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\markright{Chapter 1: Answers 3}
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\Bf{Chapter 1: Answers 3 \hfill Jack K. Cohen \hfill Colorado School of Mines}

\begin{enumerate}
\item The Alpha lettuce patch:
        \begin{enumerate}
        \item  `Best' means maximum area (usually).
        \item Intuitively a square would give the maximum area among all rectangular patches.  Thus the side ($x$) is 100/4 = 25 m and the maximum area is 625 m$^2$.
        \item Your figure should have shown a rectangle with two facing sides labeled with $x$ and the other two with $100 - 2x = 2(50 - x)$ .
        \item $x$ is one side of the rectangle, $A$ is its area.  The valid $x$-values (the ``domain'' of $A$) are $[0, 50]$ because outside this interval a side will assume a non-physical negative value.  We know that even the endpoint values 0 and 50 can't be right since then the area is zero, but it does no harm to include these and we'll see later that posing maximum or minimum problems on a \Em{closed} interval has some theoretical advantages.
                \item The  values in the middle of the most refined \Tt{Table} are:
        \begin{verbatim}        
        24.7   624.91
        24.8   624.96
        24.9   624.99
        25.    625.
        25.1   624.99
        25.2   624.96
        25.3   624.91
        \end{verbatim}
        
\item See Figure 1.
                \begin{figure}[htb]
                \epsfysize 100pt
                \centerline{\epsffile{ans3p1.eps}}
                \caption{Graph of $A(x)$ in the critical region of Problem 1.} 
                \end{figure}

\item The maximizing $x$ is half-way between the zeroes, that is $x = 25$.   One way to justify this is to multiply out the polynomial and complete the square: $A(x) = 50x - x^2 = -(x - 25)^2 + 625$.  This form of the equation shows clearly that $A \leq 625$ for all $x$ and actually achieves the maximum value $A = 625$ at $x = 25$.  Note: the ``between the zeroes idea'' is limited to quadratics---another and far more general answer to this question is that the maximum occurs where the tangent to the function is horizontal.  The pursuit and refinement of this insight is a large part of this semester's work.  At the present time, this criteria is not practical for us, since we don't yet know how to find the $x$'s that make the tangent horizontal---but we soon will!  (No credit for using this method now, if you happen to have covered this notion in high school.)

\item In ``real life'' any number of other restrictions might apply.  For example, the need for our irrigation lines to be able to reach all parts of the patch.
                
\item  Still labeling one side as $x$, we find that the adjacent side has length $P/2 - x$, so that the area is $A(x) = x(P/2 - x)$.   Using the ``half-way between the zeroes'' insight, we intuit that the maximizing $x = P/4$ and that the corresponding area is $A = P^2/16$.  Completing the square verifies this intuition: $A(x) = -(x - P/4)^2 + P^2/16$.  In most problems, we would not be able to handle the parameter so cleanly.  In the absence of some theoretical result or blinding flash of insight, we might well pick some ``typical'' $P$-values and use graphical/tabular or other methods to attempt to form a hypothesis about the general case.
        \end{enumerate}

\item From Figure 2 (you \Em{did} draw a figure, didn't you?), we see that now $A(x) = 2x(50 - x)$.   Using the ``half-way between'' insight for parabolas or some other method, find that the maximizing $x = 25$ and the maximum value is $A = 1250$ m$^2$ (exactly twice as large as for the 4-sided case).    For the general case, the formula is $A(x) = 2x(P/2 - x)$ with maximum at $x = P/4$ of size $A = P^2/8$.

\begin{figure}[htb]
\epsfysize 100pt
\centerline{\epsffile{ans3p2.eps}}
\caption{Diagram for Problem 2.} 
\end{figure}

        
\item  Note that now, we want to \Em{minimize} the perimeter.  From Figure 3, we see that the sum of side lengths is $L = x + 1000/x + x = 2x + 1000/x$. 

\begin{figure}[htb]
\epsfysize 100pt
\centerline{\epsffile{ans3p3a.eps}}
\caption{Diagram for Problem 3.} 
\end{figure}
One mode of solution is a graph.  Figure 4 shows a zoom of the critical region obtained (after some experimentation) by using the command

\Tt{Plot[2x + 1000/x, \{x, 10, 50\}, PlotRange -> \{80, 150\}]} 

\noindent Picking the point at the minimum gives $x \approx 22$.
Then the command

\Tt{Table[\{x, 2x + 1000/x\}, \{x, 22.2, 22.4, 0.05\}] //TableForm}

\noindent lets us home in on the answer:

\begin{verbatim}
22.2    89.445
22.25   89.4438
22.3    89.443
22.35   89.4427
22.4    89.4429
\end{verbatim}
Thus $x \approx 22.35$ and $A \approx 89.4427$ at the minimum.  Calculus methods let use determine that the exact answer is $x = 2 \cdot 5^{3/2} \approx  22.3606797749979$, giving a minimum of $A \approx 89.4427190999916$.

\begin{figure}[htb]
\epsfysize 90pt
\centerline{\epsffile{ans3p3b.eps}}
\caption{Plot of 2x + 1000/x in the critical  region.} 
\end{figure}

\item A diagram is shown in Figure 5.  The unknown is the radius $r$.  As shown, the quantities $h = 10$ and $s = 1000$ are given.
	\begin{enumerate}
	\item See Figure 5.

\begin{figure}[htb]
\epsfysize 90pt
\centerline{\epsffile{ans3p4.eps}}
\caption{Diagram for Problem 4.} 
\end{figure}
	\item Use the right triangle to derive: $\cos (s/r) = r / (r + h)$, 
here $h = 20, s = 1000$.
	\end{enumerate}
\end{enumerate}
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