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 $= \frac{C(10000)-C(1000)}{10000-1000}=\frac{171000}{9000}=19$
