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{SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 10 "Calculus I" }}{PARA 256 "
" 0 "" {TEXT -1 34 "Lesson 10:  Max and Min Problems 3" }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 97 "In each of t
he following problems, we first get a numerical solution and then a pr
ecise solution." }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 0 "" }{TEXT 256 9 "Example 1" }{TEXT -1 91 "\nA steel \+
pipe is being carried down a hallway 9 feet wide.  At the end of the h
all there is" }}{PARA 0 "" 0 "" {TEXT -1 97 "a right-angled turn into \+
a narrower hallway 6 feet wide.  What is the length of the longest pip
e " }}{PARA 0 "" 0 "" {TEXT -1 71 "that can be carried horizontally ar
ound the corner?   See figure below." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 222 "with(plots): with(plottools):\nL1 := line([0,0],[18,
0]):\nL2 := line([0,9],[12,9]):\nL3 := line([12,9],[12,15]):\nL4 := li
ne([18,0],[18,15]):\nbar := rectangle([5,6], [15,7], color=blue):\ndis
play([L1,L2,L3,L4, bar], axes=none);" }}{PARA 13 "" 1 "" {GLPLOT2D 
400 300 300 {PLOTDATA 2 "6(-%'CURVESG6#7$7$$\"\"!F)F(7$$\"#=F)F(-F$6#7
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G6$7&7$$\"\"&F)$\"\"'F)7$F:FG7$F:$\"\"(F)7$FEFK-%'COLOURG6&%$RGBGF(F($
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45.000000 45.000000 0 0 "Curve 1" "Curve 2" "Curve 3" "Curve 4" "Curve
 5" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 46 "We solve  this problem by \+
locating a minimum! " }}{PARA 0 "" 0 "" {TEXT -1 43 "Let A  and B  den
ote the ends of the pipe. " }}{PARA 0 "" 0 "" {TEXT -1 76 "Let L denot
e the length of the pipe, i.e. length of the segment from A to B " }}
{PARA 0 "" 0 "" {TEXT -1 34 "and passing touching the corner.  " }}
{PARA 0 "" 0 "" {TEXT -1 6 "Let   " }{XPPEDIT 18 0 "theta;" "6#%&theta
G" }{TEXT -1 32 "   (theta) be the angle shown.  " }}{PARA 0 "" 0 "" 
{TEXT -1 5 " As  " }{XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 20 "
  goes to 0 or to   " }{XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 
28 " /2, we have that L goes to " }{XPPEDIT 18 0 "infinity" "6#%)infin
ityG" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 75 "There will be an
 angle that makes L a minimum.  A pipe of this length will " }}{PARA 
0 "" 0 "" {TEXT -1 82 "just fit around the corner.  This will be the l
ength of the longest pipe that can " }}{PARA 0 "" 0 "" {TEXT -1 36 "fi
t.  In the equations below,  x =  " }{XPPEDIT 18 0 "theta;" "6#%&theta
G" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "L:= x -> 9 * csc(x) + \+
 6 * sec(x); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"LGR6#%\"xG6\"6$%)
operatorG%&arrowGF(,&-%$cscG6#9$\"\"**&\"\"'\"\"\"-%$secGF/F4F4F(F(F(
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 64 "First we do some plotting to \+
see what the graph of L looks like." }}{PARA 0 "" 0 "" {TEXT -1 68 "Re
member that csc(x) and sec(x) are not defined for all x. From the " }}
{PARA 0 "" 0 "" {TEXT -1 20 "figure we see that  " }{XPPEDIT 18 0 "the
ta;" "6#%&thetaG" }{TEXT -1 19 "  is between 0 and " }{XPPEDIT 18 0 "P
i/2" "6#*&%#PiG\"\"\"\"\"#!\"\"" }{TEXT -1 3 ".  " }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "plot(L(x), \+
x = 0.1..Pi/2-0.1, color = black);" }{TEXT -1 0 "" }}{PARA 13 "" 1 "" 
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{TEXT -1 54 "We first locate numerically  the minimum shown above. " }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "fsolve(D(L)(x) = 0, x = 0.
.3);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+w(3x_)!#5" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "D(D(L))(.8527708776);" }{TEXT -1 0 
"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+VT8@j!\")" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 15 "L(.8527708776);" }{TEXT -1 0 "" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#$\"+9Z/2@!\")" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 45 "Numerically we have a critical point at x =  " }{XPPEDIT 18 0 "
.8527708776;" "6#$\"+w(3x_)!#5" }{TEXT -1 3 " . " }}{PARA 0 "" 0 "" 
{TEXT -1 80 "Since the second derivative at the critical point is posi
tive we have a minimum " }}{PARA 0 "" 0 "" {TEXT -1 10 "when x =  " }
{XPPEDIT 18 0 ".8527708776;" "6#$\"+w(3x_)!#5" }{TEXT -1 32 ".   Numer
ically the minimum is  " }{XPPEDIT 18 0 "21.07044714;" "6#$\"+9Z/2@!\"
)" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 80 "Hence the longest p
ipe that can fit around the corner is approximately          " }}
{PARA 0 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "21.07044714;" "6#$\"+9Z
/2@!\")" }{TEXT -1 56 " feet.      Can we get a precise solution; not \+
simply a " }}{PARA 0 "" 0 "" {TEXT -1 32 "numerical estimate?         \+
    " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "D(L);" }{TEXT -1 0 "
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#R6#%\"xG6\"6$%)operatorG%&arrowGF&
,&*&-%$cscG6#9$\"\"\"-%$cotGF.F0!\"**(\"\"'F0-%$secGF.F0-%$tanGF.F0F0F
&F&F&" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 39 "From the above we see th
at L' = 0 when " }}{PARA 0 "" 0 "" {TEXT -1 3 "   " }}{PARA 0 "" 0 "" 
{TEXT -1 63 "                    6 sec(x) tan(x) = 9 csc(x) cot(x)    \+
OR    " }{XPPEDIT 18 0 "tan(x) =( 9/6) ^(1/3)" "6#/-%$tanG6#%\"xG)*&\"
\"*\"\"\"\"\"'!\"\"*&F+F+\"\"$F-" }{TEXT -1 5 "  =  " }{XPPEDIT 18 0 "
(3/2)^ (1/3)" "6#)*&\"\"$\"\"\"\"\"#!\"\"*&F&F&F%F(" }{TEXT -1 2 ". " 
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 59 "Thus th
e critical point is the unique value of    x   in  (" }{XPPEDIT 18 0 "
0,Pi/2" "6$\"\"!*&%#PiG\"\"\"\"\"#!\"\"" }{TEXT -1 8 ") where " }
{XPPEDIT 18 0 "tan(x) = (3/2)^(1/3)" "6#/-%$tanG6#%\"xG)*&\"\"$\"\"\"
\"\"#!\"\"*&F+F+F*F-" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 10 " \+
         " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 28 "Using the above diag
ram and " }{XPPEDIT 18 0 "tan(x) = (3/2)^(1/3)" "6#/-%$tanG6#%\"xG)*&
\"\"$\"\"\"\"\"#!\"\"*&F+F+F*F-" }{TEXT -1 13 "    we have: " }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "csc (x)^2  = \+
 1 + (3/2)^ (-2/3)" "6#/*$-%$cscG6#%\"xG\"\"#,&\"\"\"F+)*&\"\"$F+F)!\"
\",$*&F)F+F.F/F/F+" }{TEXT -1 8 "   and  " }{XPPEDIT 18 0 "sec(x)^2  =
  1 + (3/2) ^(2/3)" "6#/*$-%$secG6#%\"xG\"\"#,&\"\"\"F+)*&\"\"$F+F)!\"
\"*&F)F+F.F/F+" }{TEXT -1 4 " .  " }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 68 "evalf( 9 * sqrt(1 + (3/2)^ (-2/3)) + 6 * sqrt(1 + (3/
2) ^(2/3))  ); " }{TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+
9Z/2@!\")" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 69 "We see we have the s
ame estimate for the longest length of the pipe. " }}{PARA 0 "" 0 "" 
{TEXT -1 60 "We need to check that we have a minimum for the function \+
L. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 78 "Fr
om the first derivative we see that L'(x) > 0    iff    tan(x) > (3/2)
^(1/3)" }}{PARA 0 "" 0 "" {TEXT -1 24 "and  L'(x)  <  0  iff   " }
{XPPEDIT 18 0 "tan(x)  <  (3/2)^(1/3)" "6#2-%$tanG6#%\"xG)*&\"\"$\"\"
\"\"\"#!\"\"*&F+F+F*F-" }{TEXT -1 3 ".  " }}{PARA 0 "" 0 "" {TEXT -1 
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 " }}{PARA 0 "" 0 "" {TEXT -1 59 "From the numerical estimates above w
e know that c is about " }}{PARA 0 "" 0 "" {TEXT -1 5 "     " }
{XPPEDIT 18 0 ".8527708776;" "6#$\"+w(3x_)!#5" }}{PARA 0 "" 0 "" 
{TEXT -1 24 "Recall graph of tan(x). " }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 53 "aa:= plot(tan(x), x = 0 .. Pi/2 - .1, color = black):
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "bb:= plot([t,(3/2)^(1/3
), t = 0..Pi/2 - .1], color = magenta):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 45 "cc:= textplot([1.1,6,`tan(x)`], color = red):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "dd:= textplot([.5,2.5,`y = (
3/2)^(1/3)`], color = red):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
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F,7$$\"3/*)p\\3s%pD\"F,$\"35$*fmUX$43$F,7$$\"3AbUR&G$)yG\"F,$\"3<_jF#f
V)RMF,7$$\"3!p=mrK5jJ\"F,$\"3nav9f?HWQF,7$$\"3gQo-D4*)[8F,$\"3n1)f&QB@
KWF,7$$\"3i&=)ei6Yj8F,$\"3=5YQMzz`ZF,7$$\"3SK&\\,SJ!y8F,$\"3itg,&*4DB^
F,7$$\"3OAkL7jc$R\"F,$\"3cU*o\"*Q%=$e&F,7$$\"3K7L_C7549F,$\"3DQ)QAn'[I
hF,7$$\"3n'H'p<,(RU\"F,$\"3G%yC#y0whnF,7$$\"3,\"Gp3,R)Q9F,$\"3ZY7Kz6<M
vF,7$$\"3ug%R!R$GoW\"F,$\"3H=4>nNBD!)F,7$$\"3YS'4sm<[X\"F,$\"3O50MA;e$
e)F,7$$\"3=?)z`*p!GY\"F,$\"3))HYf[&\\TA*F,7$$\"3!****\\NK'zq9F,$\"3Uwd
o(4Wm'**F,-Fjt6&F\\uF)F)F)-%+AXESLABELSG6%Q!6\"F\\hl%(DEFAULTG-%%VIEWG
6$;F($\"+Fjzq9!\"*F^hl" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 
45.000000 0 0 "Curve 1" "Curve 2" "Curve 3" "Curve 4" }}}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 78 "From the above plot we see that L'(x) < 0
 for  x < c and L'(x) > 0 for x > c; " }}{PARA 0 "" 0 "" {TEXT -1 35 "
hence L has a minimum when x = c.  " }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT 257 9 "Example 2" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 89 "Find the dimensions of th
e rectangle of largest area that has its base on the x-axis and " }}
{PARA 0 "" 0 "" {TEXT -1 66 "its other two vertices above the x-axis a
nd lying on the parabola " }{XPPEDIT 18 0 "y = 8 - x^2" "6#/%\"yG,&\"
\")\"\"\"*$%\"xG\"\"#!\"\"" }{TEXT -1 14 ".  See figure." }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "f :=
 x->8-x^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fGR6#%\"xG6\"6$%)ope
ratorG%&arrowGF(,&\"\")\"\"\"*$)9$\"\"#F.!\"\"F(F(F(" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 217 "a := -2: b := 2:\nc := -1: d := 1:\nrec1
 := rectangle([a, 0],[b, f(b)], color=pink, thickness=2):\nrec2 := rec
tangle([c, 0],[d, f(d)], color=yellow):\ngraph := plot(f(x), x=-4..4, \+
thickness=3):\ndisplay(rec1, rec2, graph);" }}{PARA 13 "" 1 "" 
{GLPLOT2D 400 300 300 {PLOTDATA 2 "6'-%)POLYGONSG6%7&7$$!\"#\"\"!$F*F*
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FJ-F46&F6$\"*++++\"!\")FOF+-%'CURVESG6%7S7$$!\"%F*$FQF*7$$!3ommmmFiDQ!
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2:Fgn$\"3<q!4[Yfxs&Fgn7$$!3XLLLtK5F8Fgn$\"3n^z!>!pzQiFgn7$$!3eLLL$HsV<
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$$!3sLLL$)[`P<F^q$\"3g)*>HD(4)pzFgn7$$!3gSnmmmr[R!#?$\"3i'oOwS%)***zFg
n7$$\"3yELL$=2Vs\"F^q$\"37^]PZwEqzFgn7$$\"3)e*****\\`pfKF^q$\"3%Q)=DiQ
u$*yFgn7$$\"36HLLLm&z\"\\F^q$\"3=Fl_Dq8exFgn7$$\"3>(******z-6j'F^q$\"3
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#y'G**F^q$\"3eq(z?[8U,(Fgn7$$\"3G******H%=H<\"Fgn$\"3APjccBECmFgn7$$\"
35mmm1>qM8Fgn$\"3,m.M?3d=iFgn7$$\"3%)*******HSu]\"Fgn$\"3S5f$>uBws&Fgn
7$$\"3'HLL$ep'Rm\"Fgn$\"3.;\\dhR@J_Fgn7$$\"3')******R>4N=Fgn$\"3uOqurv
VKYFgn7$$\"3#emm;@2h*>Fgn$\"3+GB`**fb:SFgn7$$\"3]*****\\c9W;#Fgn$\"3'H
'e\"3f4`J$Fgn7$$\"3Lmmmmd'*GBFgn$\"3qYZpd%=fd#Fgn7$$\"3j*****\\iN7]#Fg
n$\"3Th3B[.#Qu\"Fgn7$$\"3aLLLt>:nEFgn$\"39_5W6N+j))F^q7$$\"35LLL.a#o$G
Fgn$!3'o$Gt**o$yv%Fhp7$$\"3ammm^Q40IFgn$!3x,[Kd!*eI5Fgn7$$\"3y******z]
rfJFgn$!3Q/%znQ*z$)>Fgn7$$\"3gmmmc%GpL$Fgn$!3y!y6\\_\"4NJFgn7$$\"3/LLL
8-V&\\$Fgn$!385NGwB.=UFgn7$$\"3=+++XhUkOFgn$!3Ij&f@(*=!GaFgn7$$\"3=+++
:o<EQFgn$!3zWNk>!H'RmFgn7$F0FY-F46&F6$\"#5FFF+F+-F?6#\"\"$-%+AXESLABEL
SG6$Q\"x6\"Q!6\"-%%VIEWG6$;FWF0%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 
1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" "Curve 3" }}}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 47 "Let AA denote the area of the rect
angle. Then  " }{XPPEDIT 18 0 "A A  =  2*x *( 8 - x^2)" "6#/*&%\"AG\"
\"\"F%F&*(\"\"#F&%\"xGF&,&\"\")F&*$F)F(!\"\"F&" }{TEXT -1 2 ". " }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "AA:= x -> 2 * x * ( 8 - x^2)
;" }{TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#AAGR6#%\"xG6\"6
$%)operatorG%&arrowGF(,$*&9$\"\"\",&\"\")F/*$)F.\"\"#F/!\"\"F/F4F(F(F(
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "D(AA);" }{TEXT -1 0 "" }
}{PARA 11 "" 1 "" {XPPMATH 20 "6#R6#%\"xG6\"6$%)operatorG%&arrowGF&,&
\"#;\"\"\"*&\"\"'F,)9$\"\"#F,!\"\"F&F&F&" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 26 "solve( 16 - 6 * x^2= 0,x);" }{TEXT -1 0 "" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6$,$*$-%%sqrtG6#\"\"'\"\"\"#!\"#\"\"$,$F$#\"\"
#F," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalf(%);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6$$!+iJ*Hj\"!\"*$\"+iJ*Hj\"F%" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "D(D(AA))(1.632993162);" }{TEXT -1 
0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$!+%z\"ff>!\")" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 47 "Numerically AA has a critical point when \+
x =   " }{XPPEDIT 18 0 "1.632993162;" "6#$\"+iJ*Hj\"!\"*" }{TEXT -1 
16 " ;   since AA''(" }{XPPEDIT 18 0 "1.632993162;" "6#$\"+iJ*Hj\"!\"*
" }{TEXT -1 7 ") < 0, " }}{PARA 0 "" 0 "" {TEXT -1 34 "we know that AA
 is a max when x = " }{XPPEDIT 18 0 "1.632993162;" "6#$\"+iJ*Hj\"!\"*
" }{TEXT -1 9 ".  Since " }{XPPEDIT 18 0 "y = 8 - x^2" "6#/%\"yG,&\"\"
)\"\"\"*$%\"xG\"\"#!\"\"" }{TEXT -1 9 " we have " }}{PARA 0 "" 0 "" 
{TEXT -1 7 "y = 8-(" }{XPPEDIT 18 0 "1.632993162;" "6#$\"+iJ*Hj\"!\"*
" }{TEXT -1 34 ")^2 .     Thus xy is numerically  " }{XPPEDIT 18 0 "5.
333333333;" "6#$\"+LLLL`!\"*" }{TEXT -1 5 "     " }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 27 "evalf(8 - (1.632993162)^2);" }{TEXT -1 0 "" }
}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+LLLL`!\"*" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 33 "Precisely, AA has a maximum when " }{XPPEDIT 18 0 "x
 = (2/3)* sqrt(6)" "6#/%\"xG*(\"\"#\"\"\"\"\"$!\"\"-%%sqrtG6#\"\"'F'" 
}{TEXT -1 6 "  and " }{XPPEDIT 18 0 "y = 8 - (4/9) *6" "6#/%\"yG,&\"\"
)\"\"\"*(\"\"%F'\"\"*!\"\"\"\"'F'F+" }{TEXT -1 2 ". " }}}}{MARK "3 0 0
" 110 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }