\documentclass{minimal}
\usepackage{amsmath,amssymb}
\def\problem#1{\par \medskip \noindent \textbf{Problem #1: }}
\begin{document}

\problem{4.7, 10} Let $x$ be the width of the box and $y$ be the height of
the box.  The volume is $x^2 y$, so $x^2 y = 32000$.  The material used in
the construction of the box can be assessed using its area.  The area of the
base is $x^2$ and the area of each side is $xy$, so the total area used is $A
= x^2 + 4xy$.  We must minimize $A$.  Rewrite it in terms of $x$ alone by
using $y = 32000/x^2 $ to eliminate $y$.  We get $A = x^2 + 4x(32000/x^2) =
x^2 + 128000/x$.  Compute $dA/dx = 2x - 128000/x^2$.  Solve $dA/dx = 0$,
i.e., $2x = 128000/x^2$ to get $x = 64000 ^ {1/3} = 40$.  Thus $y = 32000/x^2
= 32000/1600 = 20$.  Since $dA/dx = (2x^3 - 128000)/x^2$, by examining the
numerator $2x^3 - 128000$ we see that $dA/dx$ is positive for $x > 40$ and
negative for $x < 40$, and thus the critical point is a local minimum.  Since
it's the only critical point, it's the global minimum.  The box is $40 \times
40 \times 20 $.

\problem{4.7, 18} Minimize the distance $D$ from $(0,-3)$ to $(-y^2,y)$.  We
see that $D^2 = (y^2)^2 + (y+3)^2$, and we might as well minimize $E = D^2$
instead.  Simplify $E = y^4 + (y+3)^2$.  Compute $dE/dy = 4 y^3 + 2(y+3) = 4
y^3 + 2y + 6$.  Solve $dE/dy = 0$.  By inspection we see that $dE/dy = 0$
when $y = -1$.  Since $d^2E/dy^2 = 12 y^2 + 2 > 0$, we see that $dE/dy$ is an
increasing function, so has only one root.  Therefore $y=1$ is the only
critical number.  From the geometry, it must give a minimum.  The point is
$(-1,-1)$.

\problem{4.7, 30} Let $x$ be the width and $y$ be the height of the poster.
Then $180 = xy$ and we are to maximize $P = (x-2)(y-3)$.  Eliminate $y$ to
get $P = (x-2)(180/x-3)$.  Solve $dP/dx = 0$ to get $x = 2 \sqrt{30}$, so $y
= 180 / (2 \sqrt{30})$.  It's the only critical point of a positive function,
and at the endpoints of the interval we have $P = 0$, so it must be a
maximum.

\problem{4.7, 40} Using high school geometry as explained in class we see
that $AB = 4 \cos \theta$ and $BC = 4 \theta$.  The total time is $T(\theta)
= AB/2 + BC/4 = 2 \cos \theta + \theta$.  Solve $dT/d\theta = 0$ to get $\sin
\theta = 1/2$, i.e., $\theta = \pi/6$.  Comparing numerical values: $T(0) = 2$,
$T(\pi/6) \approx 2.26$, $T(\pi / 2) \approx 1.57$, we see that $\theta =
\pi/2$ gives the smallest value, so she should walk all the way.

\problem{4.7, 48} As suggested in class, we avoid trigonometry by letting $x
= RT$ be our parameter, so that $QR = QT - x$.  We must minimize the length
of the rope $L = \sqrt { x^2 + ST^2 } + \sqrt { (QT - x)^2 + PQ^2 } $.
Compute $dL/dt = x / \sqrt { x^2 + ST^2 } - (QT - x) / \sqrt { (QT - x)^2 +
  PQ^2 }$.  Setting $dL/dt = 0$ we see that $x / \sqrt { x^2 + ST^2 } = (QT -
x) / \sqrt { (QT - x)^2 + PQ^2 }$, i.e., that $RT / RS = RQ / RP$.  By high
school geometry (side-angle-side) , the two triangles are similar, and thus
$\theta_1 = \theta_2$.  Alternatively, we see that $\cos \theta_1 = \cos
\theta_2$, and thus $\theta_1 = \theta_2$, because both  $\theta_1$ and
$\theta_2$ lie in the interval $[0,\pi/2]$, on which $\cos$ is decreasing,
and thus one-to-one.

\problem{4.7, 50} As explained in class, we want to find the minimum length
the blue line can have.  There are many ways to do this problem.  Following
the book we may use the angle $\theta$ as a parameter; we find that the
length of the pipe is $L = 9/\sin(\theta) + 6/\cos(\theta)$.
Differentiating, we get $dL/d\theta = - 9 \cos(\theta) / (\sin(\theta))^2 + 6
\sin(\theta) / (\cos(\theta))^2$.  Setting $dL/d\theta = 0$ we get $ 9
\cos(\theta) / (\sin(\theta))^2 = 6 \sin(\theta) / (\cos(\theta))^2 $;
multiplying out we get $6 (\sin(\theta))^3 = 9 (\cos(\theta))^3$, or $(\tan
(\theta))^3 = 9/6$, so that $\theta = \tan^{-1}(\root 3 \of {9/6}) \approx
0.852771$.  Substituting that in the original formula for $L$ we get $L
\approx 21.0704 \textrm{ feet}$.  To see that it's an absolute minimum,
observe that it is the only critical point, and that $L$ approaches $\infty$
as $\theta$ approaches either $0$ or $\pi/2$.

One can also this one without trigonometry.  Let $A$ be the point where the
pipe touches the left wall, $B$ the point where it touches the inside corner,
and $C$ the point where it touches the upper wall.  Drop a perpendicular from
$A$ to the opposite wall to meet it at a point $D$.  Drop a perpendicular
from $C$ to the opposite wall to meet it at a point $E$.  Let $x = BE$ be our
main parameter.  Let $w = AB$, $z = BC$, and observe that $AD = 9$ and $CE =
6$.  The two triangles $ADB$ and $BEC$ are similar, so $w/9 = z/x$, from
which it follows that $w =9z/x$; this seems simpler that using the theorem of
Pythagoras to get a formula for $w$.  We want to minimize the length $L = z +
w = z + 9z/x = z ( 1+ 9/x)$.  Notice that $z = \sqrt{x^2 + 6^2}$, and that
can be used to write $L$ as a function of $x$, but let's keep the symbol $z$
around for a while.  We want to compute $dL/dx$, but first let's compute
$dz/dx = (d/dx) \sqrt{x^2 + 6^2} = (1/2) (x^2 + 6^2)^{-1/2} \cdot 2x = x /
\sqrt{x^2 + 6^2} = x/z$.  Now by the product rule $dL/dx = (d/dx) (z ( 1+
9/x)) = dz/dx \cdot ( 1+ 9/x) + z \cdot (d/dx)( 1+ 9/x) = (x/z) (1 + 9/x) + z
\cdot (-9/x^2) = (x+9)/z - (9z) / x^2 = ((x+9) x^2 - 9 (z^2) ) / (z x^2) =
(x^3 + 9 x^2 - 9 (x^2 + 6^2) ) / (z x^2) = (x^3 - 9 \cdot 6^2 ) / (z x^2)$.
Setting $dL/dt = 0$ we get $x^3 = 9 \cdot 6^2$, so grabbing a calculator, we
see that $x = \root 3 \of {54 \cdot 6} \approx 6.86829$, and $z = \sqrt{x^2 +
  6^2} \approx 9.11994$, and thus gives $L = z ( 1+ 9/x) \approx 21.0704
\textrm{ feet}$.

\problem{4.7, 52} We want to maximize the area $A = (10 + 10 \cos(\theta))(10
\sin( \theta)) = 100 (1 + \cos(\theta)) (\sin( \theta))$.  Compute $dA/dt =
100 ( ( - \sin(\theta) )( \sin (\theta)) + (1 + \cos(\theta)) (\cos(
\theta))) = 100( - ( \sin(\theta) )^2 + \cos(\theta) + (\cos( \theta))^2)$.
Writing $( \sin(\theta) )^2 = 1 - ( \cos(\theta) )^2$ we get $dA/dt = 100( -
1 + \cos(\theta) + 2 (\cos( \theta))^2)$.  Setting $dA/dt = 0$ we get $ - 1 +
\cos(\theta) + 2 (\cos( \theta))^2 = 0$.  Setting $x = \cos(\theta)$ we get
$2 x^2 + x - 1 = 0$.  Using the quadratic formula we find $x = 1/2$ or $x =
-1$.  Since $0 \le \theta \le \pi / 2$, we can rule out the solution where
$\cos(\theta) = x = -1$.  That leaves $\cos(\theta) = x = 1/2$, so $\theta =
\pi / 3$.  Since the area is $0$ when $\theta = 0$, it must be increasing
when $0 < \theta < \pi /3 $.  Since the quadratic polynomial had two roots,
$dA/dt$ changes sign as $x$ crosses $1/2$, and so $A$ is decreasing when
$\pi/3 < \theta < \pi/2$.  Thus $A$ has a maximum at $\theta = \pi/3$.

\problem{4.7, 54}  We want to maximize $ \theta = \tan^{-1}((h+d)/x) -
\tan^{-1}(d/x) $ for $x \ge 0$.  Compute $d\theta/dx = (h^2 d + h d^2 - h
x^2) / ((x^2 + (h+d)^2)(x^2+d^2)) $ and try to solve $d\theta/dx = 0$,
getting $x = \sqrt{d ( h+d )} $.  Since $\theta$ approaches $0$ as $x$
approaches $0$ or $\infty$, this single critical point must be an absolute
maximum.

\problem{4.7, 56}  (a) Let $D$ be the marked point at the right end of the main
branch, so that $a = AD$.  Thus $\sin(\theta) = b/BC$ and $\cos(\theta) = (a
- AB)/BC$.  Solving for the lengths $AB$ and $BC$, plugging them into the
formula for $R$, and adding the two results, gives the desired formula for
the total resistance. 

 (b)  Compute $dR/d\theta = b C (1 / \sin(\theta) ) ( 1
/ ( r_1^4 \sin(\theta) )  - ( 1 / ( r_2^4 \tan ( \theta ) )))$.  Setting
$dR/d\theta=0$ leads to $\cos(\theta) = r_2^4 / r_1^4 $.  

(c) In this case  $\cos(\theta) = 2^4 / 3^4 $, so we use a calculator and
find that $\theta \approx 1.37196$, which is about 79 degrees.

\end{document}