00:01
All right, so we're given this equation where p is a price and x is a quantity.
00:10
So if we want in section a, a model that expresses the revenue as a function of p, well, r equals, well, however many we have multiplied by that price.
00:23
So r of p, if we want in terms of price, we replace the x with what it is here.
00:30
And so we multiply p times that and we get negative p squared plus 500p.
00:39
Now in part b, if we want to find a domain and we assume that r is not negative, well let's go ahead and use desmos in order to graph this, negative 20x squared plus 500x.
00:54
And so i had already set my bounds so that i would fit nicely.
01:00
You may need to scroll in or scroll out in order for it to actually find this graph when you do this.
01:08
So what's the domain? well, if it's not negative, it goes from zero up to 25.
01:18
And the price that will maximize is that middle part, so 1250.
01:26
And how much are you going to make? you're going to make $3 ,125.
01:33
Now in part e, if you're going to find how many units you sell at this price, well, we take 3 ,125 and divide by 12 .5, and i got 250 units...