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MATH 180\hfill FALL 1995

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%10:00 SECTION\hfill FONG

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\centerline{ASSIGNMENT \#35}

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5.6, \#8.
The inscribed rectangle has corners at
$(9,0)$, $(x,0)$, $(x,\sqrt x)$, and $(9,\sqrt x)$.
Here $0\le x\le 9$.  In particular, the rectangle has base length
$9-x$ and height $\sqrt x$.  So the
area $A=(9-x)\sqrt x$ and the perimeter $P=2(9-x)+2\sqrt x$.
Now
$$A'={9\over2\sqrt x}-{3\sqrt x\over 2}={3(3-x)\over 2\sqrt x},
    \quad P'=-2+{1\over\sqrt x}.$$
$A$ has a critical point at $x=3$, $P$ has a critical point
at $x=1/4$, and both points are in $[0,9]$.
Since $A(0)=0$, $A(3)=6\sqrt3=10.39$, $A(9)=0$ and
$P(0)=18$, $P(1/4)=18.5$, $P(9)=6$,
$A$ and $P$ are maximal respectively at $x=3$ and  $x=1/4$.

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5.6, \#10.  The distance to be minimized is
$\sqrt{(x-a)^2+y^2}$, where $a=2$ or $\sqrt8$ and where
$y^2=1-(x/3)^2$.
Minimizing distance is the same as minimizing the square of the
distance.  So we minimize
$$f(x)={(x-a)^2+y^2}=(x-a)^2+1-(x/3)^2$$ on the interval $[-3, 3]$.
Now $f'(x)=2(x-a)-2(x/3))(1/3)=(2/9)(7x-8a)$.
Thus $f$ has a critical point at $x=9a/8$.  When $a=2$, the
critical point $x=9/4$ is in $[-3,3]$.  Since
$f(-3)=27$, $f(9/4)=1.625$, $f(3)=3$, $x=9/4$ is the global minimum.
When $a=\sqrt8$, the critical point $x=9/\sqrt8=3.182$ is
not in $[-3,3]$.  Since
$f(-3)=35.97$, $f(3)=2.03$, $x=3$ is the global maximum.

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5.6, \#12.
The length of the pipe is given by Pythogoras's theorem:
$L=\sqrt{x^2+1}+\sqrt{(4-x)^2+16}$.  This is to be minimized
for $0\le x\le 4$.
$$\eqalign{
   L'&={2x\over 2\sqrt{x^2+1}}+{2(4-x)(-1)\over 2\sqrt{(4-x)^2+16}}\cr
     &={x\sqrt{(4-x)^2+16}-(4-x)\sqrt{x^2+1}\over
       \sqrt{x^2+1} \sqrt{(4-x)^2+16}}.\cr}$$
$L'=0$ when $x\sqrt{(4-x)^2+16}=(4-x)\sqrt{x^2+1}$.  Squaring both sides
gives
$$x^2((4-x)^2+16)=(4-x)^2(x^2+1).$$
After simplification this
becomes $16x^2=(4-x)^2$.  Since $x$ and $4-x$ are positive, taking
square
roots gives $4x=4-x$ and $x=4/5$.  This is the global minimum since
$L(0)=6.66$, $L(4/5)=6.40$, $L(4)=8.12$.


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