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.