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\centerline{Mathematics 133 -- Intensive Calculus for Science 1}
\centerline{Solutions for Discussion 7}
\centerline{November 17, 2005}
\bigskip
\noindent
A)   Two non-negative numbers $x,y$ add up to 72.
What is the largest possible value of the product $P = xy$?
\bigskip
\item{} {\it Solution.}  We have $x + y = 72$, so $y = 72 - x$.
Then $P(x) = x(72 - x) = 72x - x^2$.  To maximize, we first 
compute $P'(x)$ and set equal to 0 to find critical points:
$$P'(x) = 72 - 2x = 0 \qquad \Rightarrow x = 36.$$
Then $P''(x) = -2 < 0$, so the second derivative test tells
us that $x = 36$ is a local maximum.  It must also be the
global maximum because this is the only critical point.
(Alternate method:  Note that $0 \le x \le 72$ since it is 
given that $x,y \ge 0$.  Then $P(0) = 0$, $P(36) = 36^2 = 1296$
and $P(72) = 0$.  So we can see this way that $x = 36$ gives 
the global maximum.)  The maximum value of the product
is $P(36) = 36^2 = 1296$.
\bigskip
\noindent
B)  A rectangular beam is cut from a cylindrical log of radius
30 cm.  The strength of a beam of width $w$ and height $h$
is proportional to $wh^2$.  Find the width
and the height of the beam of maximum strength.  
\bigskip
\item{} {\it Solution.}  The cross-section of the log is a circle
of radius 30, and the cross-section of the beam is a rectangle
inscribed in this circle.  So the Pythagorean Theorem shows
$w^2 + h^2 = 60^2 = 3600$.  Then it is easiest to solve this
for $h^2$ in terms of $w$ and to write the strength as a
function of $w$:  $S = kw(3600 - w^2) = k(3600w - w^3)$, 
where $k$ is the constant
of proportionality.  To maximize, we proceed as before:
$$S' = k(3600 - 3w^2) = 0 \qquad \Rightarrow w^2 = 1200,$$
so $w = \sqrt{1200} = 20\sqrt{3} \doteq 34.64$. 
This is a maximum for the strength because $S''(w) = -6kw$.
$k$ is a positive constant and our value $w \doteq 34.64 > 0$, so
$S''(34.64) < 0$.  Since this is the only critical point
for $w \ge 0$, we have a local and global maximum.
 To find the 
corresponding $h$, we substitute into the equation $w^2 + h^2 = 60^2 = 3600$:
$1200 + h^2 = 3600$, so $h^2 = 2400$ and 
$h = \sqrt{2400} = 20\sqrt{6} \doteq 48.99$.  The dimensions are
$$w \doteq 34.64 \qquad {\rm and} \qquad h \doteq 48.99.$$
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\noindent
C) Northern South Dakota
State Agricultural and Veterinary Junior College
is building a new running track for their prize-winning track team --
the ``Flying Farmers''.  The track is to be the perimeter of a region obtained
by putting two semicircles on opposite ends of a rectangle, and that
perimeter should be <EM>440 yards</EM> in length.  Due to 
financial constraints, the administration has decided to grow
sweet corn in the area enclosed by the track and sell it to the local 
grocery store to make some extra income.  Determine the dimensions 
to build the track in order to maximize the area for growing corn.
\bigskip
\item{} {\it Solution:}  Let $x$ be the length of the straight segments
of the track, and $r$ be the radius of the semicircular portions.  Then
the length of the track is the (outside) perimeter:
$$440 = 2x + 2\pi r \qquad\Rightarrow 2x = 440 - 2\pi r.$$
The area inside the track is
$$A = x(2r) + \pi r^2 = r(440 - 2\pi r) + \pi r^2 = 440r - 2\pi r^2 + \pi r^2
=440 r - \pi r^2.$$
Then $A'(r) = 440 - 2\pi r = 0$ when $r = 220/\pi$.  This is a local
and global maximum for $A(r)$ since $A''(r) = -2\pi < 0$.  The
optimum track has $r = 220/\pi$ and $x = 220 - \pi (220/\pi) = 0$ (!).
In other words the optimum track is a {\it circle}.
\bye
