College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 2 - Functions and Graphs - Exercise Set 2.6 - Page 299: 99

Answer

$(R-C)(20,000)=-200,000$ $(R-C)(30,000)=0$ $(R-C)(40,000)=200,000$

Work Step by Step

$C(x)=600,000+45x$ models the cost of the company $R(x)=65x$ models the revenue of the company. $(R-C)(20,000)=R(20,000)-C(20,000)=65\times20,000-(600,000+45\times20,000)=-200,000$ This means that if the company produced and sold $20,000$ radios, their profit will be $-200,000$ (loss). $(R-C)(30,000)=R(30,000)-C(30,000)=65\times30,000-(600,000+45\times30,000)=0$ This means that if the company produced and sold $30,000$ radios, their profit will be $0$ meaning the cost of the production will equal the revenue they gained from producing and selling $30,000$ radios. $(R-C)(40,000)=R(40,000)-C(40,000)=65\times40,000-(600,000+45\times40,000)=200,000$ This means that if the company produced and sold $40000$ radios, their profit will be $200,000$.
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