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The demand for apples is given by the equation $x p=150,$ where $x$ is the number of pounds demanded and $p$ is the price per pound. If the price is increasing at a rate of 0.25 dollar per week, at what rate is the: (a) demand changing; (b) revenue changing; when the demand is 10 pounds? (Can you solve (b) by inspection?)

(a) decreasing at $1 / 6$ lb per week(b) $\$ 0$ per week

Calculus 1 / AB

Chapter 2

An Introduction to Calculus

Section 10

Related Rates

Derivatives

Oregon State University

Harvey Mudd College

Baylor University

Boston College

Lectures

04:40

In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.

30:01

In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value (the rate of change of the value of the function). If the derivative of a function at a chosen input value equals a constant value, the function is said to be a constant function. In this case the derivative itself is the constant of the function, and is called the constant of integration.

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Given the price-demand equ…

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Price-demand. Suppose that…

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Demand Suppose the demand …

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05:07

The monthly demand for a…

06:38

05:16

The marginal price $d p / …

06:51

The quantity demanded of a…

for this problem. We've been given a function X p equals 150. And we're told that this is in relation to Apples. X is the number of pounds of apples demanded and P is the price. Well, let's write down what we know and what we're trying to find. We're told that the prices increasing Well, first of all, P is my price. So if prices increasing, that's a change, or DP DT at a rate of 25 cents per week. Okay, so there's my change of for price. It's increasing. So this is gonna be a positive 0.25 And we want to know what the rate is that the demand is changing. Will do this first piece. I'll do a first. So what demand is changing? That's DX DT. So that's what we're trying to find. And we're told we're gonna look at the point when the demand is £10. Okay, so that's the point we're looking at before we substitute that in that we're going to take our equation. This equation X p equals 1 50. That is our equation that relates our numbers to each other. We have two variables X and P, and we've got to rates DX DT is what we're looking for and DP DT is known Two variables to rates were ready to solve this problem. So we have our function. Now let's take the derivative. This is a product, so we'll use the product rule first times the derivative of the second plus the second times the derivative of the first and the derivative of a constant is just zero. Right? So I'm no, I'm gonna be wanting to look at DX DT That's what I'm trying to find. So part three, we're gonna plug in everything that we need Thio that we know to find what we are looking for So we have a plug in for X, d, p, d, t and P and solve for DX DT Well, I know that X is 10 dp DT is 0.25 I don't know P well, the good thing is, even though I don't know p I know how X and P relate to each other. So if X is 10, what value of P goes with X equal in 10? Well, if I divide by 10, that gives me P equaling 15 so I can plug 15 into my equation. Okay, so I've got 10 times 0.25 plus 15 DX DT. So let's solve. I've got 15 DX DT. I'm gonna move this to the other side. That's 10/4 or five halves. So it's become negative five halfs, and I'm gonna multiply. But both sides by 1/15, they're going to cancel, and this five will go into the 15 3 times. So DX DT is going to be a negative 1/6. So that negative means it is decreasing, so my demand is decreasing by 1/6 of a pound. So let's look apart. Be now. Now I want to know how the revenue is changing. So I want to know d r d t again at the same point when X equals 10. Well, what is revenue revenue? Is the demand times the price? Okay, but we know that that's 150. If the revenue is a constant when I go to take the derivative, it's going to be zero d r d t zero. The revenue isn't changing. There is no change so d r d t. Just by examining it, I can see that is going to equal zero

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