%% Math 131, Fall, 2003
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\begin{center}
\textbf{Holy Cross College, Fall Semester, 2003} \\
\textbf{Math 131, Professor Hwang} \\
Solutions to Worksheet on Optimization
\end{center}

\be

\item A rectangular box without a lid is to be made from a
  $24\times36$~inch piece of cardboard by removing a square from each
  corner and folding up the resulting flaps along the dotted lines.
  Find the size of the cutouts that maximizes the box's volume.

  \input{hsoln08a.eepic}
  
  \soln Let~$x$ be the size of the cutouts. Then $0\leq x\leq 12$ (the
  cutout cannot be longer than half the length of the shorter side),
  and the volume of the box is
  \[
  V(x)=x(24-2x)(36-2x)=4x(12-x)(18-x)=4(x^3-30x^2+216x).
  \]
  The volume function has two critical points, the solutions of
  $V'(x)=4(3x^2-60x+216)=0$. The quadratic formula gives
  $x=\frac{1}{3}(30\pm\sqrt{252})$; the ``$+$''~sign leads to a
  solution larger than~12, and may be discarded. The remaining
  solution, $x_0\simeq4.7085$, must be the length of cutout that gives
  maximum volume: Either make a sign diagram for~$V'$, or observe that
  the volume is zero at both endpoints, while $V(x_0)>0$.

  \hrulefill

\item A rectangular swimming pool is to be built with an area of
  1800~square feet. The pool will be surrounded by a patio of width
  5~feet on each side and 10~feet at each end. Find the dimensions of
  the pool that minimize the total area of the property required.

  \input{hsoln08b.eepic}
  
  \soln Let the pool be $x$~feet long and $y$~feet wide. We are given
  that $xy=1800$, and want to minimize $(x+20)(y+10)$, the total area
  required for pool and patio. Multiplying out and using $xy=1800$, we
  find that the total area required is $xy+10x+20y+200=10x+20y+2000$.
  Solving for~$y$ as a function of~$x$ and plugging in to the area
  expression gives the total area required as a function of the length
  of the pool:
  \[
  A(x) = 10x+20y+2000 = 10x + 20\cdot\frac{1800}{x}+2000,\qquad x>0.
  \]
  Look for critical points: $A'(x)=10-20(1800)x^{-2}=0$. There is only
  one positive critical point, $x=60$. A sign diagram for~$A'$ reveals
  that~$A$ decreases on the interval $(0,60]$~and increases
  on~$[60,\infty)$, so the critical point is an absolute minimum. In
  summary, the total area is minimized when the proportions of the
  pool alone are 2-to-1, the same as the proportions of the pool+patio
  taken together.

  \hrulefill

\item A rectangle in inscribed in a semicircle of radius~1. Find the
  dimensions that maximize the area of the rectangle.

  \input{hsoln08c.eepic}
  
  \soln The upper half of the semicircle is the graph
  $y=\sqrt{1-x^2}$. Let~$x$ be the half-width of an inscribed
  rectangle; the area is $A(x)=2x\sqrt{1-x^2}$, and the domain is the
  interval~$[0,1]$. (The area function is graphed in red on the same
  scale as the semicircle; the endpoints correspond to ``degenerate''
  rectangles that have area zero.) As usual, search for critical
  points of the area function:
  \[
  A'(x)=2x\cdot\frac{-x}{\sqrt{1-x^2}} + 2\sqrt{1-x^2}
  =\frac{2}{\sqrt{1-x^2}}(1-2x^2).
  \]
  There are two critical points: $x=1$ (where $A'$~does not exist) and
  $x_0=\frac{1}{\sqrt{2}}$. The former is an endpoint, and as remarked
  earlier corresponds to a rectangle of area zero. The latter
  corresponds to a rectangle that is twice as wide as it is high, and
  must be the absolute maximum of~$A$, since $A(x_0)>0$. (A short
  calculation shows that the maximum area is exactly~1.)

\ee

\eject

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