Single Variable Calculus, Chapter 3, 3.1, Section 3.1, Problem 43

Given the equation $c(x) = 5000 + 10 x + 0.05 x^2$, where $c$ represents the cost of producing $x$ units of
a certain commodity.



a.) Determine the average rate of change of $c$ with respect to $x$ when the production level is changed.



$
\begin{equation}
\begin{aligned}
& (i) \text{ from } x = 100 \text{ to } x = 105\\
& (ii) \text{ from } x = 100 \text{ to } x = 101
\end{aligned}
\end{equation}
$



b.) Determine the instantaneous rate of change of $c$ with respect to $x$ when $x = 100$.



a.) $(i)$ from $x = 100$ to $x = 105$



$ \quad \displaystyle \text{average rate } = \frac{c(105) - c(100)}{105-100} =
\frac{5000+10(105) +0.05(105)^2 - [5000+10(100) + 0.05(100)^2]}{105-100} = \frac{\$ 20.25}{\text{ unit}}$



$(ii)$ from $x = 100 $ to $x = 101$


$ \quad \displaystyle \text{average rate } = \frac{c(101) - c(100)}{101-100} =
\frac{5000+10(101) +0.05(101)^2 - [5000+10(100) + 0.05(100)^2]}{101-100} = \frac{\$ 20.05}{\text{ unit}}$



b.) From the definition, we can write the instantaneous rate as,



$\displaystyle c(a) = \lim\limits_{h \to 0} \frac{f(a+h) - f(a)}{h}$



$
\begin{equation}
\begin{aligned}
c(a) & = \lim\limits_{h \to 0} \frac{5000+10(a+h)+0.05(a+h)^2-[5000+10(a)+0.05(a)^2]}{h}\\
c(a) & = \lim\limits_{h \to 0} \frac{\cancel{5000} + \cancel{10a} + 10h + \cancel{0.05a^2} + 0.1ah + 0.05h^2 - \cancel{5000} - \cancel{10a} - \cancel{0.05a^2}}{h}\\
c(a) & = \lim\limits_{h \to 0} \frac{\cancel{h}(10+0.1a+0.05h)}{\cancel{h}}\\
c(a) & = \lim\limits_{h \to 0} (10+0.1a+0.05h)\\
c(a) & = 10+0.1a+0.05(0)\\
c(a) & = 10+0.1a
\end{aligned}
\end{equation}
$



The value of instantaneous rate of change of $c$ when $a = 100$ is $\displaystyle c(100) = 10+0.1(100) = \frac{\$ 20}{\text{unit}}$