Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 1 - Functions - 1.2 Representing Functions - 1.2 Exercises - Page 21: 16


The required linear function is $$f(n)=8n-175.$$ The zero profit is broken at $n=22$. The graph is in the figure below.

Work Step by Step

We know that $f(0)=-175$. Further, when $n$ increases by $1$ then $f(n)$ increases by $8$ so the slope of this linear function is $m=8/1=8$. Every linear function can be written as $$f(n)=mn+f(0)$$ which yields $$f(n)=8n-175.$$ To break $0$ profit we need that $f(n)>0$ i.e. $8n-175>0$ giving $n>175/8=21.875$. Since $n$ has to be integer they need to sell atleast $22$ tickets. The graph is obtained by drawing a line through $P(0,-175)$ and $Q(30,8\cdot30-175)=Q(30,65)$
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