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\centerline{MATH 133 -- Calculus with Fundamentals 1}
\centerline{Solution for 4.2/74 from Problem Set 8 -- Part B}
\centerline{December 5, 2015}
\vskip 10pt
\begin{itemize}
\item The formula for the concentration is given as
$$C(t) = \frac{.016 t}{t^2 + 4t + 4}.$$
\item But the problem becomes much simpler if you rewrite this as
$$C(t) = (.016) \cdot \frac{t}{(t + 2)^2}$$
\item Note:  I pulled the constant factor $.016$ out and factored the denominator
noticing that it is a perfect square.
\item  We want to find the maximum and minimum values of this function on the 
interval $[0,8]$.  
\item 
First we differentiate $C(t)$ using the quotient rule, simplify, and then set the derivative 
equal to find the critical points:
$$C'(t) = (.016)\cdot \frac{(t + 2)^2\cdot 1 - t\cdot 2(t + 2)}{(t+2)^4} = 
(.016) \cdot \frac{(t + 2)(2 - t)}{(t + 2)^4} = (.016) \cdot \frac{2 - t}{(t + 2)^3},$$
so $C'(t) = 0$ when $t = 2$.  
\item Now we compute:
\begin{eqnarray*}
C(0) &=& 0\\
C(2) &=& \frac{.032}{16} = .002\\
C(8) &=& \frac{.128}{100} = .00128.
\end{eqnarray*}
\item So the maximum value is $C(2) = .002$ and the minimum value is $C(0) = 0$.  
\end{itemize}
\vskip 10pt
\noindent
Some comments:
\begin{enumerate}
\item[(a)]  A lot of people got very tied up in dealing with the ``complicated'' decimal
number coefficients that you get if you don't start out by pulling out the constant factor
$.016$.  \emph{Don't forget the constant multiple rule for derivatives:}
$$\frac{d}{dx} (k f(x)) = k f'(x).$$
In words, \emph{the derivative of a constant $k$ times a function is the same constant
$k$ times the derivative of the function}.  You can (and usually should) do something like the 
above whenever you can pull a constant factor out of a whole function you need to 
differentiate.
\item[(b)]  If you don't factor the denominator, you will get something like this when 
you apply the quotient rule:
$$C'(t) = (.016) \cdot \frac{(t^2 + 4t + 4)\cdot 1 - t(2t + 4)}{(t^2 + 4t + 4)^2} = 
(.016) \frac{-t^2 + 4}{(t^2 + 4t + 4)^2}$$
This makes it look as though both solutions of $-t^2 + 4 = 0$, namely, $t = \pm 2$
are critical points.  \emph{However}, the root $t = -2$ \emph{is not a critical point}
because it is not in the domain of the original $C(t)$ (the function $C(t)$ has an infinite
discontinuity -- a vertical asymptote -- there).  Check the definition of critical points
on page 202 of the text book -- critical points of a function are points where the function
is defined, but where either the derivative is zero, or the derivative is undefined.
And in any case, $t = -2$ is not in the interval $[0,8]$ we are interested in, so that
should be ignored.
\end{enumerate}
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