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{SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 10 "Calculus I" }}{PARA 256 "
" 0 "" {TEXT -1 33 "Lesson 8:  Max and Min Problems 1" }}{PARA 0 "" 0 
"" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 115 "In the next t
hree lessons, we demonstrate the solution process for various optimiza
tion problems using derivatives." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT 256 9 "Example 1" }}{PARA 0 "" 0 "" 
{TEXT -1 74 "A right circular cylinder is inscribed in a sphere of rad
ius r.  Find the " }}{PARA 0 "" 0 "" {TEXT -1 51 "largest possible sur
face area of such a cylinder. \n" }}{PARA 0 "" 0 "" {TEXT -1 34 "Here'
s a diagram of the situation." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 245 "w
ith(plottools):\nr := 1: h := 1.2*r:\ncyl := cylinder([0,0,-h/2], sqrt
(r^2-h^2/4), h, shading=xyz, style=patchnogrid):\nball := sphere([0,0,
0], r, color=green):\nplots[display]([cyl,ball], scaling=constrained, \+
style=wireframe, orientation=[45,74]);" }}{PARA 13 "" 1 "" {GLPLOT3D 
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bu7&F[`uF]`u7%Fe_pF_[rFb_pF[cu7&F]`uF_`u7%F\\`pFc[rFg^pF]cu7&F_`uFa`u7
%Fc`pFg[rFawF_cu7&Fa`uFc`u7%Fj`pF[\\rFc]pFacu7&Fc`uFe`u7%FaapFjrFj\\pF
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[au7%FfapFa\\rFabpFicu7&F[auF]au7%FaapFjrFfbpF[du7&F]auF_au7%Fj`pF[\\r
F[cpF]du7&F_auFaau7%Fc`pFg[rFcqF_du7&FaauFcau7%F\\`pFc[rFccpFadu7&Fcau
Feau7%Fe_pF_[rFhcpFcdu7&FeauFgau7%F\\_pF[[rF]dpFedu7&FgauFiau7%Fa^pFgj
qFbdpFgdu7&FiauF[bu7%Fh]pFcjqFgdpFidu7&F[buF]bu7%FjuF_jqF\\epF[eu7&F]b
uF_bu7%Fc\\pF[jqFaepF]eu7&F_buFabu7%F[fpF]^rFhepF_eu7&Fj[pFcbu7%F^\\p$
!+m_$[V)F_fpF`\\pFj[p7&FcbuFebu7%Fj\\p$!+;[`s;FAF\\]pFceu7&FebuFgbu7%F
c]p$!+s&oHZ#FAFe]pFgeu7&FgbuFibu7%FawFcbsF^^pF[fu7&FibuF[cu7%Fg^p$!+&*
[#R$RFAFi^pF_fu7&F[cuF]cu7%Fb_p$!+'4]%pXFAFb_pFafu7&F]cuF_cu7%Fi^p$!+Q
2zE^FAFg^pFefu7&F_cuFacu7%F^^p$!+q0T'f&FAFawFifu7&FacuFccu7%Fe]p$!+lUF
qfFAFc]pF]gu7&FccuFecu7%F\\]p$!+7\\)>C'FAFj\\pFagu7&FecuFgcu7%F`\\p$!+
nM*oS'FAF^\\pFegu7&FgcuFicu7%F[\\p$!+)Hy@Y'FAF\\bpFigu7&FicuF[du7%F`\\
pFjguFabpF]hu7&F[duF]du7%F\\]pFfguFfbpFahu7&F]duF_du7%Fe]pFbguF[cpFchu
7&F_duFadu7%F^^pF^guFcqFehu7&FaduFcdu7%Fi^pFjfuFccpFghu7&FcduFedu7%Fb_
pFffuFhcpFihu7&FeduFgdu7%Fg^pFbfuF]dpF[iu7&FgduFidu7%FawFcbsFbdpF]iu7&
FiduF[eu7%Fc]pF\\fuFgdpF_iu7&F[euF]eu7%Fj\\pFheuF\\epFaiu7&F]euF_eu7%F
^\\pFdeuFaepFciu7&F_euFaeu7%Ffep$!+=u)z3#FgbsFhepFeiu-%&COLORG6&%$RGBG
F+$\"*++++\"!\")F+-Fd[p6#%%LINEG-%(SCALINGG6#%,CONSTRAINEDG-%+PROJECTI
ONG6%$\"#XF,$\"#uF,F\\\\p" 1 6 0 1 10 0 2 1 1 1 1 1.000000 74.000000 
45.000000 1 0 "Curve 1" "Curve 2" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
30 "Let     r  = radius of sphere " }}{PARA 0 "" 0 "" {TEXT -1 41 "   \+
       a = radius of base of cylinder " }}{PARA 0 "" 0 "" {TEXT -1 46 
"          b =  1/2 the height of the cylinder " }}{PARA 0 "" 0 "" 
{TEXT -1 9 "We have: " }}{PARA 0 "" 0 "" {TEXT -1 42 "          SA = s
urface area of cylinder = " }{XPPEDIT 18 0 "2*Pi*a^2 + 4*Pi*a*b" "6#,&
*(\"\"#\"\"\"%#PiGF&%\"aGF%F&**\"\"%F&F'F&F(F&%\"bGF&F&" }{TEXT -1 1 "
 " }}{PARA 0 "" 0 "" {TEXT -1 10 "          " }{XPPEDIT 18 0 "a^2 +b^2
 = r^2" "6#/,&*$%\"aG\"\"#\"\"\"*$%\"bGF'F(*$%\"rGF'" }{TEXT -1 23 "  \+
   (draw a picture) ." }}{PARA 0 "" 0 "" {TEXT -1 6 "Using " }}{PARA 
0 "" 0 "" {TEXT -1 9 "         " }{XPPEDIT 18 0 "a^2 = r^2  -  b^2" "6
#/*$%\"aG\"\"#,&*$%\"rGF&\"\"\"*$%\"bGF&!\"\"" }{TEXT -1 1 " " }}
{PARA 0 "" 0 "" {TEXT -1 11 "we obtain: " }}{PARA 0 "" 0 "" {XPPEDIT 
18 0 "SA(b) =  2 *Pi (r^2 - b^2) + 4*Pi sqrt( r^2 - b^2) *b" "6#/-%#SA
G6#%\"bG,&*&\"\"#\"\"\"-%#PiG6#,&*$%\"rGF*F+*$F'F*!\"\"F+F+**\"\"%F+F-
F+-%%sqrtG6#,&*$F1F*F+*$F'F*F3F+F'F+F+" }{TEXT -1 2 ". " }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 38 "We want to find a ma
ximum for SA.     " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "assum
e(r > 0);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 61 "SA:= b ->2*  P
i *(r^2 - b^2) + 4 * Pi * sqrt( r^2 - b^2) * b;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%#SAGR6#%\"bG6\"6$%)operatorG%&arrowGF(,&*&%#PiG\"\"\"
,&*$)%\"rG\"\"#F/F/*$)9$F4F/!\"\"F/F4**\"\"%F/F.F/-%%sqrtG6#F0F/F7F/F/
F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "D(SA);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#R6#%\"bG6\"6$%)operatorG%&arrowGF&,(*&%#PiG
\"\"\"9$F-!\"%*&*(\"\"%F-F,F-)F.\"\"#F-F--%%sqrtG6#,&*$)%\"rGF4F-F-*$F
3F-!\"\"F=F=*(F2F-F,F-F5F-F-F&F&F&" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 63 "simplify( -4*Pi*b-4*Pi*b^2/sqrt(r^2-b^2)+4*Pi*sqrt(r^
2-b^2)  );" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&*&%#PiG\"\"\",(*&%\"
bGF'-%%sqrtG6#,&*$)%#r|irG\"\"#F'F'*$)F*F2F'!\"\"F'F'*&F2F'F4F'F'F/F5F
'F'*$-F,6#F.F'F5!\"%" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 39 "Set numer
ator to zero and solve for b. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 51 "solve( b^2 * ( r^2 - b^2 ) = (r^2 - 2* b^2 )^2, b);" }{TEXT 
-1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6&,$*&-%%sqrtG6#,&\"#]\"\"\"*&
\"#5F*-F&6#\"\"&F*F*F*%#r|irGF*#F*F,,$F$#!\"\"F,,$*&-F&6#,&F)F**&F,F*F
-F*F4F*F0F*F1,$F6F3" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalf
(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6&,$%#r|irG$\"+%33l])!#5,$F$$!+%
33l])F',$F$$\"+@6Jd_F',$F$$!+@6Jd_F'" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 96 "We need to check which if any of the above solutions to SA' = 0
 lead to a maximal surface area. " }}{PARA 0 "" 0 "" {TEXT -1 93 "As w
e squared both sides of an equation to solve, we need to check our sol
utions. Of course, " }}{PARA 0 "" 0 "" {TEXT -1 68 "we are not interes
ted in negative solutions, as b must be positive. " }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 23 " We check whether SA'( " 
}{XPPEDIT 18 0 ".8506508084*r;" "6#*&$\"+%33l])!#5\"\"\"%\"rGF'" }
{TEXT -1 12 ") and SA'(  " }{XPPEDIT 18 0 ".5257311121*r;" "6#*&$\"+@6
Jd_!#5\"\"\"%\"rGF'" }{TEXT -1 101 ") are indeed 0.                   \+
                                                                   " }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "g:= b -> b*sqrt(r^2-b^2)+2
*b^2-r^2;" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#>%\"gGR6#%\"bG6\"6$%)operatorG%&arrowGF(,(*&9$\"
\"\"-%%sqrtG6#,&*$)%\"rG\"\"#F/F/*$)F.F7F/!\"\"F/F/*&F7F/F9F/F/F4F:F(F
(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "g(.8506508084*r);" }
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#,$*$)%#r|irG\"\"#\"\"\"$\"+:>FW*)!#5" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 17 "g(.5257311121*r);" }{TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*$)%
#r|irG\"\"#\"\"\"$!\"\"!#5" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 19 "Con
clusion:  SA'(  " }{XPPEDIT 18 0 ".8506508084*r" "6#*&$\"+%33l])!#5\"
\"\"%\"rGF'" }{TEXT -1 23 ") is NOT zero and SA'( " }{XPPEDIT 18 0 ".5
257311121*r;" "6#*&$\"+@6Jd_!#5\"\"\"%\"rGF'" }{TEXT -1 7 ") = 0. " }}
{PARA 0 "" 0 "" {TEXT -1 44 "Hence the only critical value we have is \+
   " }{XPPEDIT 18 0 ".5257311121*r;" "6#*&$\"+@6Jd_!#5\"\"\"%\"rGF'" }
{TEXT -1 3 ".  " }}{PARA 0 "" 0 "" {TEXT -1 48 "We look at SA'' to det
ermine if SA is max at    " }{XPPEDIT 18 0 ".5257311121*r;" "6#*&$\"+@
6Jd_!#5\"\"\"%\"rGF'" }{TEXT -1 22 "                      " }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "D(D(SA))(.5257311121*r);" }{TEXT 
-1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$%#PiG$!+xz1O7!\")" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 18 "We see that SA''( " }{XPPEDIT 18 
0 ".5257311121*r;" "6#*&$\"+@6Jd_!#5\"\"\"%\"rGF'" }{TEXT -1 38 ") < 0
  and therefore SA has a maximum " }}{PARA 0 "" 0 "" {TEXT -1 10 "when
 b =  " }{XPPEDIT 18 0 ".5257311121*r;" "6#*&$\"+@6Jd_!#5\"\"\"%\"rGF'
" }{TEXT -1 13 ".            " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 18 "SA(.5257311121*r);" }{TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#,$*&%#PiG\"\"\")%#r|irG\"\"#F&$\"+yz1OK!\"*" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 51 "Largest surface area is (numerically given by):   \+
 " }{XPPEDIT 18 0 "3.236067978*Pi*r^2;" "6#*($\"+yz1OK!\"*\"\"\"%#PiGF
'%\"rG\"\"#" }{TEXT -1 7 ".      " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT 257 10 "Example 2\n" }{TEXT -1 52 "Find t
he minimum distance from a point on the curve " }{XPPEDIT 18 0 "y = x^
2 + 1" "6#/%\"yG,&*$%\"xG\"\"#\"\"\"F)F)" }{TEXT -1 22 " to the point \+
(1,-1). " }}{PARA 0 "" 0 "" {TEXT -1 68 "Is there a point where the di
stance is a maximum?  If yes, find it. " }}{PARA 0 "" 0 "" {TEXT -1 0 
"" }}{PARA 0 "" 0 "" {TEXT -1 33 "Here's a diagram of the situation" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}{PARA 7 "
" 1 "" {TEXT -1 50 "Warning, the name changecoords has been redefined
\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "f2:= x -> x^2 + 1;" }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#f2
GR6#%\"xG6\"6$%)operatorG%&arrowGF(,&*$)9$\"\"#\"\"\"F1F1F1F(F(F(" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "a2:= plot(f2(x), x = -2..3, \+
color = black):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "b2:= lin
e([1,-1], [0, f2(0)], color=red, linestyle=1):" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 54 "c2:= line([1,-1], [2, f2(2)], color=red, lines
tyle=1):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "d2:= line([1,-1
], [-1, f2(-1)], color=red, linestyle=1):" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 23 "display([a2,b2,c2,d2]);" }}{PARA 13 "" 1 "" 
{GLPLOT2D 400 300 300 {PLOTDATA 2 "6(-%'CURVESG6$7S7$$!\"#\"\"!$\"\"&F
*7$$!3smm;HU,\"*=!#<$\"37-3\"\\\"[$fd%F07$$!3SL$3FH'='z\"F0$\"3%pI8\")
>&GEUF07$$!3gmmTgBa*o\"F0$\"3Y@Bk(Q`X&QF07$$!3amm\"H_\">#e\"F0$\"3\\g$
*=:+L.NF07$$!3ML$3_!4Nv9F0$\"3m[!*\\$Hgm<$F07$$!3km;/wfHw8F0$\"31%[oOh
!>%*GF07$$!3;+]PM.tt7F0$\"3u\\qqk*)QAEF07$$!3em;/,oln6F0$\"3Oe)=]SAMO#
F07$$!3%)**\\(oWB>1\"F0$\"3n!)G-29oF@F07$$!3eJLLepjJ&*!#=$\"3weY0J5_3>
F07$$!3Ulm;z/ot&)Fhn$\"3eN%)ep*z]t\"F07$$!3u)****\\P[_\\(Fhn$\"35,%H?[
(yh:F07$$!3A*****\\7)Q7kFhn$\"3&*4kl9s=69F07$$!3e*****\\i^)o`Fhn$\"3U^
ErxcC)G\"F07$$!3vlmT50A@WFhn$\"3'**G<!3>Z&>\"F07$$!3OKLLeaR%H$Fhn$\"3y
s)eVTI&36F07$$!3kJLLLo#)RBFhn$\"3'o')*4'*yua5F07$$!3f***\\PfO%H7Fhn$\"
37;0QV^6:5F07$$!3MSLLL3`lC!#>$\"3\"=,HU)yg+5F07$$\"3+L+]i!f#=$)Ffq$\"3
J'3$QV$>p+\"F07$$\"3+-+v=xpe=Fhn$\"3lko4svaM5F07$$\"3<rm;H28IHFhn$\"3;
o+!4mce3\"F07$$\"3`nm\"zpSS\"RFhn$\"3')z%\\e9(>`6F07$$\"3)GLL3_?`(\\Fh
n$\"3c_-&G9QvC\"F07$$\"3!fL$e*)>pxgFhn$\"3WaC?*R$Qp8F07$$\"3w++v$f4t.(
Fhn$\"35eG=jsB&\\\"F07$$\"3OPL$e*Gst!)Fhn$\"3YMq)R,]=l\"F07$$\"3Y+++]#
RW9*Fhn$\"3p0%p>p2i$=F07$$\"3:++DJE>>5F0$\"3z)H%f>OvQ?F07$$\"3F+]i!RU0
7\"F0$\"3s!f(=\\_hbAF07$$\"3+++v=S2L7F0$\"3ZvirN:Z?DF07$$\"3Jmmm\"p)=M
8F0$\"3[57(\\Yf+y#F07$$\"3B++](=]@W\"F0$\"3%fG1L;(zzIF07$$\"35L$e*[$z*
R:F0$\"3UJ\"=_RO:P$F07$$\"3e++]iC$pk\"F0$\"3u:j.OlQ7PF07$$\"3[m;H2qcZ<
F0$\"3M^V(\\W!*R0%F07$$\"3O+]7.\"fF&=F0$\"3vbD@%H;FV%F07$$\"3Ymm;/Ogb>
F0$\"3Bc'HmX&QC[F07$$\"3w**\\ilAFj?F0$\"3u>t4UC4d_F07$$\"3yLLL$)*pp;#F
0$\"3wowm3*edp&F07$$\"3)RL$3xe,tAF0$\"39=HZx6gmhF07$$\"3Cn;HdO=yBF0$\"
3?q\"4y]ddl'F07$$\"3a+++D>#[Z#F0$\"3c32YgNuCrF07$$\"3SnmT&G!e&e#F0$\"3
iN`L7aA&o(F07$$\"3#RLLL)Qk%o#F0$\"3=#3>.y7t?)F07$$\"37+]iSjE!z#F0$\"3)
z#[i^ie&y)F07$$\"3a+]P40O\"*GF0$\"3aiK<&fl*f$*F07$$\"\"$F*$\"#5F*-%'CO
LOURG6&%$RGBGF*F*F*-F$6%7$7$$\"\"\"F*$!\"\"F*7$$F*F*Fb[l-F[[l6&F][l$\"
*++++\"!\")Fg[lFg[l-%*LINESTYLEG6#Fc[l-F$6%7$Fa[l7$$\"\"#F*F+Fh[lF]\\l
-F$6%7$Fa[l7$Fd[lFd\\lFh[lF]\\l-%+AXESLABELSG6$Q\"x6\"Q!6\"-%%VIEWG6$;
F(Ffz%(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" "Curve 4" }}}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 94 "From the above plot we see that there is NOT going to b
e a point on the graph of  y = x^2 + 1 " }}{PARA 0 "" 0 "" {TEXT -1 
83 "whose distance to (1,-1) is a maximum.  So we search for where thi
s distance is a  " }}{PARA 0 "" 0 "" {TEXT -1 10 "minimum.  " }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 37 "We need the distance between point
s: " }}{PARA 0 "" 0 "" {TEXT -1 82 "                                  \+
                              (1,-1)    and    (" }{XPPEDIT 18 0 "x" "
6#%\"xG" }{TEXT -1 2 ", " }{XPPEDIT 18 0 "x^2 + 1" "6#,&*$%\"xG\"\"#\"
\"\"F'F'" }{TEXT -1 3 "). " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 
"" 0 "" {TEXT -1 9 "Set g2:= " }{XPPEDIT 18 0 "(x - 1)^2 + (x^2 + 1 - \+
(-1))^2 = (x-1)^2 + (x^2 + 2)^2" "6#/,&*$,&%\"xG\"\"\"F(!\"\"\"\"#F(*$
,(*$F'F*F(F(F(,$F(F)F)F*F(,&*$,&F'F(F(F)F*F(*$,&*$F'F*F(F*F(F*F(" }
{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 13 "Minimize g2. " }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "g2:= x ->  (x - 1)^2 + ( x^2
 + 2)^2;" }{TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#g2GR6#%
\"xG6\"6$%)operatorG%&arrowGF(,&*$),&9$\"\"\"F1!\"\"\"\"#F1F1*$),&*$)F
0F3F1F1F3F1F3F1F1F(F(F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 62 "Lets s
ee a plot of g2 - remember we are looking for a minimum." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "plot(g2(x), x = 0..5);" }{TEXT -1 
0 "" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6%-%'CURVES
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$\"3a+]P40O\"*[F2$\"3-l>MvWpsoF`t7$F*$\"$X(F)-%'COLOURG6&%$RGBG$\"#5!
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{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "plot(g2(x), x = 0..1);" }
{TEXT -1 0 "" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6%
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FE$\"3(o\\IKD?Y,&F27$$\"3A++D\"=lj;%FE$\"3S^#Q-+*yk]F27$$\"31++vV&R<P%
FE$\"3M#3&=&o%y<^F27$$\"3WLL$e9Ege%FE$\"3#4DK=`4'y^F27$$\"3GLLeR\"3Gy%
FE$\"3s$zq-yG&R_F27$$\"3cmm;/T1&*\\FE$\"3#fG<)>Sx5`F27$$\"3&em;zRQb@&F
E$\"3A\\mgt$y4R&F27$$\"3\\***\\(=>Y2aFE$\"3;q_*[OTgY&F27$$\"39mm;zXu9c
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_QQgFE$\"3m[z5DiP[dF27$$\"3@***\\7y%3TiFE$\"3g!\\\\48f5&eF27$$\"35****
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\"3R***\\i?=bq(FE$\"3so8;\"4'=!y'F27$$\"3\"HLL$3s?6zFE$\"3VvD.(GM)QpF2
7$$\"3a***\\7`Wl7)FE$\"3ZfI(R9lG6(F27$$\"3#pmmm'*RRL)FE$\"3!=e.7IL$)G(
F27$$\"3Qmm;a<.Y&)FE$\"3'*[=5CF$fZ(F27$$\"3=LLe9tOc()FE$\"3#o)pw4ZJqwF
27$$\"3u******\\Qk\\*)FE$\"3c#*Rw1kTcyF27$$\"3CLL$3dg6<*FE$\"3%f)G;R&H
(y!)F27$$\"3ImmmmxGp$*FE$\"3]PwfJ`\"fG)F27$$\"3A++D\"oK0e*FE$\"3v&*\\0
**=q:&)F27$$\"3A++v=5s#y*FE$\"3GzzSJvTW()F27$$\"\"\"F)$\"\"*F)-%'COLOU
RG6&%$RGBG$\"#5!\"\"F(F(-%+AXESLABELSG6$Q\"x6\"Q!6\"-%%VIEWG6$;F(Fez%(
DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Cur
ve 1" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "fsolve(D(g2)(x) = \+
0, x = 0..4);" }{TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+xV
Wp>!#5" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 29 "We have one critical va
lue:  " }{XPPEDIT 18 0 ".1969444377;" "6#$\"+xVWp>!#5" }{TEXT -1 2 "  \+
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "evalf(D(D(g2))(.1969444
377));" }{TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+MXaY5!\")
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 74 "As the second derivative is >
0 at the critical value, g2 has a minimum at " }}{PARA 0 "" 0 "" 
{TEXT -1 21 "this critical value. " }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 25 "Conclusion:  When  x =   " }{XPPEDIT 18 0 ".1969444377;" "6#$\"
+xVWp>!#5" }{TEXT -1 31 ", the distance from the point (" }{XPPEDIT 
18 0 "x,x^2+1" "6$%\"xG,&*$F#\"\"#\"\"\"F'F'" }{TEXT -1 2 ") " }}
{PARA 0 "" 0 "" {TEXT -1 41 "to the point (1,-1) is minimum.          \+
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 258 
10 "Example 3\n" }{TEXT -1 74 "A manufactureer of tin cans is to make \+
a tin can holding 80 cubic inches. " }}{PARA 0 "" 0 "" {TEXT -1 59 "a)
  What is the least amount of tin needed for such a can? " }}{PARA 0 "
" 0 "" {TEXT -1 80 "b)  What is the radius and height of the can requi
ring the least amount of tin? " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 72 
"We do part (a) first.  Let r denote the radius of the base of the can
,  " }}{PARA 0 "" 0 "" {TEXT -1 88 "h  the height of the can,  SA the \+
durface area of the can, and V the volume of the can. " }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 6 "Thus, " }{XPPEDIT 18 
0 "V = Pi  *r^2" "6#/%\"VG*&%#PiG\"\"\"*$%\"rG\"\"#F'" }{TEXT -1 4 ", \+
  " }{XPPEDIT 18 0 "h = 80" "6#/%\"hG\"#!)" }{TEXT -1 11 "    and    \+
" }{XPPEDIT 18 0 "SA =  2  * Pi*  r^2  + 2 *  Pi *  r*  h" "6#/%#SAG,&
*(\"\"#\"\"\"%#PiGF(%\"rGF'F(**F'F(F)F(F*F(%\"hGF(F(" }{TEXT -1 2 ". \+
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 7 "Using \+
 " }{XPPEDIT 18 0 "h  =  80  /  ( Pi * r^2)" "6#/%\"hG*&\"#!)\"\"\"*&%
#PiGF'*$%\"rG\"\"#F'!\"\"" }{TEXT -1 12 " we obtain: " }}{PARA 0 "" 0 
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "SA =  2 * Pi*  r^2  \+
+ 160  /  r" "6#/%#SAG,&*(\"\"#\"\"\"%#PiGF(%\"rGF'F(*&\"$g\"F(F*!\"\"
F(" }{TEXT -1 17 ".                " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 24 "We want to minimize SA. " }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "g3:= r ->  2 * Pi * r^2 + 160/r;" }
{TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#g3GR6#%\"rG6\"6$%)o
peratorG%&arrowGF(,&*&%#PiG\"\"\")9$\"\"#F/F2*&\"$g\"F/F1!\"\"F/F(F(F(
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "plot(g3(r), r = 1..5);
" }{TEXT -1 0 "" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 
"6%-%'CURVESG6$7V7$$\"\"\"\"\"!$\"3%fzrI&=$Gm\"!#:7$$\"3ALLL3VfV5!#<$
\"3&ynv?J#f,;F-7$$\"3mmmm;')=(3\"F1$\"3Uy&p7p^fa\"F-7$$\"3&****\\7z>^7
\"F1$\"3eL\"y&)H4;]\"F-7$$\"3WLL$e'40j6F1$\"33iDLVRog9F-7$$\"3<++vQ&3d
?\"F1$\"3ZLp['Gh$=9F-7$$\"3ommm6hO[7F1$\"3'pWD!zNfz8F-7$$\"3xmmm\"yYUL
\"F1$\"3uTshLE.68F-7$$\"3CLL$eF>(>9F1$\"3u;-c?ui`7F-7$$\"3kmm;>K'*)\\
\"F1$\"30WgM^2e37F-7$$\"3/++]Kd,\"e\"F1$\"3YAq*\\\"G1p6F-7$$\"3gmm;fX(
em\"F1$\"3\\K)zo`B[8\"F-7$$\"3!*****\\U7Y]<F1$\"3og<.)zol5\"F-7$$\"3QL
LLV!pu$=F1$\"3OFcc#o,H3\"F-7$$\"3ommmhb59>F1$\"31#))>8\"H5m5F-7$$\"3#*
******H,Q+?F1$\"3E;)\\`$4F^5F-7$$\"3)*******\\*3q3#F1$\"3QC=tD#=./\"F-
7$$\"3?+++q=\\q@F1$\"3'[0mW:jJ.\"F-7$$\"3mmm;fBIYAF1$\"3RH8%QIB$H5F-7$
$\"30LLLj$[kL#F1$\"31%p#R'e)zF5F-7$$\"3?LLL`Q\"GT#F1$\"3sP%[N_7*G5F-7$
$\"3o****\\s]k,DF1$\"3R!\\&)[?&zK5F-7$$\"3#HLLLvv-e#F1$\"3L6?<<@TQ5F-7
$$\"33++]sgamEF1$\"39*zXG,\"zY5F-7$$\"3!)****\\<ep[FF1$\"3a+1cT&4o0\"F
-7$$\"3;LLLe/TMGF1$\"3'H3z$=]Fp5F-7$$\"3JLL$eDBJ\"HF1$\"3ABT8MyW#3\"F-
7$$\"3immmTc-)*HF1$\"3GOY`orU)4\"F-7$$\"3Mmm;f`@'3$F1$\"3IR1jp/*o6\"F-
7$$\"3y****\\nZ)H;$F1$\"3`&R)oo3XM6F-7$$\"3YmmmJy*eC$F1$\"3M.l6Aq\"\\:
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;D&\\s34VD\"F-7$$\"3/LLL`4NnOF1$\"3#3tP)**oL\"G\"F-7$$\"3#*******\\,s`
PF1$\"3y#zA&)pq:J\"F-7$$\"3[mm;zM)>$QF1$\"3sqQ$RQn,M\"F-7$$\"3$*******
pfa<RF1$\"3=L8/`)4FP\"F-7$$\"3#HLLeg`!)*RF1$\"3p;(RpDEXS\"F-7$$\"3w***
*\\#G2A3%F1$\"3Ingnw4+R9F-7$$\"3;LLL$)G[kTF1$\"3SWV')=!*)QZ\"F-7$$\"3#
)****\\7yh]UF1$\"3B*oB_>Y;^\"F-7$$\"3Kmmm')fdLVF1$\"3I1qv2[=\\:F-7$$\"
36mmm,FT=WF1$\"3a2\"*)zsZ()e\"F-7$$\"3FLL$e#pa-XF1$\"3`\"o:d6S\"H;F-7$
$\"3!*******Rv&)zXF1$\"35**\\YP*fsm\"F-7$$\"3%GLL$GUYoYF1$\"3[Z*e(4w67
<F-7$$\"33mmm1^rZZF1$\"3AE'R'oVG`<F-7$$\"34++]sI@K[F1$\"3E&H=*)y_#)z\"
F-7$$\"34++]2%)38\\F1$\"3YFii4NKU=F-7$$\"\"&F*$\"3i'*[zEjz!*=F--%'COLO
URG6&%$RGBG$\"#5!\"\"$F*F*F_\\l-%+AXESLABELSG6$Q\"r6\"Q!6\"-%%VIEWG6$;
F(Fd[l%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 
0 0 "Curve 1" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 81 "From the plot of
 g3, there appears to be a unique minimum.  We need to precisely " }}
{PARA 0 "" 0 "" {TEXT -1 20 "locate the minimum. " }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 31 "fsolve(D(g3)(r) = 0, r = 0..4);" }{TEXT -1 
0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+]')3NB!\"*" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "D(D(g3))(2.335088650);" }{TEXT -1 
0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&%#PiG\"\"%$\"+BTF8D!\")\"\"\"
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "g3(2.335088650);" }
{TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&%#PiG$\"+,y_!4\"!\")
$\"+`#))>&oF'\"\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 53 "Thus, the \+
function g3 has a critical value at r  =   " }{XPPEDIT 18 0 "2.3350886
50;" "6#$\"+]')3NB!\"*" }{TEXT -1 2 "; " }}{PARA 0 "" 0 "" {TEXT -1 
74 "the second derivative of  g3  at the critical value is positive an
d hence " }}{PARA 0 "" 0 "" {TEXT -1 52 "g3  has a minimum at this cri
tical value.           " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 54 "Conclu
sion:  The minimum amount of tin requried is :  " }{XPPEDIT 18 0 "10.9
0527801*Pi+68.51988253;" "6#,&*&$\"+,y_!4\"!\")\"\"\"%#PiGF(F($\"+`#))
>&oF'F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 20 "We now do part (b). " 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "h3:= r -> (80)/(Pi * r^2)
;" }{TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#h3GR6#%\"rG6\"6
$%)operatorG%&arrowGF(,$*&\"\"\"F.*&%#PiGF.)9$\"\"#F.!\"\"\"#!)F(F(F(
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "h3(2.335088650);" }
{TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&\"\"\"F%%#PiG!\"\"
$\"+q%zrY\"!\")" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 84 "Conclusion:  A
t the minimum SA, the tin can has  a radius of   2.335088650  and a   \+
" }}{PARA 0 "" 0 "" {TEXT -1 33 "                    height of    " }
{XPPEDIT 18 0 "14.67179470*1/Pi;" "6#*($\"+q%zrY\"!\")\"\"\"F'F'%#PiG!
\"\"" }{TEXT -1 24 ".                       " }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 259 10 "Example 4\n" }{TEXT 
-1 48 "A triangle has two sides of length  5  and   8. " }}{PARA 0 "" 
0 "" {TEXT -1 77 "a)  What is the angle between the two sides so that \+
the the area is largest? " }}{PARA 0 "" 0 "" {TEXT -1 72 "b)  What is \+
the perimeter of the triangle when its area is the largest? " }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 76 "We do (a) first.   Let A denote th
e area of the triangle and   b denote the " }}{PARA 0 "" 0 "" {TEXT 
-1 53 "angle between the two given sides of length 5 and 8. " }}{PARA 
0 "" 0 "" {TEXT -1 5 "Then " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 29 "A = (1/2) * 5 * 8 * sin(b) = " }{XPPEDIT 18 0 "
20*sin(b)" "6#*&\"#?\"\"\"-%$sinG6#%\"bGF%" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 23 "We want to maximize A. " }}
{PARA 0 "" 0 "" {TEXT -1 7 "       " }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 22 "A4:= b -> 20 * sin(b);" }{TEXT -1 0 "" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#>%#A4GR6#%\"bG6\"6$%)operatorG%&arrowGF(,$-%$sinG6
#9$\"#?F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "plot(A4(b)
, b = 0..2*Pi, color = black);" }{TEXT -1 0 "" }}{PARA 13 "" 1 "" 
{GLPLOT2D 400 300 300 {PLOTDATA 2 "6&-%'CURVESG6#7en7$$\"\"!F)F(7$$\"3
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[E^dK,RF-$\"3+,bMM1A1wF07$$\"3ab^NehL]_F-$\"3I!ybf=%[-5!#;7$$\"35*QhW&
\\$Hf'F-$\"3PQ^yJr6D7F@7$$\"3zS11.fpPyF-$\"3\"RE9y`3>T\"F@7$$\"3upr'fw
tl7*F-$\"3qzn_2QE#e\"F@7$$\"3!QRa&4L&f/\"F0$\"3l*o@bA03t\"F@7$$\"3YjXw
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{EXCHG {PARA 0 "" 0 "" {TEXT -1 65 "From the plot there appears to be \+
one maximum between 0 and 2Pi. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 32 "fsolve(D(A4)(b) = 0, b = 0..Pi);" }{TEXT -1 0 "" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#$\"+Fjzq:!\"*" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 22 "D(D(A4))(1.570796327);" }{TEXT -1 0 "" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#$!#?\"\"!" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 47 "
The critical value of A4 between 0 and Pi is:  " }{XPPEDIT 18 0 "1.570
796327;" "6#$\"+Fjzq:!\"*" }}{PARA 0 "" 0 "" {TEXT -1 17 "Also,     A4
''(  " }{XPPEDIT 18 0 "1.570796327;" "6#$\"+Fjzq:!\"*" }{TEXT -1 9 "  \+
) < 0. " }}{PARA 0 "" 0 "" {TEXT -1 84 "Conclusion:  the area of the t
riangle is a maximum when the angle between the given " }}{PARA 0 "" 
0 "" {TEXT -1 13 "sides is     " }{XPPEDIT 18 0 "1.570796327;" "6#$\"+
Fjzq:!\"*" }{TEXT -1 22 " radians.             " }}{PARA 0 "" 0 "" 
{TEXT -1 10 "          " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 17 "We do \+
part (b).  " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 63 "Let c denote the length of the remaining side of the triangle. \+
" }}{PARA 0 "" 0 "" {TEXT -1 69 "Recall that b denotes the angle betwe
en the sides of length 5 and 8. " }}{PARA 0 "" 0 "" {TEXT -1 22 "Law o
f cosines gives: " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{XPPEDIT 18 0 "c^2" "6#*$%\"cG\"\"#" }{TEXT -1 3 " = " }{XPPEDIT 18 0 
"5^2 + 8^2  - 2 * 5 * 8 cos(b)" "6#,(*$\"\"&\"\"#\"\"\"*$\"\")F&F'**F&
F'F%F'F)F'-%$cosG6#%\"bGF'!\"\"" }{TEXT -1 2 ". " }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 42 "c:= sqrt(25 + 64 - 80 * cos(1.570796327));" }
}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"cG$\"+L6)RV*!\"*" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "P:= 9.433981133 + 5 + 8; " }{TEXT 
-1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"PG$\"+8\")RVA!\")" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 45 "Perimeter of tirangle with largest
 area is   " }{XPPEDIT 18 0 "22.43398113;" "6#$\"+8\")RVA!\")" }{TEXT 
-1 10 ".         " }}}}{MARK "4 0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 
1 1 }{PAGENUMBERS 0 1 2 33 1 1 }