College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 2, Functions - Chapter 2 Review - Exercises - Page 267: 7

Answer

(a) $C(1000)=34000$, $C(10000)=205000$ (b) The cost of printing 1000 and 10000 copies of mathematical textbooks is 34000 and 205000, respectively. (c) $C(0)=5000$. The number represents the cost of printing 0 copies of mathematical textbooks, or the minimum price for printing the mathematical textbooks. (d) Net change: $171000$; Average rate of change: $19$.

Work Step by Step

(a) $C(x) = 5000+30x-0.001x^2$ To find $C(1000)$ and $C(10000)$, plug in the $x$ values of $1000$ and $10000$, respectively, into the above $C(x)$ function. $C(1000)= 5000+30*1000-0.001\times1000^2$ $=5000+30000-0.001\times1000000$ $=35000-1000=34000$ Likewise: $C(10000)=5000+30*10000-0.001\times10000^2$ $=5000+300000-0.001\times 100000000$ $=305000-100000=205000$ (b) Since $C(x)$ represents the total cost of printing $x$ copies, we understand that the cost of printing 1000 and 10000 copies of mathematical textbooks is 34000 and 205000, respectively. (c) First, we plug in $0$ into the above $C(x)$ function to find $C(0)$. $C(0) = 5000 +30*0-0.001*0^2=5000+0+0=5000$. We already established the relationship between $C(x)$ and $x$ in (b). So likewise, this result means the cost of printing $0$ copies (or the minimum cost for printing copies) is $5000$. (d) The net change of the cost $C$ is the difference between two $C(x)$ values. Since the $x$ values are $1000$ and $10000$, we can apply the results in (a). net change $=C(x_{2})-C(x_{1})=C(10000)-C(1000)$ $=205000-34000=171000$ The average rate of change of $C(x)$ between $x=1000$ and $x=10000$ is calculated by dividing the net change of $C(x)$ to the change in $x$: avg rate of change $= \frac{C(10000)-C(1000)}{10000-1000}=\frac{171000}{9000}=19$
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