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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 "" }}}{PARA 0 "" 0 "" {TEXT 260 0 "" }}{PARA 0 "" 0 "
" {TEXT 264 11 "Example 1: " }{TEXT 269 13 "Paper Folding" }}{EXCHG 
{PARA 258 "" 0 "" {TEXT -1 367 "Problem:  A sheet of paper 4 inches wi
de by 8 inches high is folded so that the bottom right corner of the s
heet touches the left hand edge of the sheet.  The tip of the corner i
s no more than 4 inches above the bottom edge of the paper.  Then the \+
paper is creased (see figure).  Find the length L of the crease, and f
ind how to fold the paper so that L is minimum.  " }}{PARA 257 "" 0 "
" {TEXT 259 10 "Solution: " }}{PARA 257 "" 0 "" {TEXT -1 49 "Let h, x,
 and y be as shown in the diagram below." }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 88 
"A1:=plots[textplot](\{[2.6,1.9,'L'],[3.7,2,'h'],[.1,1.1,`y`],[.5,.3,`
x`],\},align=RIGHT): " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 99 "A3
:=plot(\{[[0,0],[0,8],[4,8],[4,4.45],[1.219,0],[0,0]],[[4,4.45],[0, 2.
5],[1.219,0]]\},color=blue): " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 47 "A4:=plot([[4,4.45],[4,0],[1.219,0]],color=red):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 74 "A5:=plots[polygonplot]([[4,4.45],[0
,4.45],[0,2.5]],style=patch,color=tan):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 61 "plots[display]([A1,A3,A4,A5],axes=boxed,scaling=const
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{PARA 0 "" 0 "" {TEXT -1 186 " We can see several equations relating x
, y, h, and L in the diagram.  For example, the small right triangle w
ith legs x and y has a hypotenuse which is  4-x units long.  This give
s eq1." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "eq1 := y^2+x^2=(4
-x)^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eq1G/,&*$%\"yG\"\"#\"\"\"
*$%\"xGF)F**$,&\"\"%F*F,!\"\"F)" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
71 "eq2 comes from the right triangle with hypotenuse L and legs h and
 4-x." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "eq2 := L^2 = (4-x)
^2 + h^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eq2G/*$%\"LG\"\"#,&*$,
&\"\"%\"\"\"%\"xG!\"\"F(F-*$%\"hGF(F-" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 193 "Now it is easy to work out that the tan right triangle w
ith hypotenuse h and legs 4 and  h-y is similar to the right triangle \+
with hypotenuse 4-x and corresponding legs y and x.  So we get eq3." }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "eq3 := 4/(h-y)=y/x;" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$eq3G/,$*$,&%\"hG\"\"\"%\"yG!\"\"F,
\"\"%*&F+F*%\"xGF," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "h := \+
solve(eq3,h);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"hG*&,&%\"xG\"\"%*
$%\"yG\"\"#\"\"\"F,F*!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
19 "x := solve(eq1,x); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"xG,&*$%
\"yG\"\"##!\"\"\"\")F(\"\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 42 "L := unapply(sqrt(op(2,simplify(eq2))),y);" }}{PARA 11 "" 1 "" 
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0 "" {TEXT -1 100 "So we have L expressed as a function of  one variab
le y. Examining the behavior of L as y  varies,  " }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 13 "plot(L,2..4);" }}{PARA 13 "" 1 "" {GLPLOT2D 
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{EXCHG {PARA 0 "" 0 "" {TEXT -1 56 "we see that there is a minimum len
gth crease of about   " }{XPPEDIT 18 0 "L = 5.2" "6#/%\"LG$\"#_!\"\"" 
}{TEXT -1 19 "  inches at about  " }{XPPEDIT 18 0 "y = 2.8  " "6#/%\"y
G$\"#G!\"\"" }{TEXT -1 141 "  inches.  We can get a more precise value
 with fsolve, by locating the x between 2.6 and 3 where the derivative
 of the crease function is 0." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "y1
 := fsolve(diff(L(y),y),y,2.6..3);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#
>%#y1G$\"+DrUGG!\"*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 51 "Now check \+
the value of x and L for this value of y." }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "minL := L(y1);" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#>%%minLG$\"+BC:'>&!\"*" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 21 "minx := subs(y=y1,x);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%%minxG$\"+++++5!\"*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 282 "Notice the nice integer value of   minx .  This gives rise to \+
a simple construction of the crease of minimum length:  by folding twi
ce along the bottom edge, we can mark the point  1 inch from the left \+
edge of the paper.  Then bring the corner point up to the left edge an
d crease. " }}{PARA 258 "" 0 "" {TEXT -1 76 "Exercise:   Use Maple to \+
draw the diagram showing the minimum length crease." }}}{EXCHG {PARA 
258 "" 0 "" {TEXT -1 128 "Exercise:   Suppose we wanted to minimize  L
 + y, rather than just L.  Would the  minimum occur at the same place?
  Work it out." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT 261 9 "Example 2" }}{PARA 0 "" 0 "" {TEXT -1 80 "\nA woman'
s blood pressure p ranges from a high (the systolic pressure) to a low
 " }}{PARA 0 "" 0 "" {TEXT -1 85 "(the diastolic pressure) so that at \+
time t in seconds, p is given (approximately) by " }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 39 "      p   =  P(t)  = 100 \+
+ 20 sin(5t). " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 85 "The time between two successive high pressures is the tim
e needed for one heartbeat. " }}{PARA 0 "" 0 "" {TEXT -1 18 "First, pl
ot P(t). " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
83 "a)  Find an expression for the rate at which the woman's blood pre
ssure is changing" }}{PARA 0 "" 0 "" {TEXT -1 11 "at time t. " }}
{PARA 0 "" 0 "" {TEXT -1 63 "b)  At what time is the woman's blood pre
ssure a max?  a min?  " }}{PARA 0 "" 0 "" {TEXT -1 66 "c)  What is the
 woman's max blood pressure?  min blood pressure?  " }}{PARA 0 "" 0 "
" {TEXT -1 59 "d)  How many times per minute does the woman's heart be
at? " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "restart: with(plott
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1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 27 "f:= x -> 100 + 20*sin(5*x);" }{TEXT -1 0 "" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fGR6#%\"xG6\"6$%)operatorG%&arrowG
F(,&\"$+\"\"\"\"*&\"#?F.-%$sinG6#,$9$\"\"&F.F.F(F(F(" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 5 "D(f);" }{TEXT -1 0 "" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#R6#%\"xG6\"6$%)operatorG%&arrowGF&,$-%$cosG6#,$9$\"\"&
\"$+\"F&F&F&" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 51 "a)  The womans bl
ood pressure changes at a rate of " }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "d
f/dx = 100 *cos(5x)" "6#/*&%#dfG\"\"\"%#dxG!\"\"*&\"$+\"F&-%$cosG6#*&
\"\"&F&%\"xGF&F&" }{TEXT -1 3 ".  " }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 37 "fsolve(100 * cos( 5*x) = 0, x, 0..1);" }{TEXT -1 0 "
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+aEfTJ!#5" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 36 "fsolve(100* cos(5*x) = 0, x, .5..1);" }{TEXT 
-1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+hzxC%*!#5" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "fsolve(100* cos(5*x) = 0, x, 1..2);
" }{TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+Fjzq:!\"*" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 34 "b)  Since f takes on max min when \+
" }{XPPEDIT 18 0 "df/dx = 0" "6#/*&%#dfG\"\"\"%#dxG!\"\"\"\"!" }{TEXT 
-1 14 ", we see that " }}{PARA 0 "" 0 "" {TEXT -1 57 "f has a max when
 x is about .3141592654 and 1.570796327. " }}{PARA 0 "" 0 "" {TEXT -1 
47 "Also, f has a min when x is about .942477796.  " }}{PARA 0 "" 0 "
" {TEXT -1 38 "We otained these estimates by setting " }{XPPEDIT 18 0 
"df/dx = 0" "6#/*&%#dfG\"\"\"%#dxG!\"\"\"\"!" }{TEXT -1 2 ". " }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "f(.3141592654);" }{TEXT -1 
0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"$?\"\"\"!" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 15 "f(.9424777961);" }{TEXT -1 0 "" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#$\"#!)\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" 
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{XPPMATH 20 "6#$\"$?\"\"\"!" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 64 "c)
  We see the max blood pressure is 120 and the minimum is 80. " }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 26 "1.570796327 - .3141592654;" }{TEXT -1 0 "" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#$\"+iqjc7!\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 17 "60 / 1.256637062;" }{TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#$\"+\"H[Yx%!\")" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 53 "d)  Each
 heartbeat takes approx 1.256637062 seconds. " }}{PARA 0 "" 0 "" 
{TEXT -1 49 "She has approximately 48 heartbeats per minute.  " }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 257 9 "Ex
ample 3" }}{PARA 0 "" 0 "" {TEXT -1 88 "\nA triangle is formed in the \+
first quadrant by a tangent line to the graph of the curve " }}{PARA 
0 "" 0 "" {TEXT -1 12 "            " }{XPPEDIT 18 0 "y =  e^(-x)" "6#/
%\"yG)%\"eG,$%\"xG!\"\"" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 
117 "and the coordinate axes.  What is the area of the largest such tr
iangle?  First let's see a diagram of the situation." }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 667 "f := x->exp(-x);\na1 := 1.5: a2 := 2.5:
\neqTanLine := (x,a) -> f(a) + (x-a) * D(f)(a):\nb1 := solve( eqTanLin
e(x,a1)=0, x):\nb2 := solve( eqTanLine(x,a2)=0, x):\ngraph := plot( f(
x), x = 0..5, thickness=2):\ntanLine1 := plot( eqTanLine(x,a1), x = 0.
.b1, color=blue, thickness=2):\ntanLine2 := plot( eqTanLine(x,a2), x =
 0..b2, color=green, thickness=2):\nline1 := line( [0, 0],[b1, 0], col
or=blue, thickness=2):\nline2 := line( [0, 0],[0, eqTanLine(0,a1)], co
lor=blue, thickness=2):\nline3 := line( [0, 0],[b2,0], color=green, th
ickness=2):\nline4 := line( [0, 0],[0, eqTanLine(0,a2)], color=green, \+
thickness=2):\ndisplay( graph, tanLine1, tanLine2, line1, line2,line3,
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3;nT5g&GZ1$Fbo$\"33![<tbDHd$F/7$$\"3Y++]Z`PKJFbo$\"3fUSJ7pk<IF/7$$\"3
\"pm\"z*>1*4KFbo$\"3[ar(>#\\B\"Q#F/7$$\"3[LLL=2DzKFbo$\"3:Hu4^/-7=F/7$
$\"33+vVQk=`LFbo$\"3;3b#G5>^?\"F/7$$\"3I+DccB&RU$Fbo$\"35s8iO2PUiFdz7$
$\"3++++++++NFbo$\"3))*)f5(p%****HFc[m-Fe[l6&Fg[lF(Ff[mF(F[\\l-F$6%7$7
$F(F(7$$\"+++++D!\"*F(Fd[mF[\\l-F$6%7$Fd[n7$F($\"+.SDyb!#5Fd[mF[\\l-F$
6%7$Fd[n7$$\"+++++NFh[nF(F_[nF[\\l-F$6%7$Fd[n7$F($\"+_\\(H(GF_\\nF_[nF
[\\l-%+AXESLABELSG6%Q\"x6\"Q!6\"%(DEFAULTG-%%VIEWG6$;F(F`[lFc]n" 1 2 
0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2
" "Curve 3" "Curve 4" "Curve 5" "Curve 6" "Curve 7" }}}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 87 "From the above plot we see that the triangle ch
anges with our choice of tangent lines. " }}{PARA 0 "" 0 "" {TEXT -1 
71 "We need to find a tangent line that gives  a triangle of maximal a
rea. " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 40 "Find the x-intercept of \+
a tangent line. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "solve( \+
exp(-a)-exp(-a)*(x-a)=0,x);" }{TEXT -1 0 "" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#,&\"\"\"F$%\"aGF$" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
64 "So, the x-intercept of the  tangent line when x = a is:  1 + a. " 
}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 40 "Find the y-intercept of a tange
nt line. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "tl(0,a);" }
{TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%#tlG6$\"\"!%\"aG" }}
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 57 "So, the y-intercept of the tangen
t line when  x = a is:  " }{XPPEDIT 18 0 "exp(-a)+exp(-a)*a;" "6#,&-%$
expG6#,$%\"aG!\"\"\"\"\"*&-F%6#,$F(F)F*F(F*F*" }}}{EXCHG {PARA 0 "" 0 
"" {TEXT -1 69 "Thus using the  tangent line for x = a we have that th
e triangle has " }}{PARA 0 "" 0 "" {TEXT -1 10 "height =  " }{XPPEDIT 
18 0 "e^(-a) + a* e^(-a)" "6#,&)%\"eG,$%\"aG!\"\"\"\"\"*&F'F))F%,$F'F(
F)F)" }{TEXT -1 7 "   and " }}{PARA 0 "" 0 "" {TEXT -1 62 "base = 1 + \+
a.  We can now calculate the area of the triangle. " }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 41 "A:= 0.5 * (exp(-a)+exp(-a)*a) * ( 1 + a);
" }{TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"AG,$*&,&-%$expG
6#,$%\"aG!\"\"\"\"\"*&F(F.F,F.F.F.,&F.F.F,F.F.$\"\"&F-" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "plot(A(a), a = 0..5);" }{TEXT -1 0 
"" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6%-%'CURVESG6
$7U7$$\"\"!F)$\"\"&!\"\"7$$\"+4x&)*3\"!#5$\"+vaH9bF07$$\"+uq8Q?F0$\"+S
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0$\"+qv#z(oF07$$\"+U-/PiF0$\"+&z0]1(F07$$\"+fmpisF0$\"+giH2sF07$$\"+#*
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07$$\"+#RU07$Fbo$\"+88mYPF07$$\"+?S2LKFbo$\"+-%fK`$F07$$\"+$p)=MLFbo$
\"+m\"\\yM$F07$$\"+*=]@W$Fbo$\"+](=o:$F07$$\"+]$z*RNFbo$\"+1+4!*HF07$$
\"+kC$pk$Fbo$\"+m7)[\"GF07$$\"+3qcZPFbo$\"+CZ$ol#F07$$\"+/\"fF&QFbo$\"
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*pp;%Fbo$\"+A\"\\*o?F07$$\"+ye,tUFbo$\"+gF%z$>F07$$\"+fO=yVFbo$\"+IWw9
=F07$$\"+E>#[Z%Fbo$\"+%4Qtq\"F07$$\"+(G!e&e%Fbo$\"+?cz!f\"F07$$\"+&)Qk
%o%Fbo$\"+m&=B\\\"F07$$\"+UjE!z%Fbo$\"+k`4$R\"F07$$\"+60O\"*[Fbo$\"+ly
\\.8F07$$F+F)$\"+g/$G@\"F0-%'COLOURG6&%$RGBG$\"#5F,F(F(-%+AXESLABELSG6
$Q\"a6\"Q!6\"-%%VIEWG6$;F(F_[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 87 "From the plot of the area function (area of the triangle)
, we see that there is indeed " }}{PARA 0 "" 0 "" {TEXT -1 83 "a maxim
um.  We need to precisely locate for what a values does this maximum o
ccur. " }}{PARA 0 "" 0 "" {TEXT -1 16 "At this maximum " }{XPPEDIT 18 
0 "dA/da = 0" "6#/*&%#dAG\"\"\"%#daG!\"\"\"\"!" }{TEXT -1 16 ".  So, w
e solve " }{XPPEDIT 18 0 "dA/da = 0" "6#/*&%#dAG\"\"\"%#daG!\"\"\"\"!
" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 47 "AA:= a -> 0.5 * (exp(-a)+exp(-a)*a) * ( 1 + \+
a);" }{TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#AAGR6#%\"aG6
\"6$%)operatorG%&arrowGF(,$*&,&-%$expG6#,$9$!\"\"\"\"\"*&F/F5F3F5F5F5,
&F3F5F5F5F5$\"\"&F4F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
14 "diff(AA(a),a);" }{TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,
(*(-%$expG6#,$%\"aG!\"\"\"\"\"F)F+,&F+F+F)F+F+$!\"&F**&$\"\"&F*F+F%F+F
+*(F0F+F%F+F)F+F+" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "solve(
% = 0, a);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$$\"\"\"\"\"!$!\"\"F%" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "plot(A(a), a = -2..3);" }
{TEXT -1 0 "" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6%
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7$$\"*+#pxgF-$\"+I5GQqFdo7$$\"*g4t.(F-$\"+X\"\\.=(Fdo7$$\"*!Hst!)F-$\"
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RU07\"F-$\"+gR$>L(Fdo7$$\"+?S2L7F-$\"+N6RlsFdo7$$\"+$p)=M8F-$\"+N@#[<(
Fdo7$$\"+*=]@W\"F-$\"+0f7]qFdo7$$\"+]$z*R:F-$\"+If`:pFdo7$$\"+kC$pk\"F
-$\"+q!>%[nFdo7$$\"+3qcZ<F-$\"+gN=vlFdo7$$\"+/\"fF&=F-$\"+&zN0Q'Fdo7$$
\"+0Ogb>F-$\"+:p^zhFdo7$$\"+nAFj?F-$\"+]IPgfFdo7$$\"+&)*pp;#F-$\"+l7@V
dFdo7$$\"+ye,tAF-$\"+5M1<bFdo7$$\"+fO=yBF-$\"+?Xf!H&Fdo7$$\"+E>#[Z#F-$
\"+Sw)>3&Fdo7$$\"+(G!e&e#F-$\"+-RzV[Fdo7$$\"+&)Qk%o#F-$\"+')*3Fj%Fdo7$
$\"+UjE!z#F-$\"+)RZ2T%Fdo7$$\"+60O\"*GF-$\"+/\\9-UFdo7$$\"\"$F*$\"+qa'
H)RFdo-%'COLOURG6&%$RGBG$\"#5!\"\"$F*F*Fd_l-%+AXESLABELSG6$Q\"a6\"Q!6
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45.000000 45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
34 "From the plot, we see that indeed " }{XPPEDIT 18 0 "dA/da = 0" "6#
/*&%#dAG\"\"\"%#daG!\"\"\"\"!" }{TEXT -1 15 " for a = -1,1. " }}{PARA 
0 "" 0 "" {TEXT -1 48 "We are only interested in positive values of a.
 " }}{PARA 0 "" 0 "" {TEXT -1 65 "Thus the answer is that the area of \+
the triangle is largest when " }}{PARA 0 "" 0 "" {TEXT -1 32 "the tang
ent line is for x = 1.  " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT 258 9 "Example 4" }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 83 "A cone is to be formed by revolving a
 right triangle with a hypotenuse of length  3" }}{PARA 0 "" 0 "" 
{TEXT -1 88 " feet about one of its legs.  What is the volume of the l
argest cone that can be formed " }}{PARA 0 "" 0 "" {TEXT -1 39 "in thi
s manner?  See the diagram below." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 305 "c := cone([0,0,4], sqrt(8), -4, color=green):\nt := \+
pointplot3d([[0,0,0], [sqrt(8),0,0], [0,0,4],[0,0,0]], connect=true, c
olor=black, thickness=2):\nground := plot3d(0,x=-3..3,y=-3..3,color=gr
ey,grid=[5,10], style=patchnogrid):\ndisplay([c,t,ground],scaling=cons
trained,style=wireframe, orientation=[40,60]);" }}{PARA 13 "" 1 "" 
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!*>=+DF6$\"+`Zi(>(F67$$\"+E&4Qc#F6$\"+#)es:lF67$$\"+%>5pi#F6$\"+R%p]x&
F67$$\"+bJ*[o#F6$\"+$Q!eO]F67$$\"+r\"[8v#F6$\"+#3j.7%F67$$\"+Ijy5GF6$
\"+LMEOKF67$$\"+/)fT(GF6$\"+m,([A#F67$$\"+b!)[/HF6$\"+YBH:<F67$$\"+1j
\"[$HF6$\"+Y'G*)=\"F67$$\"+`\"3u'HF6$\"+.$oO/'F07$$\"\"$F)F(-%'COLOURG
6&%$RGBG$\"#5!\"\"F(F(-%+AXESLABELSG6$Q\"b6\"Q!6\"-%%VIEWG6$;F(F^[l%(D
EFAULTG" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curv
e 1" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 73 "From the estiamted plot, \+
we see that the volume has one maximum and that " }}{PARA 0 "" 0 "" 
{TEXT -1 16 "at this maximum " }{XPPEDIT 18 0 "dV/db = 0" "6#/*&%#dVG
\"\"\"%#dbG!\"\"\"\"!" }{TEXT -1 21 ".  Hence, we solve:  " }{XPPEDIT 
18 0 "dV/db = 0" "6#/*&%#dVG\"\"\"%#dbG!\"\"\"\"!" }{TEXT -1 2 ". " }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "VV:= b -> (1/3) * Pi * (9 -
 b^2) * b;" }{TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#VVGR6#
%\"bG6\"6$%)operatorG%&arrowGF(,$*(%#PiG\"\"\",&\"\"*F/*$)9$\"\"#F/!\"
\"F/F4F/#F/\"\"$F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "d
iff(VV(b),b);" }{TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*&%#
PiG\"\"\")%\"bG\"\"#F&#!\"#\"\"$*(#F&F,F&F%F&,&\"\"*F&*$F'F&!\"\"F&F&
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "solve(%=0,b);" }{TEXT 
-1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6$*$-%%sqrtG6#\"\"$\"\"\",$F#!
\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "evalf(sqrt(3));" }
{TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+330K<!\"*" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 55 "The cone will have maximum volume \+
when b = sqrt(3) and " }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "a^2= 9 - b^2" 
"6#/*$%\"aG\"\"#,&\"\"*\"\"\"*$%\"bGF&!\"\"" }{TEXT -1 19 " = 9 - 3 = \+
6, i.e, " }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "b = sqrt(3)" "6#/%\"bG-%%sq
rtG6#\"\"$" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "a = sqrt(6)" "6#/%\"aG
-%%sqrtG6#\"\"'" }{TEXT -1 3 ".  " }}}}{MARK "61 1 0" 9 }{VIEWOPTS 1 
1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }