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A producer finds that demand for his commodity obeys a linear demand equation $p+2 x=100$ where $p$ is in dollars and $x$ in thousands of units.(a) Find the level of production that will maximize revenue. If the producer's costs are given by $C(x)=2+3 x,$ what should his level of production be to maximize profits?

(a) 25(b) $97 / 4$

Calculus 1 / AB

Chapter 3

Applications of the Derivative

Section 5

Applications II - Business and Economic Optimization Problems

Derivatives

Oregon State University

Baylor University

Idaho State University

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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The demand equation for a …

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The profit, in thousands …

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04:01

The demand $x$ and the pri…

04:20

The price $p$ (in dollars)…

09:37

01:10

Suppose it has been determ…

04:04

Maximizing Revenue The pri…

This question is about finding the production level that maximize the revenue and maximize profits. So let's first look at the first question to maximize the revenue. First, the revenue function, we use art to represent the revenue and revenue function is P times X where P is the price and acts as the quantity. And since the demand function in this question is P plus two x equals 100. Ah And Buy some alge bra. We can see that p equals 100 -2 x. And use this to equation. We can rise the revenue using just X. So to maximize our X, we should take the first derivative with respect to X. So it is equal to 100 minus four X. And to to get the maximum we need to improve, impose the first derivative of R equals zero, which means 100 -4 x equals zero. And we can solve x equals to 25. So to this is uh the first derivative equals to zero is just a necessary condition for maximum. And we have all, we should also check the second derivative and see whether It's less than or equal to one. So let's do it. We can see that the second derivative of our with respect to X is -4 which is less than zero. So uh when we produce 25 uh units of this commodity, uh the revenue achieves maximum. So in the next step, given that The cause function equals to two plus three X. We should choose the level of production which maximizes the profits. So the condition of maximizing profits is marginal revenue. Ecos marginal cost which is equivalent equivalent to the first derivative of the revenue function equals to the first derivative of the cost function. So in this question we have 100 -4 x equals two three. So to solve this, x equals to 97, divided by four, which is 24.25. So when producing 24.25 units of this commodity, uh the profits achieves maximum.

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