The price p (in dollars) and the quantity x sold of a...

The price $p$ (in dollars) and the quantity $x$ sold of a certain product satisfy the demand equation $$ x=-20 p+500 $$ (a) Find a model that expresses the revenue $R$ as a function of $p$ (b) What is the domain of $R ?$ Assume $R$ is nonnegative. (c) What price $p$ maximizes the revenue? (d) What is the maximum revenue? (e) How many units are sold at this price? (f) Graph $R$. (g) What price should the company charge to earn at least $\$ 3000$ in revenue?

Transcript

00:01 So looking here, we're given this a few variables here.

00:05 The price for a product is in p dollars.

00:08 The quantity sold is x.

00:10 So the quantity you sell will equal the price times 8 of 20 plus 500.

00:16 So it's like the more something costs, the less you would sell.

00:21 And the cheaper it is, the more you would sell, that kind of idea.

00:24 So what would it be a model that expresses revenue as a function of p? revenue.

00:30 Well, revenue would be the number of units you sell times the price.

00:37 So if i want a revenue function of p, then everything is being in terms of p.

00:43 So let's replace this x with this part.

00:47 So i'm replacing that x with that part right there, negative 20p plus 500 times p the price.

00:57 So your revenue, this is a revenue function of p price.

01:02 So if you tell me the price, i can tell you what the revenue would be.

01:07 So if we multiply it out, we're going to get a little quadratic.

01:13 So multiply that p through negative 20p squared plus 500p.

01:20 So what would be the domain here? well, you can't plug anything negative in, right? revenue is not negative.

01:28 You can't have a negative price.

01:30 So i assume that our domain here would start with parentheses zero.

01:34 Because you're, i guess you could give it away.

01:37 I don't know, it would be zero.

01:40 I guess you could include zero, but your revenue would be zero.

01:43 So i guess that could be true.

01:45 So you could do a square bracket there.

01:48 It depends on how, it just depends on how you look at it.

01:52 And then what about the upper bracket? what's the highest number you can put in? well, if you put in, let's just look at the graph of it.

02:00 So if i look at a graph of this, i would say negative 20.

02:04 I'm going to use x's here because my calculator has.

02:06 Use x is not p's squared plus 500 x and here i get a value if i plug a 25 in i get a y value of zero so if i plug 25 into this function i get zero revenue because that would be so zero zero is the first one and then we don't want any negative so i don't want anything down here so from zero to 25 those are the two values are going to give me a revenue of zero dollars.

02:42 What price p maximizes the revenue? to maximize revenue, you're looking up here at a maximum of x of 12 .5.

02:52 So a price of 12 .5 gives you a maximum revenue of so the maximum number would be $12 .5 or $12 .5.

03:01 Give you maximum revenue of $3 ,125...

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