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MATH 180\hfill FALL 1995

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%10:00 SECTION\hfill FONG

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\centerline{ASSIGNMENT \#34}

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5.6, \#1.  The fencing forms 3 sides of a rectangle with
the wall forming the 4th side.  Let the 3 sides have lengths
$x$, $y$, $x$, so that $2x+y=100$.  The rectangle then has area
$A=xy=x(100-2x)=100x-2x^2$.   The problem:  Maximize $A$
for $0\le x<\infty$.  Since $A$ is a parabola opening downwards,
the maximum occurs at the vertex.  $A'=100-4x$ is zero at
$x=25$.  The largest area that can be enclosed is $A(25)=1250$
square feet.

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5.6, \#3.  Let the bottom of the box be a square of side length $x$,
and let $h$ be the height of the box.  The box has volume
$V=hx^2$ and surface area $S=x^2+4hx$ (one bottom area and 4 side
areas).  So $S=x^2+(4V/x)$.  The problem: Minimize $S$ for
$0< x<\infty$.  Now $S'=2x-(4V/x^2)=(2x^3-4V)/x^2$
is zero at $x=(2V)^{1/3}$.  Moreover, $S'$ is negative
for $0<x<(2V)^{1/3}$ and positive
for $(2V)^{1/3}<x<\infty$.   Thus $S$ has minimal surface
area when $x=(2V)^{1/3}$ and $h=(V/4)^{1/3}$.

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5.6, \#5.  Let $q$ be the number of chairs ordered.
The revenue $R$ as a function of $q$ is
$$R(q)=\cases{
     90q  &for $0\le q\le 300$\cr
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     (90-(0.25)(q-300))q &for $300<q\le 400$\cr
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     65q  &for $400<q$\cr}$$
Here $90-(0.25)(q-300)$ is the price of a chair with a
reduction of \$0.25 for each additional chair sold over 300.
The reductions stop when $q\ge400$ with the price stabilizing
at \$65.  So the graph $R$ is the join of a line segment
of slope 90, a piece of the parabola $165q-.25q^2$, and a line
segment of slope 65.
We find the global extrema of the parabola
on $[300,400]$.  Its critical point is $q=330$.
Now $R(300)=27000$, $R(330)=27225$, $R(400)=26000$.
When $300\le q\le 400$, the minimal and maximal
revenues are \$26000 and \$27225.
When $0\le q\le 300$, the minimal and maximal
revenues are \$0 and \$27000.  When $400<q$, the minimal revenue is
\$26000; the maximal revenue can be arbitrarily large.

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5.6, \#6.
The cost $C$ of operating the boat is
the sum of two costs:  the hourly overhead \$675 and hourly cost of fuel
$kv^3$.  Here
$v$ is the velocity of the boat in units of miles/hour and
$k$ is a positive constant.  In fact, $k=0.1$ since
$F(10)=k10^3=100$ is given.
So $C(v)=675+0.1v^3$ in units of dollars/hour.  To minimize
{\it cost per mile\/}, we minimize $C(v)/v=(675/v)+0.1v^2$,
which has units of dollars/mile.
$(C(v)/v)'=-(675/v^2)+0.2v=(0.2v^3-675)/v^2$ has a zero at
$v^3=3375$, that is, at $v=15$ miles/hour.  This is a global minimum
since $C(v)/v)'$ is negative for $0<v<15$ and positive for
$v>15$.


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