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Hello.
And welcome to Bay College's online
lectures for college algebra.
I'm Jim Helmer.
And in this video we're going to be looking at section 4.4,
which deals with quadratic models.
We'll also continue in this video into
section 4.5 as well.
The first thing we want to look at is something we deal
with in economics, so an application
to quadratic models.
Now, one of these applications in
economics is demand equation.
When we talk about economics, a lot of businesses deal with
supply and demand.
Well, in the example we're going to
look at x is our supply.
And that, for this example, is the number of shirts.
So we have a supply.
Demand is defined as the relationship between the price
we need to charge and the number of product we have.
So this relationship is demand.
Now, revenue is how much money we're going to make.
And the revenue is essentially the price we charge times the
number of items we sell.
So if we look at this, this is my revenue equation--
the price we charge times the number of items.
That is our revenue.
Our relationship for our given example is a company sells
shirts, and we have a supply from 0 to 500 of these shirts.
And we're going to sell them at some price, p.
So this is our demand equation.
This is our relationship.
Now, the first part of this is to find the revenue as a
function of x.
If I sell this many shirts, how much revenue
am I going to make?
So let's find this function.
So if I want to find this function, well, of these two
equations I know p in terms of x.
I want to find R in terms of x.
I can just do a little substitution.
So I'm going to take this equation and plug it in for p.
And when I do that--
I'm just-- x times this quantity.
So I'm going to distribute the x.
Negative 1/20x squared plus 25x.
This is my revenue equation in terms of just x.
If I sell this many, this is how much money I'll make.
Now, the next part of our question asks, if I sell 20 of
my shirts, how much revenue are we going to have?
Well, I can just evaluate this function for 20 shirts.
And when I do that, I get negative 1/20 times 20 squared
plus 25 times 20.
This is going to give me 20, a negative 20, right?
25 times 20 is going to give me 500.
And 500 minus 20 gives me a revenue of $480.
Revenue is measured in dollars.
So $480 if I sell 20 shirts.
Now, as a business owner or someone who wants to maximize
our revenue, how many shirts would I have to sell in order
to find a max revenue?
Well, if we just assess for a moment, this
is a quadratic function.
And if we identify the leading coefficient to be negative, we
know it opens down.
So let me just draw a little parabola here because that
would be the shape of this function.
And because it opens down, we know we're
looking for a maximum.
Well, that's what we're looking for--
number of shirts to maximize R. Well, this value is the
vertex h, k.
And h represents an x value.
The number of shirts represents the x value.
So what this is asking me to find is the h, right?
The number of shirts to maximize our revenue.
So to do that I can use a formula that we had seen in
the previous video called the vertex formula--
negative b over 2 times a.
Well, if I do that, negative b is going to be negative 25
over 2 times a.
Well, 2 times 1/20 is negative 1/10.
Here I see a negative over a negative.
And 25 divided by 1/10 or 25 times 10, multiply by the
reciprocal.
It's just a complex fraction there.
We're going to get 250.
So I know if I sell 250 shirts, I'm going to maximize
my revenue.
Well, the next question says, find the maximum revenue.
Well, essentially that means evaluate the function for our
h value to find k, our maximum revenue, our y value.
So I'm just going to plug this in right here.
I want r of 250.
And if I do that I get negative 1/20 of 250 squared
plus 25 times 250.
And if I work this out, it's going to be negative 3,125.
This is going to be 6,250.
And if I find the difference in these two values, I'm going
to get 3,125.
So this is my maximum revenue.
This $3,125 is how much money I'm going to make if I sell
250 shirts.
And that is also my vertex, my maximum amount of money that I
can earn selling these shirts.
Now, the last part of the question says, find the price
that the business should charge in order to make this
much money, in order to make that much revenue.
So let's just go back to the beginning and say, well, this
is already solved for p.
And I know my x value to be 250.
So I'm just going to plug it up here.
250 times 1/20, negative 1/20.
And we could do a little bit of simplification here.
12.5 is when I simplify that.
Plus 25.
And I can find the difference to get $12.50.
So I'm going to charge $12.50 for each shirt that I sell and
I'm going to sell 250--
hopefully the demand is there that people want to buy 250.
That's what this equation tells me.
So we can maximize a revenue and earn $3,125.
All right.
So let's move on here.
We'll get this out of the way.
The next thing we're going to look at is section 4.5, which
deals with solving quadratic inequalities.
Now, sometimes we don't want to find an exact value.
Maybe we want to find a range of values--
something that's greater than something or something that's
less than some other item.
And let's recall--
because we've worked with inequalities before in
previous sections.
What we're going to look at here is let's just recall what
we've done before.
If I want to solve this for x, I just do a little bit of
simplifying.
I can treat this as an equality, unless I multiply or
divide by a negative value.
So I'm just going to distribute this here to get
negative 2x plus 10 less than 3x.
I'm going to add 2x's to both sides, move it down here, and
then divide by 5.
Now, 2 is less than x.
Well, that just means x is greater than 2 if you want to
see it written from left to right.
So what does this mean?
Well, if I graph it, whatever x is, it has to be
greater than 2.
And because this is a linear inequality, just a single
value to the first power, x is any value greater than 2.
And I can check my work by taking a value and plugging it
into the original equation making sure it holds true.
Let's use the value 3 as a test point.
3 minus 5 is negative 2.
Negative 2 times negative 2 is a positive 4.
If I plug in that same value here because it's the same
variable, x, 3 times 3 is 9.
Is 4 less than 9?
Yes, that's a true statement.
So I know that in this interval
this is going to work.
Well, what happens when our inequalities are not linear?
Well, we can solve it one of two ways.
And I'm actually going to solve it both ways.
First, I'm going to graph this.
This is just a parabola.
And let's find its vertex and determine any other
information.
Well, if I want to find its vertex, I can
use the vertex formula.
Negative b over 2 times a.
Well, that's negative 4.
So negative 4.
Now, what is the k value of the vertex?
I can plug that in.
Negative 4 squared is 16.
Negative 4 times 8 is negative 32 plus 15.
We're going to have 31 minus 32.
So my k value is negative 1.
So negative 4 and negative 1-- so I'm just
going to graph that.
Negative 4, negative 1 puts me right about there.
And I know it's a parabola that opens up.
And maybe I want to find a few other points.
And the key to this is actually finding the zeroes.
You know, what would actually make this equal to 0?
Well, if I do that, this actually factors to 5 and 3.
So it's negative 5 and negative 3.
Hopefully you can see the factoring that I did there
without showing work.
And we know that this is a positive coefficient.
So it opens up.
So let's just look at that for a moment.
Now, if we look at this, we can say, well, what values
make this true?
Well, there's two areas that this parabola breaks the graph
in to, right?
There is area above the x-axis and an area below the x-axis.
So what values of x am I going to choose that make this
quantity less than 0?
Well, essentially it's asking, what makes this quantity below
the x-axis?
And we can see between this 0 and this 0, this area is less
than the x-axis.
So essentially between negative 3 and negative 5.
So what makes this true?
Any value of x in between those, my graph
is below the x-axis.
So we could do it by graphing, but you still need to know
those zeroes.
Or we can just kind of see the area.
That's what we're looking for.
Any value of x in between here, we're below the x-axis.
Well, one thing that I want to show you is what if we just
didn't graph it?
What if we just did it algebraically?
Essentially now, we'll see where I factor it.
Treat this as an equality just for a moment and say, well,
what if it was equal to 0?
Let's factor this.
What are the values or the factors of 15 that sum to 8?
Well, x plus 5, x plus 3.
5 times 3 is 15.
5x and 3x is 8x.
So if I FOIL that out, I get this back.
So I know that this is true.
So what values would make it equal to 0?
Well, we can use the zero value theorem.
Negative 5 makes this factor 0.
And 0 times anything is 0.
And negative 3 makes this factor 0.
Negative 3 plus 3 is 0.
0 times anything is 0.
Now, what we can do is think of it just like we did here.
Let's write it just on a number line because I'm just
looking for x values.
So we can think of it as just a number line.
If I put a negative 5 here and negative 3 here, I can pick
some test points.
Now, picking test points instead of graphing it-- when
these equations get higher ordered polynomials, this is
actually the way to go, to break it down into what's
called linear factors.
So I can look at this and I can just pick a test point.
Well, what if I try negative 6?
If I put negative 6 in here, I get 36.
And this is going to give me negative 48 plus 15.
And I'm going to end up with a positive value.
36 and 15, well, that gives me 51.
51 minus 48 is a positive value, which is
not less than 0.
So this is not going to work.
And if I pick a value in here, maybe I choose negative 4,
negative 4 squared is 16.
Negative 4 times 8 is negative 32 plus 15.
That's going to give me that negative 1 that I found
before, that vertex, which is below that x-axis.
So this is a true statement.
So I know from negative 5 to negative 3.
If I pick a value out here, maybe I choose 0 because I
know 0 is to the right of negative 3.
That would be 0.
This would be 0.
15 is not less than 0.
So this area does not work.
So I actually find it this way too.
Now, the reason why I'm showing it to you using this
method is because when we get higher ordered polynomials, if
we can write them as linear factors, we can find it even
if it's beyond quadratic if it's higher order.
So we can graph it.
We can see what's below the x-axis because this
says less than 0.
What if it's a greater than 0?
We'd just look for the values above the x-axis.
That would be anything over here or anything over here.
All right.
Let's move on to another example.
All right.
When it comes to solving these inequalities, these higher
ordered inequalities, these quadratics, essentially we
want to treat it just like we would a normal quadratic.
We want to set it equal to 0.
Well, let's get rid of these parentheses.
I'm just going to distribute this.
And now, subtract 20 from both sides.
Get it set equal to 0.
So now, I want to say, well, this value, no matter what it
is, is greater than or equal to 0.
Well, let's think about that for a moment.
Whatever this is has to be 0 or greater.
That means it has to be positive.
So I'm just looking for positive values.
Let's take this a little bit further and factor it.
Because I could graph this and then determine what values are
above the x-axis, what values are below, and choose which
one's going to make this true.
Obviously, above the x-axis because it has to be
greater than 0.
But let's factor it because it's nice and factorable.
If we look at this, the factors of negative 20 that
have a difference of 1, well, that would be x
plus 5, x minus 4.
5 times negative 4 is negative 20.
5x minus 4x is 1x.
So we could FOIL it out and make sure
that we factored correctly.
And now that I have these in linear factors--
because if we think about it, x plus 5, that would be a
linear equation, times x minus 4 is a linear equation.
So what we can do here is use the zero factor theorem.
What would make this 0?
Negative 5.
What would make this 0?
A positive 4.
If we throw these onto a number line because we're just
looking for x values that make this a true statement because
it is only in one variable but it's a second degree, I can
simply pick some test points.
And if I do that-- let's say I choose negative 6.
Does that make this a true statement?
Well, we can work with the factored form because I just
want to find positive values greater than or equal to 0.
So if I throw in negative 6, well, negative 6 plus 5 is a
negative value.
Negative 6 minus 4 is a negative value.
A negative times a negative is a positive.
Positive values are greater than 0.
So I know this area works.
So from negative infinity-- because that's where this
points off to--
up to negative 5 and including--
because it can equal 0, right?
Our endpoint is included.
This is one of our intervals.
But we still have two other areas that our zeroes split
our number line into that I have to test.
Well, let's test 0.
If I put 0 in here, this is a positive 5 times a negative 4
is negative, negative 20.
Well, Negative 20 is not greater than 0
because it's negative.
It's less than 0.
So this area doesn't hold true.
What if I choose a value out here?
Let's say I choose 5 as my test point.
This would be a positive value of 10.
This would be a positive value of 1.
A positive times a positive is greater than 0.
So I have two areas.
And it does include 4 again because it is equal to its
endpoint all the way up to infinity.
So anything this way on my number line or this way on my
number line is going to make this a true statement.
So I have this in interval notation as my solution.
All right.
So let's look at one more example of an inequality.
All right.
Here we have a projectile is fired.
And I should check my grammar.
A projectile is fired straight up from the ground at 80 feet
per second.
The height, s of t in feet at any time, t, which is in
seconds, is modeled by this equation right here.
So let's pay close attention to that right there.
Find the interval of time that the projectile is greater than
96 feet, when it's higher than 96 feet.
Well, if we think about this, essentially
we're given the equation.
This is good information, 80 feet per second.
And we can see that it is in this equation.
If you take physics, you'll see how this relates.
This negative 16, well, that's a coefficient due to gravity.
Gravity is pulling down on this object as it's being
fired into the air.
So let's see what we can build with this information.
We're already given an equation, but we're told find
the interval of time that the projectile is
greater than 96 feet.
s of t is the height measured in feet.
So this, its units, are feet, its height.
Well, guess what?
We were given a height, 96 feet.
We want to know, whatever t is, how long, what time
interval of time that this is greater than 96 feet.
My equation has to be greater than 96 feet.
So now we have this.
We can solve it using the methods that we
had just seen before.
Well, let me first assess this.
Say, OK, I have a quadratic.
And I want to get it equal to 0 or in this case an
inequality with 0 on this side.
I'm just going to subtract this 96 from both sides.
So 0 is less than negative 16t squared plus 80t minus 96.
So with this here, maybe I can do a little factoring.
Well, if I assess this, negative 16, 80, and negative
96, they are all factors of 16.
Maybe you don't see that.
Maybe you take it a little bit at a time.
Maybe say, hey, these are all even.
Maybe you factor out a 2, and you'd see, oh,
they're still even.
Let's factor out another 2.
And then, you could do another 2 and finally another 2, and
you'll get this down to this form right here.
0 is less than t squared-- and I'm going to factor out a
negative 16 because I want this first
coefficient be positive--
minus, let's see, 5t plus 6.
So if I factor out a negative 16, well--
if you see here, you don't see that negative 16 in here
because essentially if I have a constant out here, if I
divide one side or the other well, 0 divided by anything
isn't going to change the equation.
So I can essentially just not worry about that factored
value because it didn't have any variable in it.
So now, I can say, hey, let's factor this just like we did
in the previous example.
Well, the factors of 6 that have a difference of 5 or a
positive 6 that sum to a negative 5.
Well, that means they're both negative in order to give me
this sum with that positive.
x minus 2.
Oh, it's t that is my variable.
Let's stay consistent and t minus 3.
Negative 2 times negative 3 is a positive 6.
Negative 2t and negative 3t is negative 5t's.
So it does work.
And now, I can just use that zero value theorem.
T is either 2 or a 3.
And now, let's think about this for a moment.
Find the interval of time that the projectile is
greater than 96 feet.
If we just use a little bit of critical thinking for a moment
here, if an object is fired up, eventually it's going to
come down, all right?
Kind of parabolic, and if we think about this, it's
negative 16t squared.
Well, squared represents that parabola.
So at what time is that object above 96 feet?
Some time that it's going up and then sometime that it's
falling down.
Well, we can see we do have two values of time.
Let's put it on a number line and pick some values here.
Now, I want it to be greater than 0.
So let's just use this equation here.
If I choose 0, well, 0 minus 2 is a negative value--
0, and negative.
A negative times a negative is a positive value.
Well, one thing we have to consider is hey, for a moment
that works, doesn't it?
If we think about this, we have to realize that is that
value in my domain?
At time 0 it hasn't been fired yet.
It wouldn't make sense.
So if we continue on here, we want to say, all right, well
let's pick a value in between here.
Let's say, 2 and 1/2.
Well, 2 and 1/2 minus 2 is 1/2, which is
just a positive value.
2 and 1/2 minus-- oh.
You know what?
I made an error here.
And I want to pay close attention to this.
When I factored out that negative 16, if you divide by
a negative, what did I--
my error that I made here was when I divide through by a
negative, I got to change the sign.
It's not an equality, right?
It's an inequality.
We have to remember that.
So I got my sign in the wrong direction.
And I apologize for that.
We all make mistakes, right?
But I could see that as I was working through this, well, it
wouldn't make sense that this area worked because we haven't
fired it yet.
So let's continue on.
Let's use our test points now that we've
discovered our error.
All right.
If I use this test point of let's say 0, right, which
technically isn't in my domain because it can't equal 0.
But I just want to show that this interval isn't true.
So 0 minus 2 is a negative value.
0 minus 3 is a negative value.
A negative times a negative is a positive value.
Positive values are not less than 0.
So this area doesn't work, which it shouldn't because it
hasn't reached this height yet.
Well, what if I try a value out here?
What if I do 4?
4 minus 2 is positive.
4 minus 3 is positive.
A positive times a positive is not going to be less than 0.
So this area doesn't work.
So we can see, well, we do have this interval
between 2 or 3.
And let's just make sure that this works, 2 and 1/2.
2 and 1/2 minus 2 is a positive.
2 and 1/2 minus 3 is a negative.
Well, a positive times a negative is a
value less than 0.
So my interval of time is from 2 to 3 because it asks greater
than 96 feet.
So it doesn't include my endpoints because it's not
equal to 96 feet.
So here we have 2 to 3 seconds.
Any time in between there, this projectile is going to be
above 96 feet.
All right.
So this has been section 4.4 and 4.5.
Thank you for watching.