Cornell University
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
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Cornell University
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
Polynomial functions are useful when solving problems that ask us to find things like maximum income, revenue or production quantities. Finding maximum and minimum values of polynomial functions help us solve these types of problems. When setting up these functions, we first determine what the problems is asking us to maximize and then set up the function accordingly.
So we've got some word problems
I want to talk about.
And these are specifically optimization
problems, and that just means that we're
looking for the maximum or minimum
value of some function.
And in this case we'll deal with
polynomial functions mostly.
Let's read this one.
It says: "A farmer currently has 200
crates of apples and can harvest an
additional 10 crates per day.
The current price of apples is $120 per crate
and is expected to drop $4 each day.
When should the farmer sell
to maximize her income?"
Okay.
Well, let's back up for a second.
Income.
That's going to be our function.
And that's what we want to maximize.
We want to maximize income.
And we want to maximize over
time, when should she sell.
So -- and the variable is going
to be the number of days.
So income. Income is going to be the number
of crates she sells, times the price
per crate.
Now, first of all, I need to get a function
for each of these guys, the number
of crates.
And the problem says that she
has 200. 200 crates.
And that she can harvest an
additional 10 per day.
Let me write that as plus 10X. The minute
I do, I've identified the variable
X as the number of days.
So here X is the number of days she waits.
So, for example, if she doesn't wait at all,
0 days, then she'll have 200 crates.
Now, the price per crate, this guy, that's
also going to be a function of X.
It's going to be -- well now the price
is $120 and the price drops by
$4 each day. So minus 4X.
And that means that the income function,
let's call it I of X, is the number
of crates, 200 plus 10X times the
price per crate, 120 minus 4X.
Now, this is a quadratic function.
And the great thing about quadratic functions
is we know exactly where the maximum
is going to occur.
It's going to occur at the vertex.
So we're going to make use of that fact.
Now, here if you look at this equation, I
can actually tell where the X intercepts
of its graph would be.
So I'm actually going to
graph this function.
1X intercept is going to be when 200
plus 10X equals 0. 200 plus 10X
equals 0. That means 10 X equals
negative 200, so X equals negative
20.
So one intercept is going
to be at negative 20.
Let me mark that.
Negative 10.
Negative 20.
That's the intercept.
And then another one will
come from this factor.
When does 4X equal 120?
When X equals 30.
So there's another intercept.
Positive 30.
Now, one thing I need to think about is
what values of X makes sense in this
case?
We're not just dealing with a function
in the abstract sense.
We're dealing with a function that describes
income for this farmer after she's
waited X days.
X can't be negative.
So we should probably specify that X
is greater than or equal to 0. All
right.
Now let me draw a rough graph
of this quadratic.
It's going to be a downward
opening quadratic.
We can tell that because the leading coefficient
would be negative 40 X squared.
So let me just draw this.
And because we're not really concerned
with the negative part of this graph,
right, that would represent a negative
number of days, which doesn't make
sense, I'm just going to dash
this part of the graph.
We're also not really concerned
with this part.
Now, why would you think that is?
Here, the income's negative, right?
I'm sure she wants to stay away from this,
and of course it doesn't actually
make any sense.
It represents a time when the
price is actually negative.
That doesn't make sense.
We'll just stay between 0 and 30.
And remember the maximum is going to
occur right where the vertex is.
Now, the vertex, how are
we going to find that?
Well, parabolas have the marvelous quality
that the vertex is exactly halfway between
the intercepts.
The vertex, we'll call it X max, the
X coordinate where the maximum occurs
is going to be the average
of these two.
Negative 20 plus 30 over 2. That's
10 over 2 or 5. And that's it.
The problem only asked for us to find the
number of days that she should wait.
So that her income would be the maximum.
5's the answer.
So she should wait five days.
Now, if you're curious about how much she
would make, you can easily calculate
that by plugging 5 back
into the function.
And even though the problem doesn't
call for it, I'm kind of curious.
Let's find income of 5. We have 200 plus
10 times 5, and we have 120 minus
4 times 5. So that's 250. 100.
25,000.
She'll make $25,000.
Unit
Polynomial and Rational Functions