Download the App!

Get 24/7 study help with the Numerade app for iOS and Android! Enter your email for an invite.

Get the answer to your homework problem.

Try Numerade free for 7 days

Like

Report

Each month an automobile dealer makes a profit of $\$ 200$ on each car that she sells if not more than 50 cars are sold. For every car above 50 that she sells her profit per car is decreased by $\$ 2 .$ How many cars should the dealer sell monthly to maximize her profit?

75

Calculus 1 / AB

Chapter 3

Applications of the Derivative

Section 5

Applications II - Business and Economic Optimization Problems

Derivatives

Campbell University

University of Michigan - Ann Arbor

University of Nottingham

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.

01:38

An automobile manufacturer…

03:17

An electronics store is se…

01:13

Profit. A seamstress decid…

01:52

An artist is planning to …

00:50

Using a profit equation, y…

01:31

Translate to an equation a…

Hello. So here we let the rebate the X. So the number of cars sold per month is given by 2000 plus 200 times X over 50 which is eager to 2000 plus four X. And the profit per car is then going to be 1000 minus X. Okay so then the total monthly profit p groups, a total monthly profit is going to be equal to 2000 plus four X times 1000 minus X. Which gives us four times 500 1000 plus 500 X minus X. Squared. Um We then differentiate disrespect with respect to X and get that the derivative, the p the acts that's equal to the um We differentiate our equation here and get the derivative is then going to be equal to four times 500 -2 x. And then for the maximum profit we just take our derivative here and set it equal to zero. So if this is equal to zero we get that four times 500 minus two X. Is equal to zero, which gives us that 500 minus two X. Is equal to zero which gives us that two X. Is equal to 500. So therefore x. Um is eager to 500 over two which is two. So therefore X. Is equal to 2 50. And we have the required solution to the maximum profit is going to be $250

View More Answers From This Book

Find Another Textbook

Numerade Educator

02:24

Sketch the graph of the function defined by the given equation.$$f(x)=3^…

01:23

Write the given expression in logarithmic format.$$16^{3 / 4}=8$$

02:15

If the cost equation in Exercise I is $C(x)=0.5 x^{2}+x+1,$ what price shoul…

04:36

$$5 s^{2}\left(v^{3}-1\right)=7 . \text { Find }(a) d s / d v;(b) d v / d s$…

02:25

01:00

Are there functions that are their own inverse?

00:39

Write the given expression in exponential format. $$\log _{b} 1=0$$

Use your calculator to compute the expression given.$$(-5.23)^{5}$$

01:15

Write the given expression as a sum of logarithms.$$\ln \sqrt{x} y^{2}$$…

05:51

Find the elasticity of demand for each of the following demand equations.