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10) A company manufactures and sells x web cameras per month. The monthly cost and price-demand equations are respectively, Price = p = 500 -.025x, 0

10) A company manufactures and sells x web cameras per month. The monthly cost and price-demand equations are respectively, Price = p = 500 -.025x, 0

Algebra: Structure And Method, Book 1
Algebra: Structure And Method, Book 1
(REV)00th Edition
ISBN:9780395977224
Author:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Publisher:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Chapter2: Working With Real Numbers
Section2.3: Rules For Addition
Problem 8P
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A company manufactures and sells x web cameras per month. The monthly cost and price-demand equations are respectively,

Price= p=500- 0.025x, 0< x < 20,000 

C(x)=350x+50,000

 

a.) Find the maximum revenue using calculus. Show all work. remember R=xp

 

b.) How many web cameras should the company manufacture to maximize its profit?  

How much should the company charge for each web camera? 

10) A company manufactures and sells x web cameras per month. The monthly cost and price-demand equations are respectively, Price = p = 500 -.025x, 0<x < 20,000 C(x) = 350x + 50,000. %3D a) Find the maximum revenue using calculus. Show all work. Remember R = xp %3D b) How many web cameras should the company manufacture to maximize its profit? How much should the company charge for each web camera? Use next page if needed
Transcribed Image Text:10) A company manufactures and sells x web cameras per month. The monthly cost and price-demand equations are respectively, Price = p = 500 -.025x, 0<x < 20,000 C(x) = 350x + 50,000. %3D a) Find the maximum revenue using calculus. Show all work. Remember R = xp %3D b) How many web cameras should the company manufacture to maximize its profit? How much should the company charge for each web camera? Use next page if needed
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