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Finding Maximum and Minimum Values - Concept
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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.

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