Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 1 - Precalculus Review - 1.2 Linear and Quadratic Functions - Exercises - Page 18: 26

Answer

The required equation is $N(P)=\dfrac{-50}{3}P+\dfrac{2000}{3}$ number of tickets when price is $P$ dollars per ticket. The decrease in number of tickets $\Delta N =\dfrac{-250}{3}\approx -83.33$. Here, a negative sign indicates a decrease.

Work Step by Step

Let the required function be $N(P)=m\cdot P + b$. Since, $N(10)=500$ and $N(40)=0$, so, we get $500=10m+b$ and $0=40m+b$. Subtracting the equations gives $500=-30m$, or, $m=\dfrac{-50}{3}$. Substitute the value of $m$ in second equation, to get $0=40\cdot \dfrac{-50}{3}+b$ , or, $b=\dfrac{2000}{3}$. Therefore, the required equation is $N(P)=\dfrac{-50}{3}P+\dfrac{2000}{3}$. The slope of the linear function is $\dfrac{\Delta N}{\Delta P}=\dfrac{-50}{3}$. When $\Delta P= 5$ dollars, then, $\dfrac{\Delta N}{5}=\dfrac{-50}{3} \Rightarrow \Delta N=-\dfrac{250}{3}$. Negative sign indicates decrease.
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