=−5
 Now we will work on the blue, horizontal line. It has a yintercept of b=−3 and, since it is horizontal, a slope of m=0. Thus it has the equation y=−3. [We can also see this by the same logic as we used above: y is fixed, while x is allowed to "roam freely".]
 Finally, we finish with the green, diagonal line. It has a yintercept of b=−2. It crosses the xaxis from there by going up 2 while going backwards by −1. Since slope is rise over run, we have
giving the line equation y = −2x −2.m = 2 −1= −2,  Now that we have equations to describe all of the lines, we need to figure out the appropriate inequality for each. For the red, vertical line, we see that it is dashed, so we will use either < or > . Notice that the solution set is always right of the red line, so the x value must be greater, giving us
For the blue, horizontal line, we see that it is solid, so we will use either ≤ or ≥ . Notice that the solution set is always above the blue line, so the y value must be greater, giving usx > −5
For the green, diagonal line, we see that it is dashed, so we will use either < or > . Notice that the solution set is always above the green line, so the y value must be greater, giving usy ≥ −3 y > −2x −2

 We will approach this problem by the method of linear programming. We have various constraints on our variables, and the maximum/minimum to the objective function must come from a point within the space determined by those constraints. Specifically, from the theory of linear programming, we know the max/min must occur at one of the vertices (corners) of the space determined by the constraints.
 While it is not strictly necessary to graph the constraints, it can sometimes be helpful to assist us in understanding the problem. Graphing them carefully isn't the important part: just a rough sketch so that we can better understand what we're doing. Here's a graph of the constraints and the space they determine:
 We know that the minimum and maximum of the objective function (z) occur at the vertices of the constraint space (at the corners). Thus, we want to solve for what those points are. Find these intersections by solving each pair of constraints as system of equalities.
First, we see where the lines y=0 and y=x intersect:
Next, where the lines y=0 and y = −2x + 12 intersect:0 = x ⇒ y = 0
Finally, the last pair of y=x and y=−2x+12:0 = −2x + 12 ⇒ x = 6 ⇒ y = 0 x = −2x + 12 ⇒ x=4 ⇒ y = 4  Thus, our three vertices are located at the points
Now plug each of these points into the objective function z to find the value at those points:(0, 0) (6, 0) (4, 4) (0, 0) ⇒ z = 3(0) + 2(0) = 0 (6, 0) ⇒ z = 3(6) + 2(0) = 18
Therefore we see that plugging in (0, 0) gives the smallest number, while (4, 4) gives the largest number.(4, 4) ⇒ z = 3(4) + 2(4) = 20

 First, let's begin by naming variables:
Once we have our variables in place, we can come up with an equation for the thing we're trying to minimize. We're trying to minimize the price (P) of the ore purchase. The total price isc = kilograms of commonium r = kilograms of rarite
However, from the problem, we have constraints on the amount of commonium and rarium we must purchase: we need enough of the ores to hit our required totals of elements A and B.P = 5c + 20r  Working on the constraints, we know that we need at least 50 kg of type A. Thus, it must be that the total amount of A obtained from the ores is greater than or equal to 50 kg:
Similarly, we need at least 4 kg of type B, so the total amount obtained from the ore must be greater than or equal:50 ≤ 0.9c + 0.2r
Those are all the constraints that the problem directly gave us, but there are two very important, implied constraints: we cannot have a negative amount of ore. While it was never directly stated, the idea is obviousit is impossible to have negative kilograms of ore. However, it is important to state this mathematically so we can work with it as a constraint:4 ≤ 0.1c + 0.8r c ≥ 0 r ≥ 0  Since we have a set of linear constraints and an objective function (the price, P) that we are trying to minimize, we do this with linear programming. We need to find the vertices to the space of possible choices.
To help us visualize the problem, let's graph the space of possible choices. To make it easier to graph, put the constraints in a form similar to slopeintercept (let's arbitrarily choose c to be the independent variable [what x usually is]):
50 ≤ 0.9c + 0.2r ⇒ 0.2 r ≥ 50 − 0.9 c ⇒ r ≥ −4.5c + 250 4 ≤ 0.1c + 0.8r ⇒ 0.8 r ≥ 4 −0.1c ⇒ r ≥ −0.125c+ 5  Looking at our visual reference, we see that there are only two vertices: where the red line (from r = −4.5c + 250) intersects the vertical axis and where it intersects the horizontal axis.
The intersection for the vertical axis will occur at c=0:
The intersection for the horizontal axis will occur at r=0:r = −4.5(0) + 250 ⇒ r = 250 (0) = −4.5 c + 250 ⇒ 4.5 c = 250 ⇒ c ≈ 55.556  Thus we have two vertices to check in our price function:
c=0, r=250:
c=55.556, r=0:P = 5(0) + 20 (250) = 5000
Thus, it clearly is much less expensive to purchase nothing but commonium ore.P = 5(55.556) + 20 (0) = 277.78
 Begin by naming variables:
Once we have the variables in place, we can create an equation to describe the thing we're trying to maximize. The amount of income (I) the factory brings in is based on the number of each type of widget produced:a = number of Type A b = number of Type B
From the problem, we have constraints on the number of widgets that can be produced. We must mathematically name these to go on.I = 135a + 55b  We know that each widget produced uses up some amount of raw material. We only have 1000 units to work with, so the total amount of raw material used must be less than or equal to that:
Similarly, there are only 440 hours of production labor to go around, so the total amount used must be less than or equal:1000 ≥ 5a + 2b
Those are all the constraints that the problem directly gave us, but there are two very important, implied constraints: we cannot produce a negative number of widgets. It is important to state this mathematically so we recognize it as one of our constraints:440 ≥ 2a + b a ≥ 0 b ≥ 0  Since we have a set of linear constraints and an objective function (the price, I) that we are trying to maximize, we do this with linear programming. We need to find the vertices to the space of possible choices.
To help us visualize the problem, let's graph the space of possible choices. To make it easier to graph, put the constraints in a form similar to slopeintercept (let's arbitrarily choose a to be the independent variable [what x usually is]):
1000 ≥ 5a + 2b ⇒ 2b ≤ 1000 −5a ⇒ b ≤ −2.5a + 500 440 ≥ 2a + b ⇒ b ≤ 440 −2a ⇒ b ≤ −2a + 440  Looking at our visual reference, we see that there are a total of four vertices.
b = −2a + 440 and a=0:
b = −2.5a + 500 and b = −2a + 440:b = −2 (0) + 440 ⇒ b = 440 ⇒ a = 0
b = −2.5a + 500 and b=0:−2.5a + 500 = −2a + 440 ⇒ a=120 ⇒ b = 200
a=0 and b=00 = −2.5a + 500 ⇒ a = 200 ⇒ b = 0  Finally, we check all of our vertices using the income function:
a=0, b=440
a = 120, b=200I = 135(0) + 55 (440) = 24 200
a = 200, b=0I = 135 (120) + 55(200) = 27 200
a=0, b=0I = 135(200) + 55(0) = 27 000
Thus, of all the vertices, the one that gives us the maximum income is a=120, b=200.I = 135 (0) + 55(0) = 0
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.
Answer
Systems of Linear Inequalities
Lecture Slides are screencaptured 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 RefresherSolutions
 RefresherNegative Multiplication Flips
 RefresherNegative Flips: Why?
 RefresherStick 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 LineNot Solutions
 Solid LineAre Solutions
 Test Points for Shading
 Graphing a System
 Solutions are Best Found Through Graphing
 Linear ProgrammingIdea
 Linear ProgrammingMethod
 Example 1
 Example 2
 Example 3
 Example 4
 Intro 0:00
 Introduction 0:04
 Inequality RefresherSolutions 0:46
 Equation Solutions vs. Inequality Solutions
 Essentially a Wide Variety of Answers
 RefresherNegative Multiplication Flips 1:43
 RefresherNegative Flips: Why? 3:19
 Multiplication by a Negative
 The Relationship Flips
 RefresherStick 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 LineNot Solutions
 Solid LineAre 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 ProgrammingIdea 19:52
 Use a Linear Objective Function
 Variables in Objective Function have Constraints
 Linear ProgrammingMethod 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
Precalculus with Limits Online Course
Transcription: Systems of Linear Inequalities
Hiwelcome 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 signsleft 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 thiseverything 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 operationaddition, subtraction, multiplication, and divisionon 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 cause0153
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 with0172
negative anything that you are multiplying on both sidesyou 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 happenwhy 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 ab was originally more right than awe 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 trueit checks out: (3)^{2} is greater than 0, because 9 is greater than 0.0312
So, that is great; but 3 > 0that 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 things0330
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 stuffnot because you wouldn't get it0367
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 everything0395
just using these basic operationsaddition, 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 yhow 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 goesit 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 suddenlyit 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 yvalue for a given xvalue, 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 varietyit is this large area of possibilities.0602
This brings up a question: what does the line become if the inequality is strictless than or greater than0609
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 notstrict 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 notstrict 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 pointit is completely your choice0729
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 line0898
(if it is a strict inequality, less than or greater than); and you use a solid line if it is a notstrict inequality0904
(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 linedashed 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 bottomright 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 arewe 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 equalityif 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 togetherthat 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 overlap1158
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 portionsome 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 themthings 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 function1309
(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 ≤ 8we 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 ≥ 3we 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 yintercept 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 solutionsand 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 exists1421
if there is some optimum maximum or minimumif we can maximize or minimize our objective function1427
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 yintercept is right here at 4.1583
So, its yintercept 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 yvalues are greater than1688
(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 sidewe 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 dowe 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 tickmark 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 yintercept 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 yintercept 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 with2045
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 programmingwhat we learned about thatthe critical ideawas that we know2187
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 imageone 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 + 8/3if that was the linewould 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 inwe 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 buyS 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 moneywe 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 is2699
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 p2724
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 buyhow much N and S we have to go around2735
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 numbersthe 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 caseif they only have 100 spots to fit cars intowell 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 Nicescan 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 cars2968
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 45degree 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 yintercept 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 verticesthe 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 pointswe 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 nowif 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 Superfinesit could sell nothing but Superfinesand 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 Superfineswe spend all of our money on Superfinesthen 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 latergoodbye!3611
3 answers
Last reply by: Professor SelhorstJones
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