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{SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT -1 0 "" }{TEXT 274 0 "" }{TEXT 
275 23 "Calculus IV with Maple\n" }{TEXT 276 32 "Copyright 2002, Dr. J
ack Wagner\n" }{TEXT -1 30 "j.wagner@intelligentsearch.com" }}}{EXCHG 
{PARA 4 "" 0 "" {TEXT -1 61 "\nLesson 4: Constrained Optimization with
 Lagrange Multipliers" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 5 "" 0 "" {TEXT -1 128 "Topics: The method of Lagrange multiplie
rs for solving optimization problems that have constraints is illustra
ted geometrically." }}{PARA 5 "" 0 "" {TEXT 277 0 "" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 198 "The problem of optimization, just as in the si
ngle variable case, involves maximizing or minimizing some function su
bject to a set of constraints.  For ease of visualization we are going
 to work in " }{XPPEDIT 18 0 "R^3;" "6#*$%\"RG\"\"$" }{MPLTEXT 0 21 1 
" " }{TEXT -1 270 "with two constraint equations, but the ideas and me
thods are identical for higher dimensions and larger constraint sets. \+
 Formal proofs of the method of Lagrange multipliers abound.  What we \+
present here is a geometric picture and geometric explanation of the p
rocedure." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT 258 11 "Example 4.1" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 45 "We suppose that we are required to optimi
ze  " }{XPPEDIT 18 0 "F(x,y,z) = y+2*z-x^2;" "6#/-%\"FG6%%\"xG%\"yG%\"
zG,(F(\"\"\"*&\"\"#F+F)F+F+*$F'F-!\"\"" }{TEXT -1 61 "  subject to two
 constraints: F is restricted to the sphere, " }{XPPEDIT 18 0 "x^2+y^2
+z^2 = 25;" "6#/,(*$%\"xG\"\"#\"\"\"*$%\"yGF'F(*$%\"zGF'F(\"#D" }
{TEXT -1 29 "  and  also to the cylinder, " }{XPPEDIT 18 0 "x^2+y^2 = \+
9;" "6#/,&*$%\"xG\"\"#\"\"\"*$%\"yGF'F(\"\"*" }{TEXT -1 0 "" }
{MPLTEXT 0 21 6 ".\n   \n" }{TEXT -1 86 "First, examine a plot of the \+
two constraints together with several level sets of F.  \n" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 79 "restart:with(VectorCalculus):with(L
inearAlgebra):with(plots): with(plottools): " }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 19 "F := y + 2 * z-x^2;" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 177 "Now we create plot struc
tures for several level sets of F.  The multiple 5 in the expression 5
 * k is necessary to separate the sets by a sufficient amount to assur
e visibility." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 153 "P1 := k -
> implicitplot3d(F = 9 * k, x = -10..10, y = -10..10, z = -10..10, axe
s = boxed, style=patchnogrid, numpoints=2000):\nS1 := seq(P1(k), k = 1
..2):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 74 "Now we define the constraints and create plot structures \+
for their graphs." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 126 "P2 :=
 sphere([0,0,0], 5, color=blue):\nP3 := cylinder([0,0,-8], 3, 16, capp
ed=false, style=wireframe, color=red, thickness=3):\n" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 118 "By inspection of the constraint equation
s they intersect where z  =  \2614. This intersection is plotted as a \+
spacecurve." }{MPLTEXT 0 21 1 " " }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 166 "S2 := spacecurve(\{[x, sqrt(9-x^2), 4], [x, -sqrt(9-
x^2), 4], [x, sqrt(9-x^2), -4], [x, -sqrt(9-x^2), -4]\}, x = -3..3, nu
mpoints = 1000, color = black, thickness = 2):" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 68 "display(S1, P2, P3, S2, orientation=[-75,65], \+
scaling=constrained);\011" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 204 "The
 cylinder and sphere intersect in two circles, S2.  At an optimal poin
t for F, one of the level sets must be tangent to one of the circles i
n S2.  Looking ahead we reproduce the plot of the level set, " }
{XPPEDIT 18 0 "F^(-1);" "6#)%\"FG,$\"\"\"!\"\"" }{TEXT -1 4 "(11)" }
{MPLTEXT 0 21 3 " , " }{TEXT -1 82 "the maximum of F restricted to the
 surfaces, together with the tangents to S2 and " }{MPLTEXT 0 21 1 " \+
" }{TEXT -1 0 "" }{XPPEDIT 18 0 "F^(-1);" "6#)%\"FG,$\"\"\"!\"\"" }
{TEXT -1 4 "(11)" }{MPLTEXT 0 21 2 ", " }{TEXT -1 112 "and the gradien
ts of F , the cylinder and the sphere as well as the normal plane to S
2 at the point of tangency." }{MPLTEXT 0 21 1 "\n" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 25 "Function to be opt
imized." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "F := y + 2 * z-x
^2:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 12 "Constraint 1" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "C1 :
= x^2 + y^2 + z^2 = 25:     \011\011" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 12 "Constraint 2" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 20 "C2 := x^2 + y^2 = 9:" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 33 "Plot of intersecti
on of C1 and C2" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 141 "Sa := s
pacecurve(\{[x, sqrt(9-x^2), 4], [x, -sqrt(9-x^2), 4]\}, x = -3..3, \n
                 numpoints = 1000, color = black, thickness = 2):  " }
}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 41 "
Plot of level set of inverse image of 11." }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 94 "Sc := implicitplot3d(F = 11, x = -3..3, y = -3..5, \+
z = 0..6, color = blue, style = wireframe):" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 42 "Positive part of inters
ection of C1 and C2" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "e :=
 <x, sqrt(9-x^2), 4>;     \011  " }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 20 "de[x] := diff(e, x);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 33 "de[x = 0] := subs(x = 0, de[x]);\n" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 12 "Tangent to e" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "T := <0, 3, 4> + s * de[x]
; " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "Sb := spacecurve(eval
m(T), s = -4..4, color = red, thickness = 3):\n" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 41 "Plot of level set of
 inverse image of 11." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 94 "Sc
 := implicitplot3d(F = 11, x = -3..3, y = -3..5, z = 0..6, color = blu
e, style = wireframe):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "r
 := [x, y, z]:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "gF := Gra
dient(F, r);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "gF := subs(
x = 0, gF);\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 
"" {TEXT -1 37 "Gradient vector of F at P = [0, 3, 4]" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "G1 := <0, 3, 4> + s * evalVF(gF, <r
>);\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 74 "Sd := spacecurve(e
valm(G1), s = -1.5..1.5, color = black, thickness = 3):\n" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "gC1 := Gradient(lhs(C1), r);\n" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "gC1 := subs(\{x = 0, y = 3, \+
z = 4\}, gC1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "gC2 := Gr
adient(lhs(C2), r);\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "gC
2 := subs(\{x = 0, y = 3, z = 4\}, gC2);\n" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 27 "Gradient vector of C1 at \+
P." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "G2 := <0, 3, 4> + s *
 evalVF(gC1, <x, y, z>);   \n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 27 "
Gradient vector of C2 at P." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
56 "G3 := <0, 3, 4> + s * evalVF(gC2, <x, y, z>); \011\011       \n" }
}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 26 "
Plane spanned by G3 and G2" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
58 "G4 := <0, 3, 4> + evalVF(s * gC1 + t * gC2, <x, y, z>);\011 \n" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 84 "Se := spacecurve(\{evalm(G2
), evalm(G3)\}, s = -0.3..0.4, color = red, thickness = 3):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 86 "Sf := plot3d(evalm(G4), s = \+
-.3..0.4, t = -.3..0.4, color = green, style = wireframe):" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 22 "Plane ort
hogonal to G1" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "G5 := (<x,
 y, z>-<0, 3, 4>).evalVF(gF, <x, y, z>) = 0; \n" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 78 "Sg := plot3d((11-y)/2, x = -2..2, y = -1..6, c
olor = pink, style = wireframe):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 75 "display(Sa, Sb, Sc, Sd, Se, Sf, Sg, axes = framed, or
ientation=[-130,50]);\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 105 "Refer
ring to this illustration, we will identify each function, its graph a
nd its plot by the same name. " }{MPLTEXT 0 21 1 " " }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 55 "By the implicit function theorem, the level set
s of F, " }{TEXT 265 1 "\{" }{TEXT -1 1 " " }{XPPEDIT 18 0 "F^(-1);" "
6#)%\"FG,$\"\"\"!\"\"" }{TEXT -1 3 "(k)" }{TEXT 266 1 "\}" }{TEXT -1 
789 ", yield surfaces. To satisfy the constraints, these are required \+
to intersect the graphs of C1 and C2. It is clear that to satisfy both
 constraints simultaneously, the graphs of the level sets must  inters
ect their intersection, one of the two circles, S2.  In addition, it i
s  evident that the extreme values of F will occur at those points whe
re a level set of F is tangent to the graph of  S2. This is analogous \+
to the situation in linear programming where the optimized linear func
tion must be tangent to the feasible set.  Call such a point of tangen
cy P. At every point the tangent line to S2 must lie in the tangent pl
ane to the graphs of both C1 and C2, and, therefore, in the intersecti
on of those tangent planes.  In particular, this is true at P. Since, \+
at P there is a graph of" }{MPLTEXT 0 21 1 " " }{TEXT -1 0 "" }{TEXT 
268 1 "\{" }{TEXT -1 1 " " }{XPPEDIT 18 0 "F^(-1);" "6#)%\"FG,$\"\"\"!
\"\"" }{TEXT -1 3 "(k)" }{TEXT 269 1 "\}" }{TEXT -1 40 " tangent to S2
, for some value of k, say" }{TEXT 267 1 " " }{XPPEDIT 18 0 "k[0];" "6
#&%\"kG6#\"\"!" }{TEXT -1 2 ", " }{MPLTEXT 0 21 1 " " }{TEXT -1 80 "at
 this point the tangent line to S2 lies in the tangent plane to the gr
aph of  " }{TEXT 270 1 "\{" }{XPPEDIT 18 0 "F^(-1);" "6#)%\"FG,$\"\"\"
!\"\"" }{TEXT -1 1 "(" }{XPPEDIT 18 0 "k[0];" "6#&%\"kG6#\"\"!" }
{TEXT -1 1 ")" }{TEXT 271 1 "\}" }{TEXT -1 1 " " }{MPLTEXT 0 21 2 ". \+
" }{TEXT -1 602 "At each point on C1 and C2, its gradient is orthogona
l to its tangent plane.  Because the tangent line to S2 lies in the in
tersection of the tangent planes of C1 and C2, both gradients must be \+
orthogonal to the tangent line to S2.  That is, at P on S2, the gradie
nt vectors to C1 and C2 lie in a plane orthogonal to the tangent to S2
, i.e. the normal plane to S2 at P.  Since the gradient vectors are in
dependent any vector in that plane is a linear combination of the two \+
gradients.  But the gradient of F, at P, is orthogonal to its tangent \+
plane at that point, i.e. the plane tangent to the graph of" }
{MPLTEXT 0 21 1 " " }{TEXT -1 0 "" }{TEXT 272 1 "\{" }{XPPEDIT 18 0 "F
^(-1);" "6#)%\"FG,$\"\"\"!\"\"" }{TEXT -1 1 "(" }{XPPEDIT 18 0 "k[0];
" "6#&%\"kG6#\"\"!" }{TEXT -1 1 ")" }{TEXT 273 1 "\}" }{TEXT -1 1 " " 
}{MPLTEXT 0 21 2 ". " }{TEXT -1 327 "At P where F is just tangent to S
2, the tangent plane of F must contain the tangent line to S2.  The gr
adient of F, therefore, is orthogonal to that tangent line and  so lie
s in the normal plane at P. The result of this is that the gradient of
 F is a linear combination of the gradients of the two constraint surf
ace equations." }{MPLTEXT 0 21 2 "  " }{TEXT -1 1 " " }{XPPEDIT 18 0 "
grad(F) = lambda[1]*grad(C[1])+lambda[2]*grad(C[2]);" "6#/-%%gradG6#%
\"FG,&*&&%'lambdaG6#\"\"\"F--F%6#&%\"CG6#F-F-F-*&&F+6#\"\"#F--F%6#&F16
#F6F-F-" }{TEXT -1 243 "   This yields three equations in five unknown
s.  The lambdas are the Lagrange multipliers.  Together with the two c
onstraint equations, we have five equations in five unknowns.  Impleme
nting this in Maple is a lot simpler than explaining it.\n" }{MPLTEXT 
0 21 1 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "r := [x, y, z]
:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "gF := Gradient(F, r);
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 80 "gC := lambda[1] * Gradi
ent(lhs(C1), r) + lambda[2] * Gradient(lhs(C2), r);  \011\011 \n" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "for k to 3 do eq||k := gF[k]
 = gC[k] od;       " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 48 "Note the u
se of the concatenation operator \"||\"." }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 0 "" }{MPLTEXT 1 0 72 "sol := solve(\{eq1, eq2, eq3, C1, C2\}
, \{x, y, z, lambda[1], lambda[2]\});\011" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 62 "Now we substitute each of these sets of values in turn in
to F." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "V := evalf(seq(sub
s(sol[k], F), k = 1..6));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 74 "From
 whence it is clear that F is maximized at [0, 3, 4] and minimized at \+
" }{XPPEDIT 18 0 "[`+/-`*sqrt(35)/2, -1/2, -4];" "6#7%*(%$+/-G\"\"\"-%
%sqrtG6#\"#NF&\"\"#!\"\",$*&F&F&F+F,F,,$\"\"%F," }{TEXT -1 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 100 "In order to understand this very \+
important method more thoroughly we will dissect a second example. " }
}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 257 11 
"Example 4.2" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 53 "Find the maximum \+
and minimum values of the function: " }{OLE 1 4185 1 "[xm]Br=WfoRrB:::
wk;nyyI;G:;:j::>:B>N:F:nyyyyy]::yyyyyy::::::::::::::::::::::::::::::::
:::::::::::::::::::::::::::::::::::::::fyyyyya:nYf::G:jy;:::::::::::::
::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::JcvG
YMt>^:fBWMtNHm=;:::::::n<Ejyyiy=?Z:VZHQ:>:;`:Z@[::JjpvpLu=XK[Aj;JZQZ:B
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;F:f:vYxY:B::::::F:;jnJ:j;<ja:MLqnFYM:>::::::::N<<jyyiy=J:B::::::V:;J<
J:JyyI?:Jyyyyg:n:;JyK>j>J?>:Q:SJ:f;;JAjA>:[Z:F<N<;jCJDjDJE>:wAyA::::::
:::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::F:DJ<J:
:::::::::F:wyyAb<vYxy;tV;J:>R<:TNEj`@Pt\\Pd`QrPPJPMQ<MJPnrPqjLqnxPq^;<
J\\HOJPnr:v;fbk;n_>WdG_dnBwmcf]]:>::::::V:>::::::::::^=NZ:vn`AG>:^wgGC
KYXM<;j;@j:HZDFZLVjBc:?fHDJ=q]<>kob>AFXHjAr<]R:?B\\_;Ob:GvDDk;aa=v;::b
:K>da=v;::DKKDJ>U]LVJ<J:<::::::::::jysyAB:^:F:;jysy?:;:GX>Vi<JYI:=j>r:
yX:N:m_:>:C:?R:ElrfH=MtFGYMq>>Wlj:gmlJ::::::>^:NZ:vYxI:;J:<::::::C:C>;
:?ja^G>d:NHEms>@[;;B:::::::JFNZ;V:;J<jysy;Z::::::j=B:ky:B::::::::::::j
ysy:>:<:::::::::::::::::::vYxI:;Z::::::::j:f:<Z<vyyyYC:WS:mj<jRDj:B:E:
Qb:>Z:f:^[<>b<>^JV<n^@v;sjyyiyAZ:fc:><<JZyIewyyyy;y=fx;j?<:G;Sj`@@RS;B
:[l<B:Mb:b:E:cb:cqKPjDjwEJKxI;B:>l;B:;R:;b:<Kd`ppPps<LaPpu<Lcxpp@Pq<j:
^:;jP@:SB:;B:_c<B:AB:N`lN@is:Ix:sW:E:cJJAJcAjDjw?jx]:JBC:;R:D::::C:Uk:
^:>x;B:U:_c<B:C:oSGecG?X:Or;ew:Hj<<:c:GlDjw;<:[n:>Z;b::a=:J<jP>:C:[q:v
[:>:_;<j<J^\\qvdQw>^>F`=Fu=Vq=r:<:c:ob:c:qAB:>L;B:;R:D:ETWAEUMES;sNmMk
xPpFZ:^:f?=J<JrAJ:<j@JSd:<J<jXhlOploIjUHjWI:m?G;IU:UTRc=PpdPog@j<@Z?BK
aTMP]:V\\QGghGbHWdCS>EBOMUUEuV:bU@k;PntH@Z\\wGHk\\:GdV=uTcUW_mBTj\\Pqk
JJTNs@@J<PntH@fbr_hlGF_B:KRBYSKYFKYSJ[Dw[r=ItQ=DN_yAwRJ[Dwer==D:cYKcYC
cy]?tQ=\\JrJyTJvk?_xdveB_xIHb:B><NtyrZxQ^LYBxjZdliLniPNtyo\\RJYB==j:FZ
<NZ<NZ:N:=R:=j;@j:<J<b<=B:?B;AFL\\N;LJ;e]tNZ>njwRI?B;KVGPM;Lj;]arW:sJ:
hj:@j:N:KfFD=Ob:CjOV_<^Z=HImu:l;:::J>Q=bXdq;^=:bF:N\\\\f_<^:gKUDJ<HB<:
QB:n>^;yayI:<J:JB<Z:VY[B:I:;JXEZ:>Z:FZ<bBaTXaEWEUU^:<jP>:C:[q:<j;<Z:>:
:::::::::::::::::::Z<>Z:B::::::::::2:" }}{PARA 0 "" 0 "" {TEXT -1 32 "
subject to the two constraints, " }{TEXT 256 1 " " }{OLE 1 4181 1 "[xm
]Br=WfoRrB:::wk;nyyI;G:;:j::>:B>N:F:nyyyyy]::yyyyyy:::::::::::::::::::
::::::::::::::::::::::::::::::::::::::::::::::::::::fyyyyya:nYf::G:jy;
::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:::::::::JcvGYMt>^:fBWMtNHm=;:::::::n<Ejyyiy=?Z:VZHQ:>:;`:Z@[::JjpvpLu
=XK[Aj;JZPZ:B:F:YLpfF>:::::::::J?NZ;vyyyyy=J:B:::::::c:;:=j[vGUMrvC?Mo
J::::::::JCNZ;F:f:vYxY:B::::::F:;jnJ:j;<ja:MLqnFYM:>::::::::N<<jyyiy=J
:B::::::V:;J<>:;:wyyN::wyyyq<:v:<J>j>J?>:Q:S:UJ:n;v;;JBB:]:_J:V<^<nYvY
::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
:=Z<^:>::::::::::=Jyyy;d:yayYZHQ:>:;`:Z@o<fc[_hb_ds?h_?^on_j>^?GhoGfnG
gioGSZ:^bM?^?GH:Y:ETV:WCKaTMcDGLyVDurB:;::::::A:;::::::::::s:?B:IgsC<;
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KFF>::::::jysyAB:^:F:;jysy?:;:Gr<Vi<JsH:=j>r:mW:N:uN:>:C:?R:f:FZ<vyyyy
:^:n_;F=E:]c:=Z:f:V[<J:<j<J@DJZDZDV<<JMTjA^=yyyxy\\FHemj^HMmqnG;KaFFJu
fF>::::::;C:?B:yay=J:<J:<::::::C:;E:<:N:YLpJbNh:fFaMR>@>Z::::::::kJ;@j
;>:C:yayA:<::::::GB:^o=Z:::::::::::::yay=J:B:::::::::::::::::::jysy:>:
<::::::::fc:><<JZyIewyyyy;X<fx;j?<Z:n>^;UTRZ_?Z:>DEZ:F[<J:Dj<JDDJtG[;f
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v`=^q;f:^<;W:`jDjw?jx]:JBC:;R:DZJ:::^:f?=J<JrAZ:F[:>Z:N`DZ:F:<JRr:EB:;
JDJ`f<VYZ:JBAj:>Z;b::a=:J<jP>:C:[q:^;;B:_;<j;<:kUXomRHj@HZ=fZ:JDjoAJDj
wE:B:>L;B:;R:D:ETWAEUMES;sNmMkxPpFZ:^:f?=J<Jr?Z:^[:JSd:<j;jOplXV_=Ng<r
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\\wGHk\\:GdV=uTcUW_mBTj\\PqkJJTNs@@J<PntH@fbr_hlGF_B:KRBYSKYFKYSJ[Dw[r
=ItQ=DN_yAwRJ[Dwer==D:cYKcYCcy]?tQ=\\JrJyTJvk?_xdveB_xIHb:B><NtyrZxQ^L
YBxjZdliLniPNtyo\\RJYB==j:FZ<NZ<NZ:N:=R:=j;@j:<J<b<=B:?B;AFX\\q;^=;r<=
R:=J;>klbFN[<^Z@X?Qc:CrZmmvHZV:::b>?SE?B;AfXdQ;LZwRI?B;KFF<=QB:<JMJ@vY
xy:B:;:[B:<jw?<I:;JXEZ:>Z:FZ<bBaTXaEWEUU^:<jP>:C:[q:<j;<Z:>:::::::::::
::::::::::::::::::::::::::::::::::::::::::4:" }{TEXT -1 4 "and " }
{OLE 1 4181 1 "[xm]Br=WfoRrB:::wk;nyyI;G:;:j::>:B>N:F:nyyyyy]::yyyyyy:
::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
fyyyyya:nYf::G:jy;::::::::::::::::::::::::::::::::::::::::::::::::::::
:::::::::::::::::::::::::::JcvGYMt>^:fBWMtNHm=;:::::::n<Ejyyiy=?Z:VZHQ
:>:;`:Z@[::JjpvpLu=XK[Aj;JZPZ:B:F:YLpfF>:::::::::J?NZ;vyyyyy=J:B::::::
:c:;:=j[vGUMrvC?MoJ::::::::JCNZ;F:f:vYxY:B::::::F:;jnJ:j;<ja:MLqnFYM:>
::::::::N<<jyyiy=J:B::::::V:;J<>:;:wyyN::wyyyq<:v:<J>j>J?>:Q:S:UJ:n;v;
;JBB:]:_J:V<^<nYvY::::::::::::::::::::::::::::::::::::::::::::::::::::
:::::::::::::::::::=Z<^:>::::::::::=Jyyy;d:yayYZHQ:>:;`:Z@o<fc[_hb_ds?
h_?^on_j>^?GhoGfnGgioGSZ:^bM?^?GH:Y:ETV:WCKaTMcDGLyVDurB:;::::::A:;:::
:::::::s:?B:Igs;L:<:^aYHCcbiM<;j;@j:HZDFZLVjrs:YZEF\\;NZBP@N[<njeb:KvF
Dk;aa=v;::p]<>kkZVHC>::::::vYxy;<:C:=J:vYxY;J:jt]:qe:nf=j:F;HjSA:?JGK:
^:NZ;j<j:Djyyyy=J<JQ@jFf:F`<F:<j<j?D:;B:E:Sb:;d:;C<aZ:n^@v;sjyyiyiB]mt
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:<:::::::>=?R:AJ:^:vYxY:B::::::n:;JJI:<::::::::::::vYxI:;Z::::::::::::
::::::::yay=J:B::::::::j`<JBB:;xyfpyyyyAvAjtA:QB:n>^;UTRZ_?Z:>DEZ:F[<J
:Dj<JDDJtG[;f<Vy<N^y]:JBAZ:>Z;>Z<B>cTTUUSaEBWTSiEB_tUUURWE:=J<jP@:OB:;
JSd:<J;<J:N`lfv=^q;f:^<;W:Ut:eJ:VY;fyB:>L<J:@Z<B>:::C:Uk:^:>x;B:O:_c<B
:?J:V`kFx=r:EB:;JDJ`f<VYZ:JBG:;R:DZJ:a=:J<jP>:C:[q:N;N@B:?J:>iswu=r:<:
c:QY:c:qi:Z:JB?Z:>Z;b:<k\\PqklpnLPJHovNfi_G=B:C:Uk:^:>X;B:O:_c<B:?J:V_
wvx=r:o>G;OS:UTRc=fdOo];fZ;R;Lncp^oB:arOMeU=dMaTL@k<LonpplHA:t`KVZ?gh]
;bRY]MfBJ]DqjhPtPQsF\\@fboWF>^@Oh[;^Z?gh];EDXceV=MC<J>@lQPnAMnQ@NbLYBx
j]xoZ<OsyKI@NbLyDxjZ<JtQNtQLtyR[xoZB>Hni@>YVKciDyDLcy]=D:<[:_xIHby?cnA
\\IFbdve>we?_xYgB@nAlj:F:=b:?b:?B:?j:@j:VZ;FZ:^:dj:<J;Lj;]arW:sJ:hj:@j
:N:KfFD=Ob:Cb[QkODJ<\\rS<jSDJ<HrvFi=BA:JS:RIpM;LJ>A]k>Z:V[::^;yayI:<J:
JB<Z:VY[B:IZ:>:sg:<J:<j:DZ\\VdsWhngfgK<B:UK:^:>X=B:AB:<J::::::::::::::
:::::::::::::::::::::::::::::::::::::::::6:" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 63 "We will start by subjecting H to only the first constra
int, c1." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{MPLTEXT 1 0 42 "re
start: with(plots):with(VectorCalculus):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 87 "c1 := x^2 + y^2 - z = 0: c2 := x^2 + 3 * y^2 = 1:  H \+
:= 5 * x^2 + y^2 + 5 * z^2-6 * z:\011" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 15 "r := [x, y, z]:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 21 "gH := Gradient(H, r);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 28 "gc1 := Gradient(lhs(c1), r);" }}}{EXCHG {PARA 0 "" 0 
"" {TEXT -1 37 "Set up the three equations containing" }{MPLTEXT 0 21 
1 " " }{XPPEDIT 18 0 "lambda[1];" "6#&%'lambdaG6#\"\"\"" }{TEXT -1 0 "
" }{MPLTEXT 0 21 1 "." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "eq
 := seq(gH[k] = lambda[1] * gc1[k], k = 1..3);" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 0 "" }{TEXT 259 0 "" }{TEXT -1 103 "Solve the system of \+
equations consisting of the constraint equation and the three equation
s containing " }{XPPEDIT 18 0 "lambda[1];" "6#&%'lambdaG6#\"\"\"" }
{TEXT -1 0 "" }{MPLTEXT 0 21 1 "." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
182 "Note how the use of backward quotes on the left of the assignment
 operator permits the creation of more suggestive notation. The subscr
ipt selector must be placed outside the quotes." }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 47 "sol_1 := solve(\{eq, c1\}, \{x, y, z, lambda[1
]\});" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 55 "Now we compute values of H for each of these solutions." 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "V1 := evalf(seq(subs(sol_
1[k], H), k = 1..3));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 84 "We note a maximum value for H of 0 and a \+
minimum of -1.25.  Compare the graphs of   " }{XPPEDIT 18 0 "H^(-1);" 
"6#)%\"HG,$\"\"\"!\"\"" }{TEXT -1 4 "(0) " }{MPLTEXT 0 21 1 " " }
{TEXT -1 3 "and" }{MPLTEXT 0 21 1 " " }{TEXT -1 0 "" }{XPPEDIT 18 0 "H
^(-1);" "6#)%\"HG,$\"\"\"!\"\"" }{TEXT -1 7 "(-1.25)" }{MPLTEXT 0 21 
1 " " }{TEXT -1 21 "with the graph of c1." }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 120 "PH[0] := implicitplot3d(H = 0, x = -1..1, y = \n-1
.5..1.5, z = 0..2, numpoints = 2000, style=patchnogrid, shading=zhue):
\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 75 "Pc1 := plot3d(x^2+y^2
, x = -1..1, y = -1..1, color = red, style=wireframe):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "display(PH[0], Pc1, axes=framed, or
ientation=[60,60]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 
260 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 305 "You will be able to c
onvince yourself, by rotating the plot on the screen, that H is tangen
t to c1 at its lowermost extreme. Nonetheless, this graph does not rep
resent a solution to the optimization problem. There are many points o
n c1 that yield larger values of H than 0. An obvious choice is [1, 1,
 2]." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "subs(\{x = 1, y = 1
, z = 2\}, H);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 405 "This is just one of many instances in which a solut
ion to an equation is not necessarily the solution to the problem. Rem
ember, we set up equations based on conditions the solution must meet.
 It's the \"if\" side of a proposition. If X is a solution then it wil
l satisfy this equation. However, the \"only if\" side, \"If X is a so
lution to the equation then it is a solution to the problem.\" may not
 be true. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 33 "Now the same thing for H =  -1.25" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 128 "PH[-1.25] := implicitplot3d(H = -1.25, x = -1
..1, y = \n-1.5..1.5, z = 0..2, numpoints = 2000, style=patchnogrid, s
hading=zhue):\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "display(
PH[-1.25], Pc1, axes=framed, orientation=[180,80] );" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 92 "Here the graph of H is tangent to the graph of \+
c1 at each end of the ovoid graph of H. This " }{TEXT 261 2 "is" }
{TEXT -1 88 " a solution. There are no points on c1 that will make the
 value of H smaller than -1.25." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 
"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 54 "Now we turn our attention to
 optimizing H on c2 alone." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
28 "gc2 := Gradient(lhs(c2), r);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 32 "Set up the equations containing " 
}{XPPEDIT 18 0 "lambda[2];" "6#&%'lambdaG6#\"\"#" }{TEXT -1 1 "." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "eq2 := seq(gH[k] = lambda[2]
 * gc2[k], k = 1..3);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 74 "Solve the system consisting of the constr
aint and the equations containing" }{MPLTEXT 0 21 1 " " }{XPPEDIT 18 
0 "lambda[2];" "6#&%'lambdaG6#\"\"#" }{TEXT -1 0 "" }{MPLTEXT 0 21 1 "
." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "sol_2 := solve(\{eq2, \+
c2\}, \{x, y, z, lambda[2]\});" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 37 "Compute the value of H at each point.
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "V2 := evalf(seq(subs(so
l_2[k], H), k = 1..3));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 53 "Now we compare the plot of H = 3.2 to the
 plot of c2." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 116 "PH[3.2] :=
 implicitplot3d(H = 3.2, x = -1..1, \ny = -1..1, z = 0..2, axes = fram
ed, numpoints = 2000, color = blue):\n" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 3 "c2;" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 92 "Pc2 := implicitplot3d(c2, x \+
= -1..1, \ny = -1..1, z = 0..2, color = red, style = wireframe):\n" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "display(PH[3.2], Pc2, axes \+
= framed);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 206 "H is tangent to c2 at its outermost extreme. Again, this
 plot does not represent a solution to the optimization problem. It is
 certainly clear that many points on c2 will yield values for H larger
 than 3.2." }{MPLTEXT 0 21 1 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 133 "PH[-1.466666667] := implicitplot3d(H = -1.466666667, x = -1..
1, \ny = -1..1, z = 0..2, axes = boxed, numpoints = 2000, color = blue
):\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "display(PH[-1.46666
6667], Pc2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 94 "Now the graph of \+
H fits just inside the graph of c2.  This is a minimum. There is no le
vel set" }{MPLTEXT 0 21 1 " " }{OLE 1 4181 1 "[xm]Br=WfoRrB:::wk;nyyI;
G:;:j::>:B>N:F:nyyyyy]::yyyyyy::::::::::::::::::::::::::::::::::::::::
:::::::::::::::::::::::::::::::fyyyyya:nYf::G:jy;:::::::::::::::::::::
::::::::::::::::::::::::::::::::::::::::::::::::::::::::::JcvGYMt>^:fB
WMtNHm=;:::::::n<Ejyyiy=?Z:VZHQ:>:;`:Z@[::JJeWrLu=XK[Aj;J:T:<:=ja^GE=;
:::::::::N;?R:yyyyyy:>:<::::::JDJ:j:VBYmp>HYLkNG>::::::::N<?R:=:EJ:vYx
Y:B::::::F:;jnJ:j;<ja:MLqnFYM:>::::::::N<<jyyiy=J:B::::::V:;J<J:JyyI?:
Jyyyyg:n:v:<J>JyK?j?J@j@>:W:YJ:><<jBJC>:a:c:e:gJ:v<nYvY:::::::::::::::
::::::::::::::::::::::::::::::::::::::::::::::::::::::=Z<B:<::::::::::
=Jyyy;d:yayYZHQ:>:;`:Z@o<fc[_hb_ds?h_?^on_j>^?GhoGfnGgioGSZ::RSPN]v;fb
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@[;;B:::::::JFNZ;V:;J<jysy;Z::::::j>J<IZ:B::::::::::::jysy:>:<::::::::
:::::::::::vYxI:;Z::::::::J=EKKpL;LJ>A]k>Z:N[<^Z<@?Kc:CrZ[E:[e:CbZMkNr
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Z:FZ<vyyyy:^:n_;F=E:]c:=B:<j<j?D:;B:E:Sb:;d:;S;aZ:n^@v;sjyyiyAZ:fc:><<
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;b::::^:f?=J<JrAZ:N[:>:_;<J;<J:>_cnj;r:EB:^<nBe:qAB:>L=B:;R:DZJ:a=:J<j
P>:C:[q:N;N@B:?J:>cjGb=r:EB:^<cq?AJDjwO>B:>L;B:;R:DZJfboWFrNNfi_G=B:C:
Uk:^:>X;B:MJ:N`DZ:F:<jP<JXAJAKMjB@j`@Pt:eTOER:ER:PZ>[oB:a:t>:EuV:bU@;e
=Z\\wGH;J]Zo[@:>^@SR:C:j\\jj:Y;Bw[j]BN_A[lDB:cYK:B>H^;kAQNte<xb:B><NtG
by?C[r==dDZYgB@>B:=b:?J;<:=j:<j;@j:^:dj:<Z>VjB;>ZEF:N:Gf>?[>:;B:Ob:>?r
Z[MrbZM;;j?<:G;Sjysy=Z:>:>\\:B:qQBv:;J:^q<B:<j:DZ\\VdsWhngfgK<B:US:;J<
JrGZ:VZ:B:;:::::::::::::::::::::::::::6:" }}{PARA 0 "" 0 "" {MPLTEXT 
0 21 1 " " }{TEXT -1 51 "| p<-1.466666667, which intersects the graph \+
of c2." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 108 "Now we put it all together.  Since both constraints appl
y, we need to look at the intersection of c1 and c2." }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 28 "E := eliminate(\{c1, c2\}, z);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 64 "He
re we have z in terms of x and y. Now we need y in terms of x." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "solve(E[2, 1], y); " }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 81 "Th
is gives two parameterizations which, together, cover the line of inte
rsection." }{MPLTEXT 0 21 1 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 72 "L1 := [t, 1/3 * sqrt(-3 * t^2 + 3), t^2 + (1/3 * sqrt(-3 * t^2
 + 3))^2]:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 73 "L2 := [t, -1/
3 * sqrt(-3 * t^2 + 3), t^2 + (1/3 * sqrt(-3 * t^2 + 3))^2]:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 111 "Const := spacecurve(\{L2, L
1\}, t = -1..1, color = black, thickness = 3, numpoints = 1000, axes =
 framed):\nConst;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 90 "Now we will \+
optimize H subject to both constraints.  Set up the three equations co
ntaining" }{MPLTEXT 0 21 1 " " }{XPPEDIT 18 0 "lambda[1];" "6#&%'lambd
aG6#\"\"\"" }{TEXT -1 4 " and" }{MPLTEXT 0 21 1 " " }{XPPEDIT 18 0 "la
mbda[2];" "6#&%'lambdaG6#\"\"#" }{TEXT -1 0 "" }{MPLTEXT 0 21 1 "." }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 77 "eq3 := seq(gH[k] = evalm(la
mbda[1] * gc1[k] + lambda[2] * gc2[k]), k = 1..3);" }}}{EXCHG {PARA 0 
"" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 103 "Solve the system
 of five equations consisting of the two constraints and the three equ
ations containing" }{MPLTEXT 0 21 1 " " }{XPPEDIT 18 0 "lambda[1];" "6
#&%'lambdaG6#\"\"\"" }{TEXT -1 4 " and" }{MPLTEXT 0 21 1 " " }
{XPPEDIT 18 0 "lambda[2];" "6#&%'lambdaG6#\"\"#" }{TEXT -1 0 "" }
{MPLTEXT 0 21 1 "." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "sol_3
 := solve(\{eq3, c1, c2\}, \{x, y, z, lambda[1], lambda[2]\});" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 66 "He
re are the values of H corresponding to each of these solutions." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "V3 := evalf(seq(subs(sol_3[k
], H), k = 1..4));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 
"" 0 "" {TEXT -1 41 "Since the largest value of H is 4 we plot" }
{MPLTEXT 0 21 1 " " }{OLE 1 4173 1 "[xm]Br=WfoRrB:::wk;nyyI;G:;:j::>:B
>N:F:nyyyyy]::yyyyyy::::::::::::::::::::::::::::::::::::::::::::::::::
:::::::::::::::::::::fyyyyya:nYf::G:jy;:::::::::::::::::::::::::::::::
::::::::::::::::::::::::::::::::::::::::::::::::JcvGYMt>^:fBWMtNHm=;::
:::::n<Ejyyiy=?Z:VZHQ:>:;`:Z@[::JJeWrLu=XK[Aj;J:T:<:=ja^GE=;:::::::::N
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::::::::::::::::::::::::::::::::::::::::::::=Z<B:<::::::::::=Jyyy;d:ya
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FGYMq>>Wlj:gmlB:;:::::JJ<J;vYxI:;Z::::::J<B:CS:>Z:J;vCS=[LsfFaMR>@>Z::
::::::kJ;@j;>:C:yayA:<::::::M:Gv:B::::::::::::jysy:>:<::::::::::::::::
:::vYxI:;Z::::::::J=EKKpL;LJ>A]k>Z:N[<^Z<@?Kc:CB\\cKTDJ<DrNF?>ZZ@:;:::
:jysyAB:^:F:;B:yayQ:<J:jiPZ:Vi<J`H:=j>r:GW:<J;nK>:C:?R:=B:EJ:<j:Djyyyy
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yYJM:eY:<j?<:G;Sj`@@RS;B:[l<B:Mb:b:E:cb:cQVDjDjwEJKxI;B:>l;B:;R:;b:;B>
cTTP@:nG=J<B:US:F[:>Z:N`DZ:F:;B:]C:sW:E:cJJAj`EjD<:qQ:uI<:[V:R:D::::C:
Uk:^:>x;B:SB:;B:_;<j;<:KCGMK??RHj<<:c:GlDjw;<:[n:<J:@Z<:VH::C:UK:^Z:Jr
G:MJ:N@B:=J:<J^>Z=B:^<cq?AJDjwO>B:>l;>:@Z<j\\PqkZMKkxPpFZ:^:f?=J<Jr?Z:
F;;JSd:<j:B:UC:sW:W>G;]R:UTRc=PpdPo<@j<@Z?BKRG<jCZH;jlHA:t`Kjt:bRY]M:n
BRWb;:;cK@@J<:fBFFjQ:LYBxj]BN_A[lDB:cYK:B>H^;kAQNte<xb:B><NtGby?C[r==d
DZYgB@>B:=b:?J;<:=j:<j;F:CZDFZ:B;A>J:hj:J;njLN>LZkJ:<J?D:KKNB\\c;F?>:Q
B:n>^;yayI:<J:JB<Z:VY[j=J:^q<B:<j:DZ\\VdsWhngfgK<B:US:;J<JrGZ:VZ:B:;::
::::::::::::::::::::::::2:" }{TEXT -1 65 " together with the curve of \+
inteersection of the two constraints." }{TEXT 262 0 "" }{TEXT -1 0 "" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 122 "PH[4] := implicitplot3d(
H = 4, x = -2..2, y = -2..2, \nz = -2..2, numpoints = 2000, color = tu
rquoise, style = wireframe):\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 37 "display(PH[4], Const, axes = framed);" }}}{EXCHG {PARA 0 "" 0 
"" {TEXT -1 34 "This is the plot of the maximum H." }{MPLTEXT 0 21 1 "
 " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 135 "The sol_3 corresponding to \+
-2.050000001, is imaginary. The smallest value of H corresponding to a
 real solution is, -1.111111111.  Plot" }{MPLTEXT 0 21 1 " " }{OLE 1 
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;>:C:[q:<j;<Z:>:::::::::::::::::::::::::::::::::::::::::::::::::::::::
::::::::::::::::::::::::::::::4:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 145 "PH[-1.111111111] := implici
tplot3d(H = -1.111111111, x = -1..1, \ny = -1..1, z = 0..1.5, numpoint
s = 2000, color = turquoise, style = wireframe):\n" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 48 "display(Const, PH[-1.111111111], axes = fra
med);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 34 "This is the plot of the \+
minimum H." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 199 "It is clear that r
estriction of H by two constraints is the same as restriction by a sin
gle constraint represented by the line of intersection of the two grap
hs representing the constraint equations." }{MPLTEXT 0 21 1 " " }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 227 "The power and usefulness of Maple
 in presenting these plots is apparent.  The question as to whether a \+
particular solution really represents a maximum or minimum is frequent
ly answerable by inspection that can then be verified." }{MPLTEXT 0 
21 1 "\011" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 263 0 "" }}
{PARA 0 "" 0 "" {TEXT 264 8 "Practice" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 120 "1.What are the radii of the largest and smallest  sphere
s, centered at the origin, \ntangent to the ellipse described by " }
{OLE 1 4657 1 "[xm]Br=WfoRrB:::wk;nyyI;G:;:j::>:B>N:F:nyyyyy]::yyyyyy:
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fyyyyya:nYf::G:I:wAyA:::::::::::::::::::::::::::::::::::::::::::::::::
::::::::::::::::::::::::::::::NDYmq^H;C:ELq^H_mvJ::::::::gj<vyyuyj<j;t
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::::::::::::::::::::::::::::::4:" }{TEXT -1 4 " ?\n\n" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 211 " 2. Use the method of Lagrange multiplie
rs to find the radius of the sphere, centered \nat (a, b) and tangent \+
to the line Ax + By + Cz = K, and thereby derive a formula for \nthe d
istance from a point to a line. \n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 277 "3. For each of the following functions, and sets of constraint
s:\n      a) Plot the sets of constraints. \n      b) Find the maximum
 and minimum of the functions subject to the indicated constraints.\n \+
     c) Plot the inverse image of the solutions together with the cons
traints." }}{PARA 0 "" 0 "" {TEXT -1 24 "                     1. " }
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{TEXT -1 12 "            " }{XPPEDIT 18 0 "10*z-5*x^2-3*y^2 = 20;" "6#
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-1 21 "                     " }{XPPEDIT 18 0 "(x-z)^2+(z-2)^2 = 1;" "6
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