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Algebra 2: 3.4: Linear Programming — Transcript

by Mrs. Allen · 2,926 words · 412 segments · language en · Watch on YouTube

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0:00welcome to the 3.4 lesson video today
0:02we're going to be learning about linear
0:04programming so I'm going to begin by
0:06telling you what that means and our goal
0:09is to solve systems of inequality
0:12specifically using linear programming so
0:15let's talk about what this process is
0:17linear programming is a method for
0:19finding a minimum or maximum value of
0:22some quantity given a set of constraints
0:25and another word for constraints um is
0:28limitations so you can see right here in
0:31this example that we have a system of
0:33linear inequalities and because the
0:37domain and range are limited by these
0:39inequalities these are called
0:41constraints as you can see in the graph
0:43at right the region that is formed by
0:46the linear inequality boundary lines and
0:48the shading overlapping of the shading
0:50that's called the feasible region and
0:52the feasible region is the graph of the
0:54system and it contains all the points
0:56that satisfy all of the constraints or
0:58all of the linear inequality
1:01an objective function is one more term
1:02that we're going to be discussing today
1:04before we get to examples an objective
1:06function is an equation that model the
1:08quantity that we are minimizing or
1:10maximizing so for example two examples
1:13of minimizing something would be
1:14minimizing cost or minimizing time and
1:17two examples for maximizing something
1:19would be maximizing profit or maximizing
1:22items produced so you can see at right
1:25we have an example equation for an
1:27objective function and that is 2x + y
1:30equals c where C is some positive number
1:32and you can see right there that we have
1:34the feasible region in the yellow um
1:37surrounded by the boundary lines of the
1:40inequalities and we have two red
1:43objective functions and the only
1:45difference between them is the value of
1:48C the graphs of the objective function
1:51for the various values of c would be
1:53parallel lines and when the lines are
1:55closer to the origin those represent the
1:58smaller values of c and those further
2:00away from the origin represent bigger
2:03values of c so you can see right here
2:05that the line 2x + yal 7 is intersecting
2:10the feasible region at the coordinate
2:12point
2:1323 and the other objective function 2x +
2:18y = 17 is intersecting the feasible
2:21region at the coordinate 73 these
2:24vertices of the feasible region
2:26represent the least and the greatest
2:29values for the objective function
2:31basically you need to realize that the
2:33domain is going to be restricted
2:35typically for real world situations and
2:37we've actually been working with this
2:38already this year um when we are talking
2:41about uh for example the number of
2:43people that variable would not be
2:45negative so that would be restricting
2:47the domain which means we have some
2:49constraints that we're dealing with and
2:52as in the previous lesson the most
2:54reasonable method to determine the
2:55common Solutions of a set of
2:57inequalities is to graph so that's what
2:58we're about to do in our key concept box
3:01here we're going to be talking about the
3:02vertex principle of linear programming
3:04this is something important and you want
3:05to jot this down in your notes so if we
3:08are having a maximum or minimum value of
3:11the linear objective function it's going
3:13to occur at one or more vertices of the
3:15feasible region and we actually looked
3:17at that above when I was talking about
3:19this example right here the maximum and
3:21minimum version values would be at the
3:24vertices I'm assuming you remember what
3:26vertices or the word vertex in singular
3:29form means
3:30a Vertex is when two lines come together
3:33to form a corner so now let's dive into
3:37example one and learn how to solve a
3:39problem using linear programming by
3:41testing in the objective function all of
3:43the vertices of the feasible region so
3:46here in example one we need to figure
3:48out what point in the feasible region
3:49maximizes the function P for the
3:52objective function P equal 2x + y so you
3:55can see at right that we're given the
3:57constraints so take a moment to jot down
3:59these four
4:00constraints these are all linear
4:02inequalities and as usual we're about to
4:05graph um and as in the past we typically
4:08solve for y to put it in slope intercept
4:10form to graph easily so you can see what
4:12I did was I just solved for y so I
4:14subtracted the X and divided by two so
4:17I'm getting Y is less than or equal to
4:19-2x + 5vs and then I just put it in
4:22decimal form just to show you that five
4:24halves is the same thing as 2.5 so now
4:26what I'm going to do is graph that
4:28function
4:31and I'm going to go to 2 and a half on
4:33the Y
4:34AIS and then the slope is one2 so I'm
4:37just going to go down one and over
4:46two and then connect with a solid
4:48boundary line I'm also going to be
4:50shading below because the inequality
4:52symbol is less than or equal to
5:00as you can see the skills that we
5:02practice in Chapter 2 are necessary for
5:05this problem so now I'm going to do a
5:07similar thing and solve for y for the
5:08second inequality when solving for Y for
5:10the second inequality I get Y is greater
5:13than or equal to x - 2 so I'm going to
5:15go to minus two on the Y AIS and the
5:18slope is one over one so that means
5:20we're going up one over
5:22one this is also going to be a solid
5:25boundary line because we have a line
5:26underneath the inequality symbol and
5:29going to be shading above for the third
5:32inequality it is just X is greater than
5:35or equal to Z which means it's a
5:36vertical line going through the origin
5:38and all values that are greater than
5:40equal to zero are solutions and for the
5:42last inequality it is uh Y is greater
5:44than equal to Zer which means it's a
5:46horizontal line starting at the x axis
5:48and going above so now I'll show you the
5:50overlap okay now that we have all of the
5:52linear
5:53inequalities graphed you can see the
5:56overlap is that quadrilateral looking
5:58shape that that is outlined in the thick
6:01black pen and now we're given four
6:05different vertices those are all
6:07vertices of this feasible region now
6:10what we're going to do is use these
6:13vertices and evaluate the function P at
6:17each of these vertices to see which
6:19value we get for the biggest that will
6:21be the maximum
6:22value okay so I erased some of the
6:25highlighting and drawing on the graph so
6:27it's easier to focus on what need to
6:29deal with so the vertices are
6:32highlighted in the purple and then you
6:35probably remember from geometry class
6:36last year that typically we label
6:38vertices with letters to identify them
6:42so I wrote down the C of the vertices Q
6:45R st and now we're going to be
6:48evaluating the function p p = 2x + y and
6:53evaluting just means plugging in the
6:54values for the equation so we're going
6:57to be using
6:59the
7:02equation p = 2 * 0 + 0 and that equals
7:080 Let's try the next one 2 * 0 + 2.5
7:14that equals
7:152.5 2 * 3 + 1 that equal 7 and 2 * 2 + 0
7:24that equals 4 so we wanted to figure out
7:27what point in the Fe I region which is
7:30this quadrilateral shape here maximizes
7:33the function p and we can see that the
7:36coordinate s because we substitute that
7:39value when we substitute that value in
7:40for the objective function we find out
7:43that P equals 7 which means value for
7:47this
7:51function now we're about to take a look
7:53at a real life example problem and
7:56discover the process of how to use
7:58linear programming to maximize
8:00profit here's example two you are screen
8:03printing t-shirts and sweatshirts to
8:05sell at the pulk county Blues Festival
8:08and are working with the following
8:09constraints you have at most 20 hours to
8:12make shirts you want to spend no more
8:15than $600 on supplies and you want to
8:18have at least 50 items to sell we need
8:21to figure out how many t-shirts and how
8:22many sweatshirts should you make to
8:24maximize your profit and how much is the
8:27maximum profit so let's take a moment
8:30and look at the Givens and then identify
8:33the
8:34variables I'm going to let x equal the
8:36number of t-shirts made and y equal the
8:39number of sweatshirts made so let's use
8:41this table to organize the information
8:44that we have let's focus in on the first
8:47row for the minutes it says here in the
8:49picture that it takes 10 minutes to make
8:52one colored t-shirt so I'm going to
8:55write
8:5610x and for sweatshirts it takes takes
9:0030 minutes to make so 30
9:03Y and our first constraint said we have
9:06most 20 hours hopefully you're realizing
9:09that the units are different um minutes
9:11versus hours so in order to make the
9:14total into minutes we just need to
9:16multiply 20 * 60 and that gives us 1,200
9:24minutes so here is our first row
9:27complete and the symbol that want to use
9:30is going to be less than or equal to
9:33because we do not want to go over the 20
9:35hours or also known as the 1200 minutes
9:39now let's take a look at the number of
9:42t-shirts and
9:45sweatshirts it says that we want to have
9:48at least 50 items to sell so that tells
9:50me that we need to have 50 at the total
9:55um for the minimum actually so that
9:56means we want a greater than or equal to
9:58because 50 is the smallest amount of
10:00items that we want to sell and then from
10:03there we don't know the exact number so
10:05that's why we have the variables X and Y
10:08the cost it says here in the picture
10:11that supplies cost $4 for one t-shirt so
10:13we're going to have
10:154X and then for the sweatshirt it says
10:18supplies cost $20 so 20
10:22Y and it told us that we do not want to
10:26spend more than $600 so we have 600 over
10:29here
10:29and then we're going to have a less than
10:31or equal to because we do not want to
10:33surpass that
10:37$600 so I wrote down the three
10:40constraints that we have so far in
10:42inequality form and I just Ed the same
10:44values that we were talking about in the
10:45table there are just two more
10:47constraints that we need to write down
10:50and they should make sense to you it's X
10:53is greater than or equal to
10:55Zer and Y is greater than or equal to
10:58zero and that's because since X and Y
11:01are the number of t-shirts and
11:02sweatshirts respectively we cannot have
11:04a negative number so that's why we're
11:05starting at zero and going beyond that
11:08now let's talk about how to come up with
11:10the objective function for the profit so
11:13taking a look at the tags right here up
11:16in the picture we see that the profit is
11:18$6 for a T-shirt and the profit for a
11:20sweatshirt is $20 so that's why I have
11:226X + 20 y so that is going to
11:27be the object Ive function and I know
11:31it's going to be for profit because it
11:32talks about in the question how many
11:35t-shirts how many sweatshirts can we
11:37make to maximize The Profit so that's
11:39going to be our objective
11:48function okay now that we made our
11:51objective function if you are listening
11:53if you could just put a smiley face
11:55right next to the equation that we just
11:57wrote down that will help me Det
11:59determine who is actually listening
12:01through and watching the videos
12:04thoroughly now let's learn how to use
12:06this objective
12:07function before graphing the feasible
12:10region I would just like to talk about
12:12how this function is an a discrete
12:15function and that's because the number
12:17of t-shirts X and the number of
12:18sweatshirts y can only be whole numbers
12:21so remember discreet just means we don't
12:23have all real numbers as Solutions only
12:25whole numbers work for this
12:27situation so we're about to graph the
12:29constraints that we formed earlier to
12:31form the feasible region so feel free to
12:34pause here to start setting up your
12:36graph you can see that I used the
12:39variables that we defined earlier for
12:40the X and Y
12:42labels and you can also see that I'm
12:44going up on the x axis I'm going up by
12:4615s but I'm only labeling every other
12:49and then on the y- axis I'm going up by
12:51threes but again I'm only labeling every
12:53other when graphing the constraints my
12:56suggestion is to either put the
12:58inequality in intercept form or to find
13:00the X and Y intercepts plot those and
13:02connect I actually use the second
13:05process I just talked about you can see
13:07right here I pretended that it was an
13:10equation at first and I found the X and
13:12Y intercepts plotted those and connected
13:14those and we're going to be shading
13:16below for this region because it's less
13:19than or equal to in the original
13:20inequality for the next inequality X
13:22Plus Y is greater than or equal to 50 I
13:25have this in the blue I found out that
13:27the X and Y intercepts are both at 50 50
13:30um I can easily graph this on the xaxis
13:32but on the y- AIS it's off the graph so
13:33that's why indicated as an arrow now
13:36let's take a look at the third one for
13:38the inequality 4x + 20 Y is less than
13:41equal to 600 I know that I'm shading
13:42below and for the x intercept I found
13:45out that it's 150 and for the Y
13:46intercept I found it's at 30 so I
13:48graphed this in the green the X and Y
13:50intercepts and I connected them the last
13:53two constraints are actually just the X
13:55and Y axis so that just tells us that we
13:58are working in the first quadrant only
14:00so now I have the feasible region shaded
14:03remember the feasible region is just
14:05where all of the individual linear
14:08inequality shading
14:09overlap you may be wondering how do we
14:12find the coordinates of each
14:14vertex if they are hard to read off the
14:17graph what we want to do is solve the
14:19system of equations related to the lines
14:21that intersect to form the vertex so
14:24using that process that we have learned
14:26earlier in this chapter so now I have
14:29labeled the feasible region with the
14:30vertices a b c d the vertex a is the
14:34coordinate 50 and that is just the x
14:37intercept of this equation here in the
14:40blue and the D the vertex D the
14:43coordinate is 120 and that is also the x
14:47intercept of the other equation as shown
14:50right here for the coordinates of
14:54vertices B and C we would need to solve
14:58the systems that I have listed right
14:59here and in order to find out which
15:02systems I'm talking about I just
15:05followed the color coordinating that I
15:06was using so the blue and the green
15:09equations would be the intersection at
15:12vertex B and then the red equation which
15:16I show in the black up here as well as
15:19the green equation that would solve for
15:22the intersection point at vertex C so
15:25just as in the previous problem now I'm
15:27going to sub to each of these
15:29coordinates in for our objective
15:31function P = 6X + 20 y to see which
15:36coordinate maximizes the
15:44equation okay so I'm going to start off
15:46by plugging in 50 for the equation so 6
15:49* 50 + 20 * 0 and that gives us
15:56300 next coordinate 6 6 * 25 + 20 * 25
16:04and that equals
16:07650 for coordinate C we have 6 * 75 + 20
16:13* 15 and that equals
16:17750 and the last one p equals 6 *
16:23120 + 20 * 0 and that equals 720 so like
16:29I said I'm just using the coordinates of
16:31the vertices that I just discussed in
16:33step two and substitute those in for the
16:36objective function equation P = 6X + 20
16:40and now we can see which coordinate
16:42maximizes the profit in this
16:45situation so you can see that the
16:47greatest value is at coordinate C
16:51because we have P equals
16:53750 so now let's discuss that like we
16:56were just talking about at the
16:57coordinate C 75 15 the function is
17:00maximized so that means we can maximize
17:02profit by selling 75 t-shirts and 15
17:05sweatshirts and I'm just getting that
17:07from the coordinate itself X comma Y and
17:09the maxim maximum profit at this
17:12coordinate is
17:14$750 so that completes our real life
17:21example okay that's it for today's
17:24lesson there were only two problems just
17:26because there were longer in length L
17:29and difficulty so now here's the lesson
17:32check I am just looking for you to do
17:34the evens so two four and
17:40six so take a moment to jot down these
17:43three systems of inequalities and
17:46constraints and make sure you show your
17:48work and graphs for each of these I'll
17:50see you soon

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