Math Matters: Why Do I Need To Know This?
Bruce Kessler, Department of Mathematics
Western Kentucky University
Episode Six
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Surface area – Home improvement projects with paint
Objective: To illustrate how area formulas are relevant to our everyday lives by
examining a real-life example of their use, specifically in calculating how much
paint is needed in a home-improvement project.
Hello, welcome to this week’s show of “Math Matters: Why do I need to know this?,”
where we try to show you some of the math that you’re learning in these entry level math
courses and how they kind of apply to the real world. I’ve got some neat stuff to show you
today, so I’m actually kind of giddy about the stuff I’m going to show you today. I think it’s
really kind of slick. Let’s get going.
The first thing that I want to show you today is an application of surface area and how it
applies to home improvement projects, and particularly painting home improvement projects.
There’s all kinds of things that, where the concept of area is kind of useful, wallpapering for
example, but it takes a little more knowledge than just the area of the wall because there’s
other issues involved such as you know is there a design on the wallpaper and these kinds of
things, and so do you have to worry about losing part of your wallpaper at the top and the
bottom of the wall, things like this. Or, after you cut a section off, you can’t use it somewhere
else, so area is a thing to worry about when you’re doing that, but it’s not the only thing to
worry about. However, if we talk about painting, painting, paint is one of these things that I
can kind of smear around and there’s very little waste involved. There’s a little bit when you
think about rinsing out your brushes and your rollers, these kinds of things, you do end up
wasting a little bit of paint, but not a ton.
So really, when we’re painting, an area measure is pretty useful to us. I’m just basing
this on some gallons of paint that I had out in my garage. A gallon will cover from 250 to
400 square feet of the material depending on the coarseness of the surface and if it’s very
coarse, it will probably take closer to 250 and we also have to take into account that a second
coat is usually going to be needed on newer surfaces. Now what I want to look at today is a
case where suppose you’ve built a storage shed and it’s covered in exterior paneling that you
intend to paint. (Figure 1) Now exterior paneling is very rough, so I’m going to go with that
250 number in this and you may even find that we may need to scale this down a little bit,
but we’re going to work on the premise that I can cover 250 square feet of this paneling with
paint and really, I’ve got a nice picture of my shed here, I don’t need to worry about the roof
so I’m actually going to take the roof off of this, let’s go back and take a look at things. I’ll
take the roof off, that’s aluminum roofing, I’m not going to worry about that and I’ll give
you the dimensions of all the different things we’re looking at. (Figure 2)
The question I would like to try and answer with you guys is how many gallons of paint
do I need to buy in order to coat this twice, in order to put two coats on the exterior paneling
of this shed and it’s an important question, I mean you can always go back and buy more
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Figure 1, Segment 1
Figure 2, Segment 1
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paint, but if you’re doing this as a contract or something like that, you may want to, you
probably have to give an estimate, so you want to know up front how much paint do I need,
these kind of things, or how much do I need to pay for?
Alright, let’s take this house apart and do things a piece at a time. There are actually four
surfaces, four flat surfaces that I need to apply paint to. I’m painting the outside I should
say so I’m going to take it apart and look at a piece at a time. This is the back wall and it
is, the whole shed is 12 by 16. The base of this short wall is 12 and it goes up eight feet, let
me kind of get my pointer going, up to here and that’s 12 feet up to that point. But let me
just deal with this 12 by eight business. That’s a rectangle. It doesn’t look like a rectangle
that’s because there’s dimension involved, but it’s flat, if I looked at it flat straight on, that’s
a rectangle and the area of a rectangle is length times width so that is 12 times eight and
then I’ve got this triangle on top here that I want to deal with. If the whole thing is 12 feet,
I’ve accounted for eight of those. It’s four feet up to there and the area of a triangle is half
base times height, so that’s one half, the base is 12 so height here is four so that works out
to be 120 square feet on the back wall. I’m just painting that other surface on the back wall,
that’s what I’m painting. Now if we look at the front wall, that’s the same except that it has
a door. A three by seven opening for the door so we’re going to subtract that area. That’s
seven times three, 21 and so the front wall then is 99 square feet. So that’s got the front, got
the back.
Figure 3, Segment 1
Now let’s move on to the sides. The right hand side is simply a rectangle. There’s no
window there. It’s 16 by eight so that is 128 square feet if I’ve done my calculations correctly
there. (Figure 4) And then I’ve got the opposite wall that would also be 128 and then, but
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I’ve got that window there which is a rectangle, 3 times 3.5. So if I’ve done all my arithmetic
correctly, that works out to be one 117 12 square feet. (Figure 5) Now what I want to do is
add all that up, get the total area so this is my original kind of construction. I add all those
things up and I get 464.5 square feet total and then I want to do two coats, so I’ll double that
and then that comes up to 929 square feet and then I need to do something I did a few weeks
ago and that is convert that into gallons. So I’m drawing an equivalence here between one
gallon and 250 square feet. I multiply by one, I get the square feet to cancel out and so it
works out that it is about 3.7 gallons of paint to cover that. So you’d end up, you’d probably
end up making a purchase of four gallons of paint to cover up the outside of that. (Figure 6)
Figure 4, Segment 1
Hopefully this is a good demonstration of how kind of take the known area formulas for
like rectangles, and triangles, and things like that and adapt them to find the area of things.
You know those ends were pentagons. They were five-sided things, but yet we were able to
kind of slice them and use a rectangle on the bottom and a triangle on the top. So that’s a
good trick, that’s a great trick for calculating area. We also did the subtract trick where I
had the area of the whole thing and then I had a portion of that I wanted removed and so I
subtract that. It’s very important when you’re doing real world applications that you’re able
to adapt your formulas to kind of fit the situations.
Let me give you one weird example. On this campus, we have a planetarium, the Hardin
Planetarium and that is basically shaped as a hemisphere and I thought about trying to get
a picture of it and squishing it in there, but I’ll just do this. Half a sphere, the other half
of a sphere, that’s actually the shape of the Hardin Planetarium. Now, and you can check
your books for this, the area for the surface outside the Hardin Planetarium or for a sphere
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Figure 5, Segment 1
Figure 6, Segment 1
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is 4πr2 , and since we’re doing a hemisphere just and I’m just talking about this business
right here, the outside curvy part. For the hemisphere, it would be half of that and now the
part I’m not taking into account is okay the circular base. That actually doesn’t need to be
painted. Suppose we measure the circumference of this thing at 300 feet and the reason I
say circumference is this is solid. I can’t measure through it so I have measure around it.
But that’s okay, the circumference I can, from the circumference I can calculate then the
radius. 2πr is the circumference, the distance around is 300. So then solving for “r,” you
get this little measure and then I can plug that into the area formula that I talked about. Do
the arithmetic, you end up with a π 2 in the denominator here that will cancel out and it
works out to be about 14,300 square feet. If you do that same kind of calculation it ends up
being about 57.3 gallons of paint. It would take a lot of paint. I bet that is a fairly accurate
measure. I didn’t go measure the Hardin Planetarium, but I bet you that’s pretty close to
what’s going on. (Figure 7)
Figure 7, Segment 1
Alright, I flashed a lot of stuff by you very quickly so let me flash up some summary pages
that kind of talk about the things that I mentioned.
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Summary page 1, Segment 1
Summary page 2, Segment 1
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Summary page 3, Segment 1
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Complex numbers – Video-game design
Objective: To illustrate how complex numbers can be used in a real-world situation, specifically to manage the translation and rotation of points in the plane
is a computer animation.
The next application I would like to show you involves complex numbers and I’ve got to
tell you, this was a real challenge because complex numbers, by definition, are not real values.
They’re kind of imaginary in the sense that you can’t have a bank account that has a complex
number in it, for example. So tying this into the real world was kind of a challenge. Now
it does come into play. There are a lot of applications where you can use complex numbers
to kind of sidestep difficulties that you have with real numbers and you get a real number
answer and voila! There you go, you’ve solved your problem. But I really wanted to show
you some way that we could apply this to the real world and I’m going to do this by showing
you an example from video game design, okay?
The idea behind imaginary numbers is that you’ve got this crazy thing that you could
square and you get a negative one, which is pretty bogus. I mean you can’t take a real
number, square it and get a negative number. But let’s imagine that you can, so here’s this
“i” for imaginary and sometimes
√ you’ll see this notation. It’s not technically correct, but you
know we’ll say “i” is equal to −1. The technical part is that you’re not supposed to have
negatives under that radical but that’s okay, I won’t call the math police on you or anything
like that. Complex numbers, we actually take a real number and we add a multiple of this
“i” together and call that a complex number. The reason you learn it in college algebra is
because that is something that pops up when you start using the quadratic formula and that
certainly would be an application of this. But I don’t want to get into that. (Figure 1)
I want to get into the equivalence of complex number to points in the plane. Now you
can do a lot of things in the plane where we more around and so forth. Well, a complex
number, if you think about plotting the real number part of the complex number along the xaxis, and the imaginary part along the “y”, there’s an equivalence between complex numbers
and points on the plane. One plus three “i” would be the point (1, 3) in the plane and two
minus “i” would be the point (2, −1), and so forth. So I can go back and forth between the
two. (Figure 1) The advantage of complex numbers is I have operations. I can add, subtract,
multiply, and divide them in an obvious way. You just simply collect your like terms. If
I’m adding these two complex numbers, I collect the real parts to get four and I collect the
two “i” and the four “i” to get six “i”. If I’m subtracting, well distribute your negative and
collect your like terms. So 1 − 3 and then 2i − 4i and then do the same thing. (Figure 2)
To multiply them, you just use the FOIL method and the same kinds of things you’ve
always done. First would be 3, outer would be 4i, inner would be 6i, and then last is 2i times
4i, which would be 8i2 . But now remember the “i” squared business is − 1. So that is 10i in
the middle and − 8 right here, so that’s a − 5 + 10i. Now that’s stuff that I can’t do with
points in the plane. I can’t do the operations on them. So there’s advantages to doing things
in complex numbers and then applying them to points in the plane. (Figure 2)
That’s what I’m going to do with you guys when I start talking about video game design.
The thing I’ve got to convince you on is that these operations move points around. So for
example, when I add things, I move points around. Translations: left and right, up and
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Figure 1, Segment 2
Figure 2, Segment 2
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down. Now if I start with three “i”, 3 + i, excuse me that’s (3, 1), the point, and I add two
to that, it takes me to 5 + i, which is two units over. That’s a translation of two to the right.
If I then add 3i to it, well that’s 5 + 4i and that moves me four units, excuse me, I mean
three units up and then if I subtract 4, that’s going to move things four units to the left and
if I subtract 7i, that’s going to move me seven units down. That goes to 1 − 3i and that is
right there. So this is a method, this simple addition and subtraction of complex numbers to
move points around. If I add, if I add real numbers, I move left and right, let me do it so
that you understand it. Right if it’s a positive number, left if it’s a negative number. If I
add and subtract purely imaginary numbers, then I go up and down. It’s a good way to move
stuff around. (Figure 3)
Figure 3, Segment 2
Alright, same thing with multiplication. If I multiply by a complex number that is on the
unit circle, and that just simply means that if I square the first and the second part, I get
one. I want that so that I don’t scale things. I just want to move things. If I multiply by “i”,
“i” here is 90 degrees above this horizontal. So if I multiply two by “i”, that constitutes a 90
degree, counterclockwise rotation. If I do it again, I get another counterclockwise rotation of
90 degrees. It takes me over to negative two. If I change the thing I’m multiplying by, and
then go to this thing which is 45 degrees, that has coordinates √12 and √12 . If you square this
and you square that then you get half plus a half is one so I think you believe me there. If I
multiply this out, I get this point which is right there. And so that is a 45 degree rotation.
So, again, I can rotate things by multiplying by a complex number that’s on the unit circle
like that. (Figure 4)
Now, here’s the thing that happens with video games a lot of the time, you’ll have what
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Figure 4, Segment 2
they call I think a first player game and it looks like you’re moving through the scenery, but
you’re not. You’re standing still and the scenery is moving around you. You’re sitting still,
your gun or whatever it is right here and things are moving around you. So that involves a
lot of translations and rotations to shoot things or whatever you’re doing. Now that can be
kind of tedious and they have software to do this, but I’m saying that you can actually do
this with complex numbers. I’ll show you an example.
Here is my game field, which is pretty dull, I know but you know I’ve got to sleep sometime,
I can’t spend infinite time on these things, but this is the player and here is the game field.
It has a rectangle, it has fence, a circle, and a triangle and what I’d like to do is take this
player through a walk and what I’ve done is calculated, I’ve moved actually the scenery around
this player by using repeated applications of either adding or multiplying complex numbers.
(Figure 5)
So here we go, I’m going to take you through this real quick. We’re going forward, I’m
actually adding a negative complex number, that’s where I want the scenery to come down.
Then I’m rotating to my left, I’m rotating the scenery actually to the right, so this is the
number that I multiply by. Now I’m sliding to the left, okay, so I added so I moved the
scenery over to the right. Now I’m traveling in an arc. I’m moving the scenery back and
then rotating the scenery to the left a little bit and then I’m alternating between those things
and the last thing I’ll do is I’m rotating 120 degrees to the right. What I’ve done actually is
I’ve applied this mapping to just points, to just the corners of things, moved them around.
Let me just show you that again because I think it’s cool. You know all I’ve done here is I’ve
taken the corners of things and moved them around by taking the complex numbers to kind
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Figure 5, Segment 2
of do those complications quick and dirty and then animated it. I’ve taken several slides and
then you could do it with pencil and paper where you kind of flip through a notebook to get
the slides. (Figure 5)
So there you go. A nice real world application of complex numbers and how you can use
them for something I think is pretty cool. I know that a lot of the kids I talk to that want
to be computer scientists want to be video game designers. Well, there you go, there’s an
application of it. Let me slide, put up some summary slides that kind of talk about the things
that I mentioned very quickly and let you absorb some of that information.
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Summary page 1, Segment 2
Summary page 2, Segment 2
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Inequalities – Optimization problems with constraints
Objective: To show how linear inequalities can be used in a real-life situation
where we want to find the optimal solution to a problem subject to linear constraints.
The last thing I’d like to talk to you today are inequalities and how we can use them to
solve some real world kind of problems as far as finding optimal solutions. A lot of times we
like to do things in such a way that we for example, maximize profit, minimize waste and we
have to work within certain constraints. That can be a difficult thing to figure out by trial
and error. I mean you can just try everything that is possible, but never know that you’ve
got the best answer. Well here’s a way using inequalities that we can go through and find the
best answer to get the optimal answer to problems like that.
Here’s the example I’m going to work with you. Let’s say you’re helping the band boosters
or some other civic club and you’re helping out by baking cookies or brownies for them to
sell at their next event. And you’ve told them “Well, I can devote ten hours to the project
and then I’ve got to do my other stuff ” and so that’s a constraint, you’re only going to spend
ten hours on the baking. And then the other constraint that we have to kind of worry about
is that you only have eight trays that you can provide full of cookies or brownies, whichever
it is. Now, there’s certain things to take into account here. Let me kind of point to them. I
can cook, I can bake cookies faster than I can bake brownies. I can fill a tray of cookies in
about 45 minutes, although it takes me about two hours to fill a tray with brownies. So you
know the question here is, which is better? Should I try to just make, in two hours I could
make five trays of brownies, should I do that? It turns out I could profit by 18 dollars on
each tray of brownies. That’s more than ten dollars than I profit on each tray of cookies.
Maybe I should just do that. On the other hand, and I could definitely fill all my trays here
with cookies in the ten hours allotted. So the question then becomes okay what mix of trays
of cookies and trays of brownies would be best to produce the most profit and you’re going to
want to work on this, that’s what you’re going to do. The biggest bang for your effort, you
want to get the money out of it. (Figure 1)
So let’s talk about how we set this up. We’re actually going to set up the constraints.
We’re going to use a graph now and we’re going to graph, I’m going to let “x” be one thing,
the number of trays of cookies. I’ll let “y” be the number of trays of brownies and let’s
actually kind of graph the set of valid answers that I could look at. So my constraints, the
obvious constraints on “x” and “y”, “x” for the number of trays of cookies, “y” for the
number of trays of brownies is that they both have to, I can’t have negative trays of cookies or
brownies. So I’m talking about working here in this first quadrant, definitely things have to
be here. (Figure 2) But if I start to look at the other constraints, the time constraint is that I
have a maximum of ten hours to spend on this project. I can spend, it takes me 45 minutes
to fill a tray with brownies, so that’s three fourths of an hour. I’m doing things in terms
of hours, and it takes me two hours to construct, or put together, bake a tray of brownies.
So depending on what “x” is, however many trays of cookies I have, I’m going to multiply
by three fourths, however many brownies I have, I’m going to multiply by two. And now
let’s put this on our graph. I actually need to solve for “y” so I’ll bring all the other stuff
over, divide by two real quick, and then I’ll graph that and because this is “y” less than this
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Figure 1, Segment 3
amount, I’ll graph this line and then I’ll take the bottom half or the bottom part of my graph,
so I’ve actually eliminated a lot of the solutions I had earlier, okay? Definitely things have
to fall within this triangle. The valid answers have to fall inside this triangle. (Figure 3)
The other constraint is the number of trays. Well “x” plus “y” cannot be bigger than eight.
I’ve got eight trays, that’s it, that’s all I’ve got to work with so I’ll graph this as well. “y”
is less than or equal to eight minus “x”. I’ll graph eight minus “x” which is actually right
through here and again I’ll take things under it, but again I want the overlap, what’s already
shaded so watch this, I’ll put the graph. Here you’ve got some overlap, yeah, we’ll probably
go back and show this. So I’ve eliminated now, let me show you the part, this part was in
there, but now it’s eliminated. So what we have here, this light blue business, that’s the set
of valid answers. (Figure 4)
My answer has to come out of this and what I would claim to you folks is that to find the
optimal answers that will maximize profit, I just need to look at these corners of the polygon.
It’s the quadrilateral, okay? And let me show you why that’s the case. If you think about
the profit that we’re going to make on this, it would be a certain amount for every tray of
cookies, and a certain amount for every tray for every tray of brownies. Well, that, if you
graph that in three space, you’ve got “x” is something, “y” is something and you take that
amount and graph that on the z-axis. You kind of think about here’s everything happening
in the plane and then the amount I get for “x” and “y”, put that on the z-axis, so you get
a three-dimensional graph. Well, there’s different things that could happen. You can have
things where it only grows in the “x” direction. In this case, that would be the highest point.
That’s directly above this corner of the polygon. (Figure 5) Or it could only grow in the “y”
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Figure 2, Segment 3
Figure 3, Segment 3
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Figure 4, Segment 3
direction, okay, in which case this would be the highest point on the polygon. (Figure 6) Or,
as is the case in our problem, it could go grow a little bit in each direction and it’s kind of
slanted there in which case this is the highest point on the polygon and it’s kind of hard to
see that the way things, let me move the graph up just a little bit so you can see, change the
perspective on it, see that that is now the highest point on it, on the graph, okay? (Figure 7)
So what we’ve got to do now is take our profit function, our profit ten dollars for every
tray of cookies and 18 dollars for every tray of brownies. And you know, I don’t really want
to worry about what the graph looks like, but I just want to check the corners. I know that
this is a plane in three space, whatever, but I just have to check the corners. Even if it’s
just along, you know if one of these lines happened to be less than level, I could still check
the endpoints and get the same values, so it’s sufficient just to check the corners. So that’s
what I’m going to do. The only constraint that I really have here is that the corners may
not be whole numbers and I kind of need to think about doing whole number trays of things.
So instead of (4.8, 3.2), that’s where these things actually kind of hit each other, I’m going
to look at whole number answers just inside the polygon, like (4, 3) and (5, 3) are inside the
polygon – (4, 4) is not. (4, 4) would be about right here and that’s actually outside the blue
region, so we just check these amounts. At (0, 0), I get zero, that makes sense. If I don’t do
any trays, I don’t get any money. If I do all cookies, I get $80. If I mix them up and do four
and three, I get 80, I mean $94 profit. If I do five and three, that is this point closest to that,
I get $104 and if I do all brownies I just get $90. So the maximum answer, the maximum
mix is do five trays of cookies, do three trays of brownies. That will maximize the amount of
profit that I can get from this endeavor. (Figure 8)
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Figure 5, Segment 3
Figure 6, Segment 3
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Figure 7, Segment 3
Figure 8, Segment 3
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Neat little application of inequalities and just then just plug it into a formula just to get
your maximum answer. I’m going to flash up some of the graphs that kind of show you the
kinds of things that I was doing as a summary and we’ll come back in just a moment and
wrap things up.
Summary page 1, Segment 3
Closing
Oh, we’re back! I hope you’ve enjoyed the things we’ve talked about today. I’m really
kind of proud of some of those things, especially the complex number thing. I know that I
talked about things very fast and I have to, it’s kind of a short program. But these, all these
episodes are downloadable as computer files on our web page and we’ll be flashing that up in
just a few minutes. With that, I am done. Thanks for watching us this week and join us
next week, thanks.
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Summary page 2, Segment 3
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