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\markright{Chapter 3: Worksheet 7}
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\Bf{\Star \Star\ Chapter 3: Worksheet 7  \hfill Jack K. Cohen \hfill Colorado School of Mines}

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\Cb{Maximum-Minimum Problems, Dimensional and Plausibility Checks}
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\noindent \Bf{Suggested Problems}
Section 3.6: Ponder the ``5 step'' procedure for solving problems given on page 113.  Also read carefully the discussions on page 119 on checking results by plausibility and dimensional checks.  Look at the Examples \Em{carefully}, especially Example 6.  Make sure you understand the basics by doing some ``warm-up'' problems, say 1, 3, 5.  After doing the problems given below, other interesting ones are 27, 31, 33, 35, 41, 45, 51.  In all cases, make an effort to use letters instead of specific numbers.  Note: problem 48 is of scientific interest and will probably be covered in class.

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\noindent \Bf{Example}
(3.6.2) Find the maximum possible area of a rectangle of perimeter 200 ft.
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\noindent Let's use this easy problem to illustrate using the 5-step process and checks.  First off, to enable getting a more general result we replace the specific number 200 by $P$, the perimeter.
\begin{description}
\item[Find quantity to be maximized or minimized]  They tell us right off: \Em{maximize} the {area} (would it make sense to \Em{minimize} the area?).  As part of this step, we should also usually make a sketch and define any variables needed:
	\begin{figure}[htb]
	\epsfysize 80pt
	\centerline{\epsffile{ws7ex.eps}}
	\caption{Sketch for Example} 
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Let $x$ be the width and $y$ the length as shown in Figure 1.
\item[Express the dependent variable as a function of the independent variable]  From the figure, $A = x y$.  We need an auxiliary equation to eliminate one of $x$ and $y$.  Here, we use that the perimeter is fixed, $P = 2x + 2y$, and solve for $y = P/2 - x$ to get $A = x(P/2 - x)$.  Physical area must be positive, so this last expression for $A$ shows that we can confine consideration of $x$-values to $(0, P/2)$, that is $0 < x < P/2 = 100$.   The endpoints $x = 0, x = P/2$ give zero area, which isn't physical, but does no harm to include these points  since we know that 0 is not the maximum value we seek.  The benefit of including the endpoints in the domain is that we can apply the Closed Interval Extremum theory:  the maximum will occur either at the endpoints (not in this case though!) or at places where $A(x)$ is not differentiable (in this case there are no such points---$A$ is a polynomial, so differentiable everywhere) or at places where the derivative $A'$ vanishes (yes!).
\item[Apply calculus to find the critical points]  $A' = 0$ gives $x = P/4$.  Since, as already discussed, $A$ is differentiable at all points, this is the \Em{only} critical point.
\item[Identify the extrema]  The only candidates are the endpoints where $A(0) = 0, A(P/2) =  0$, and the critical point where $A(P/4) = P^2/16 > 0$.   So we see that $x = P/4$
is the value that gives the maximum.
\item[Answer the question posed in the problem]  Yes, don't lose out after having done all the hard stuff!  The maximum area is \fbox{ $A = P^2/16 = 2500$ sqft.}
\item[Plausibility]  That $x$ is 1/4 of the perimeter ($P/4$) is so plausible that most of us didn't need calculus to find it out.  That we can enclose an area of  2500 ft$^2$ with 200 ft of fence seems reasonable too.  We'd be suspicious of an answer like 4 ft$^2$ or like 400,000 ft$^2$.
\item[Dimensional checks]  Since $x = P/4$, the dimensions of $x$ are those of perimeter, namely length---this is correct.  Possible notation: $[P] = L$ (where $L$ stands for length).  Similarly, $[A] = [P^2/16] = [P^2] = L^2$---correct.  See the following table for some useful examples of dimensions using  this bracket notation.
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\begin{tabular}{||l|l||}	\hline
\Bf{Quantity} & \Bf{Dimensions} \\ \hline
perimeter & $[P] = L$	\\
area & $[A] = L^2$	\\
volume & $[V] = L^3$	\\
velocity & $[v] = L/T$	\\
acceleration & $[a] = L/T^2$	\\
force & $[F] = ML/T^2$ \\ \hline
\end{tabular}

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\noindent Realize that \Em{dimensions} are \Em{not} the same thing as \Em{units}: feet, inches, cm, km \Em{all} have dimensions of length ($L$).  It is true that you should convert to consistent units in final numerical answers (eg. 6 ft / 3 in = 24 in).   Although this is important, it is a less fundamental matter than making sure your answer has the correct dimensions.

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\noindent  In the following variants of the text problems, parameters are often introduced in place of specific numbers---proceed as far as seems practical with the parameters. Have a strong reason if you do simplify to specific numbers and thus lose the chance to derive general results and rely on dimensional checks; try to use \Mma\  if you run into excruciating algebra.

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\item (3.6.7) The sum of two positive numbers is $S$.  What is the smallest possible value of the sum of their squares?  Also obtain numerical values for $S = 48$ and  make a confirming \Tt{Plot} for this case.  Also discuss getting the \Em{largest} possible value for the sum of squares.  Are your answers plausible?  Pure numbers don't have dimensions, so technically you can't use dimensional checks here---can you think of a way around this?

\item(3.6.9) Repeat last problem for sum of \Em{cubes}.

\item $[{\bf Optional}]$ There is a general conclusion valid for \Em{any} power lurking in the last two problems, can you find it?  Check out your result for cube roots (3.6.39).

\item (3.6.11) A farmer has $P$ yards of fencing with which to build a rectangular corral having two internal dividers both parallel to two of the sides of the corral.  What is the maximum total area of such a corral?  Also obtain numerical values for $P = 600$.

Dictionary note:  a \Em{corral} is a fenced area used to contain farm animals.

Mathematical note:  a surprising feature of this problem is that it doesn't matter if the two internal dividers are equally spaced or not, so don't assume this.

\item (Example 6) In the cow application of the law of reflection, redo the calculation more simply by observing that $\alpha = \beta$ implies that $\cot \alpha = \cot \beta$.
Also use this method to get the general result by using letters instead of numbers.

\item (3.6.15) The volume of 1 kg of water at a temperature $T$ between 0$^\circ$C and 30$^\circ$C is approximately \[V = a - bT + c T^2 - d T^3  \]  cubic centimeters.  At what temperature does water have its maximum density?   The empirical values for the positive constants $a, b, c, d$ are $a = 999.87, b = 0.06426, c = 0.0085043, d = 0.0000679$.

\item (3.6.19) Three large squares of tin, each of edge length $s$ m, have four equal small squares cut from their corners.  All twelve resulting small squares are to be the same size.  The three large cross-shaped pices are then folded and welded to make boxes with no tops, and the twelve small squares are used to make two cubes.  How should this be done to maximize the total volume of all five boxes?  Evaluate your answer for the numerical value $s = 1$.  Suggestion: use \Tt{Plot} to confirm your numerical answer.

% Logistic growth
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