Applying Quadratic Functions 3

U10_L2_T1_we3 Applying Quadratic Functions 3

Suppose marketing experts have determined the relationship between the selling price of an item

and the cost of an item can be represented by the linear equation q = -30s + 800,

where q is the quantity sold in a year and s is the selling price.

If the cost to produce the item is $20,

so the cost to produce an item is $20,

what is the selling price that optimizes the yearly, the yearly, profit?

So what's the profit going to be? So let me write this down.

So a yearly profit is going to be the quantity, is going to be, the quantity that we sell in a year,

it's going to be the quantity that we sell in a year,

times, times, the price that we sell it at, price, times the price that we sell it at,

minus the cost of us actually producing that item, and in this case they tell us it is $20.

So for example, if we sell two items, if q is 2 and we sell them for $25,

we're gonna make $5 on each item, 'cause it cost each of them cost us 5, $20 to produce,

so 25 minus 20 will be 5.

If we sell two items at that price it'll be 2 times 5 or we'll have a profit of $10.

So what, how can we figure out how to maximize this profit?

Well they gave us the quantity as a function of selling price,

so we could, we could express the entire profit as a function of selling price.

So we could say, we can substitute q is equal to -30 s plus 800 right over, right over here.

And let's be very clear what this is telling us.

This is telling us that if the selling price increases, then this will become a larger negative number

so we're going to sell fewer, we're gonna sell a smaller quantity.

And actually if you believe this, and if you actually made the selling price zero,

if you just gave away this product, it tells us that we would sell at most 800.

So it might not be a perfect model but let's just use this for,

you know some marketing experts have told us this, so let's just use it.

So if we substitute -30 s plus 800 for q, we get -30 s plus 800 times s minus 20,

times, and this is in a different shade of yellow, times s minus 20.

This is profit as a function of selling price.

And now we can just, let me be very careful here, let me be very care--

This is just, this is q right here, and so this whole thing is q.

Wanna make sure we're multiplying this whole expression times this entire expression right over there,

and so let's do that.

So this is going to be equal to,

this is going to be equal to -30 s.

So let me just distribute it out. This is going to be -30 s,

times s minus 20, times this whole thing, we're taking this whole term,

we're first multiplying it times -30 s.

And then we're gonna take this whole term and then multiply it by 800, s minus 20.

And so this gives us, this is equal to -30 s times s, we have to distribute again,

-30 s squared, -30 s times -20, is going to be positive, positive, positive 600 s.

And then we have 800 times s, so that's plus 800 s,

and then 800 times -20, so that is -8 times 2 is 16,

and we have one, two, three zeros, one, two, three, zeros.

And if we simplify we can add these two terms right over here.

We get -30 s squared plus 1400 s minus 16000.

So we now, we've now expressed, we've now expressed, our profit as a function of selling price.

And this is actually going to be a downward opening parabola,

and we can tell that because the coefficient on the second degree term,

on the quadratic term, is negative.

So if we were to graph this, if we were to graph this...

So over here--let me draw a better graph than that.

Over here, this axis right here is going to be the selling price,

And this is profit which is a function of selling price.

This graph, this equation right over here, is going to look like this, is going to look something like this.

We already saw the selling price--let me write, just write it this way.

So let me just--is going to look something like this.

I don't know what the exact equation is gonna look like, but it's gonna be downward opening.

And what we wanna do is maximize the profit.

We wanna find this maximum point right over here.

You could do it with calculus, if you had, if you had calculus at your, at your, at your fingertips.

Or you could just recognize this is the vertex of the parabola.

And you could, you could figure out the vertex by putting in the vertex form,

but the fastest way is to just know that the normally the x coordinate,

or the s coordinate, the s coordinate, of the vertex is going to be -b over 2 a.

And if we wanna figure out what -b over 2 a is,

we just take the--this is the b right over here so it's a -b

so it's -1400 over 2 a, over 2 times -30,

which is equal to -1400 over -60.

Negatives cancel out, we could divide the numerator and the denominator by 10.

So this is the same thing as a 140 over 6.

We can divide the numerator and the denominator by 3, or by, by 2.

And you get 70, you get 70 over 3.

And then we can just divide that, so 3 goes into 70,

3 goes into 7 two times, 2 times 3 is 6.

Subtract, you get a difference of 1, bring down the zero, 3 goes into 10 three times,

3 times 3 is 9, subtract, you have, bring down, you get a 1.

Now we're in the decimals, we bring down another zero, it becomes a 10 again,

3 goes into 10 three times, I think you see where this is going.

It's 23.3 repeating times. If we just keep doing this, we'll just keep getting more, more, more threes.

Or if we just wanted to round to the nearest penny, since we're talking about selling something,

this optimal profit, this optimal profit, will happen at a selling price of $23 and 30,

$23 and 33 cents.

That will optimize the yearly profit.