porting goods store that sells exercise machines has fixed costs of $3,200 per day, and each exercise machine they sell costs them $100. c(x)= 100x +3200 R(X) = -2x^2 +300x How many exercise machines sold in a day would maximize the revenuefor the store? What is the maximum revenue

Calculus: Early Transcendentals
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ISBN:9781285741550
Author:James Stewart
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Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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A sporting goods store that sells exercise machines has fixed costs of $3,200 per day, and each exercise machine they sell costs them $100.

c(x)= 100x +3200 R(X) = -2x^2 +300x

How many exercise machines sold in a day would maximize the revenuefor the store?

What is the maximum revenuethe store can earn in a day?

Explain specificallywhy the number of machines the store should sell to maximize their revenueis not the same as the number of machines they should sell to maximize their profit.

Expert Solution
Step 1

The cost function represents as , cx=100x+3200.

The revenue function represents as,Rx=-2x2+300x.

1. The number of exercise that sold in a day that would maximize the revenue is ,

Take the derivative of the revenue function for maximization, 

dRdx=-4x+300.

-4x+300=0-4x=-300x=3004=75.

Thus the number of exercise that sold in a day that would maximize the revenue is 75 units.

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