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If the cost equation in Exercise I is $C(x)=0.5 x^{2}+x+1,$ what price should be charged to maximize profit?

60,4

Calculus 1 / AB

Chapter 3

Applications of the Derivative

Section 5

Applications II - Business and Economic Optimization Problems

Derivatives

Harvey Mudd College

University of Michigan - Ann Arbor

University of Nottingham

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.

02:17

Given cost and revenue fun…

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Question Given the cost fu…

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Find the maximum profit if…

02:59

Maximum Profit A commodity…

01:12

Profit The demand function…

02:55

For the cost function in e…

04:18

The cost of producing a pl…

So given the price equation for the demand equation, P plus two X equals 100 your cost equation you want to find where the profit is maximized and that happens when marginal cost equals marginal revenue. So first we have to turn your demand equation into a revenue equation. We do that by multiplying your price times your demand. So first we wanna get this so the prices on one side. So we get 100 -2 x. And then to make it revenue, we multiply that by X. And then we want to find the derivative of that function. So 100 it's two X. We want to find the derivative to that with respect to X. So using the exponent role, oh that should be squared, we get 100 -4 x. And then we want to find your marginal cost. So exponent rule here we get one. Thanks plus one. Now when you set those two equal to each other 100 -4 x equals X plus one. We solve for X. So we get your exes on one side and your constants on another and we get 19.8. Now we want to plug this X value back into your demand equation. So that way you can get the actual price. So I'm not going to write it but you would put price equals, You would actually plug it back into this equation. Sorry about that price equals 100 miles two times 90.8 which will give you 60.4

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