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{SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT 256 25 "Several Variables, Part
 2" }}{PARA 256 "" 0 "" {TEXT -1 10 "March 2009" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 59 "This is a continuation of the Several Variables worksheet. " }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{SECT 1 {PARA 3 "" 0 "
" {TEXT -1 44 "Stationary Points, Extrema and Saddle Points" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "restart; with(plots):" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 172 "We search for stationary points, and onc
e they are found, we will decide whether we have a maximum, a minimum \+
or a saddle point.  The example throughout is the function    " }
{XPPEDIT 18 0 "f(x,y) = x^3+y^3-3*x*y;" "6#/-%\"fG6$%\"xG%\"yG,(*$F'\"
\"$\"\"\"*$F(F+F,*(F+F,F'F,F(F,!\"\"" }{TEXT -1 52 " .\nFirst we defin
e f once and for all ( or until we " }{TEXT 263 7 "restart" }{TEXT -1 
1 ")" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "f := (x,y) -> x^3 + y^3 -3*
x*y;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 33 "The first partial derivat
ives are" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "D[1](f)(x,y);\nD[2](f)(
x,y);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 95 "We have to set both quan
tities to zero, and then solve for x and y.  This can be done with the
 " }{TEXT 257 5 "solve" }{TEXT -1 8 " or the " }{TEXT 258 6 "fsolve" }
{TEXT -1 93 " command.  The key is to enclose the two equations and th
e two unknowns with curly brackets. " }}{PARA 0 "> " 0 "" {MPLTEXT 1 
0 48 "solve(\{D[1](f)(x,y)=0, D[2](f)(x,y)=0\} , \{x,y\});" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 42 "The first two pairs contain our solutions
." }}{PARA 0 "" 0 "" {TEXT -1 5 "With " }{TEXT 259 6 "fsolve" }{TEXT 
-1 11 " we obtain " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "fsolve(\{D[1]
(f)(x,y)=0, D[2](f)(x,y)=0\} , \{x,y\});" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT 260 6 "fsolve" }{TEXT -1 177 " as such will always return just t
his answer, but we have options to find other solutions.  For one, we \+
can we can specify a starting point for the search of a stationary poi
nt:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{MPLTEXT 1 0 57 "fsolve(\{D[1](f
)(x,y)=0, D[2](f)(x,y)=0\} , \{x=0.8,y=1.2\});" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 118 "Secondly, we can specify a search region; for examp
le we can look for solutions in the rectangle [0.5 , 2] x [-2 , 2]:" }
}{PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "fsolve(\{D[1](f)(x,y)=0, D[2](f)(x
,y)=0\} , \{x=0.5..2,y=-2..2\});" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
103 "If we pick a rectangle without a stationary point, maple cannot c
reate one for us.  We get the response" }}{PARA 0 "> " 0 "" {MPLTEXT 
1 0 65 "fsolve(\{D[1](f)(x,y)=0, D[2](f)(x,y)=0\} , \{x=-2..-0.01,y=-1
..1\});" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 85 "The classification of the stationary poin
ts we can do step by step.  First we define " }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 62 "A := D[1,1](f)(0,0);\nB := D[1,2](f)(0,0);\nC := D[2,
2](f)(0,0);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 23 "and then we calcul
ate d" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "A*C-B^2;" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 83 "In this case we see that we have a saddle point
 at the origin, since d is negative." }}{PARA 0 "" 0 "" {TEXT -1 67 "A
s an alternative, we could calculate d as a function of x and y.  " }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 61 "d := (x,y) -> D[1,1](f)(x,y)*D[2,2]
(f)(x,y)-D[1,2](f)(x,y)^2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
7 "d(x,y);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 23 "At the origin we ob
tain" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "d(0,0);" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 44 "while at the other stationary point we have " }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "d(1,1);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 14 "Combined with " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "D[1,
1](f)(1,1);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 48 "we see that we hav
e a relative minimum at (1,1)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 142 "We conclude this portion with a contour \+
plot and with a graph of the function.  The relative minimum and the s
addle point are clearly visible." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 
120 "contourplot(f,-2..2,-2..2,contours=[-2, -1, -0.9, -0.5, 0, 1, 2],
numpoints=2000,filledregions=true,coloring=[red,blue]);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 58 "plot3d(f,-2..2,-2..2,view=[-2..2,-2
..2,-2..2],axes=boxed);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 3 "
" 0 "" {TEXT -1 20 "Lagrange Multipliers" }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 21 "restart;\nwith(plots):" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT 266 8 "Problem:" }{TEXT -1 41 " Find the minimum value of the fu
nction  " }{XPPEDIT 18 0 "f(x,y,z) = 2*x^2+y^2+3*z^2;" "6#/-%\"fG6%%\"
xG%\"yG%\"zG,(*&\"\"#\"\"\"*$)F'F,F-F-F-*$)F(F,F-F-*&\"\"$F-*$)F)F,F-F
-F-" }{TEXT -1 30 "   subject to the constraint  " }{XPPEDIT 18 0 "2*x
-3*y-4*z = 49;" "6#/,(*&\"\"#\"\"\"%\"xGF'F'*&\"\"$F'%\"yGF'!\"\"*&\"
\"%F'%\"zGF'F,\"#\\" }{TEXT -1 33 ".\nThis is Example 2 on page 723.\n
" }{TEXT 267 9 "Solution:" }{TEXT -1 43 " First we define the repectiv
e functions   " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "f := (x,y,z) -> 2
*x^2 +y^2 +3*z^2;\ng := (x,y,z) -> 2*x -3*y - 4*z;" }}}{EXCHG {PARA 0 
"" 0 "" {TEXT -1 133 "The Lagrange multiplier methods leads to the fol
lowing system of equations.  We abbreviate it as LMS  (Lagrange Multip
lier System).  " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 136 "LMS := \+
\nD[1](f)(x,y,z)=lambda*D[1](g)(x,y,z),\nD[2](f)(x,y,z)=lambda*D[2](g)
(x,y,z),\nD[3](f)(x,y,z)=lambda*D[3](g)(x,y,z),\ng(x,y,z) = 49;" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 123 "This is the resulting system give
n in the book.  We solve the problem with the fsolve command.  Notice \+
the braces on LMS,  " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "fsolve(\{LM
S\});" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 39 "This is the solution fou
nd in the book." }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 16 "# --------------" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
268 8 "Problem:" }{TEXT -1 94 "  Find the highest and lowest point on \+
the curve defined by the intersection of the surfaces  " }{XPPEDIT 18 
0 "x^2+y^2 = z^2;" "6#/,&*$)%\"xG\"\"#\"\"\"F)*$)%\"yGF(F)F)*$)%\"zGF(
F)" }{TEXT -1 7 "  and  " }{XPPEDIT 18 0 "x+2*z = 4;" "6#/,&%\"xG\"\"
\"*&\"\"#F&%\"zGF&F&\"\"%" }{TEXT -1 4 "    " }}{PARA 0 "" 0 "" {TEXT 
269 11 "Solution:  " }{TEXT -1 214 "This leads to a problem with two c
onstraint equations, and thus two multipliers.  First wedefine the per
tinent functions.  The height is given by the z-coordinate, which lead
s to the definition of f as given below." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 73 "f := (x,y,z) -> z;\ng := (x,y,z) -> x^2+y^2-z^2;\nh :
= (x,y,z) -> x + 2*z; " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 56 "A graph
 helps to understand the problem a little better." }}{PARA 0 "> " 0 "
" {MPLTEXT 1 0 127 "implicitplot3d([g(x,y,z)=0, h(x,y,z) = 4],x=-6..2,
y=-4..4, z=0..8, color=[red,blue], axes=normal, numpoints=2000,style=h
idden);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 217 "LMS := \nD[1](f
)(x,y,z) = lambda*D[1](g)(x,y,z) + mu*D[1](h)(x,y,z), \nD[2](f)(x,y,z)
 = lambda*D[2](g)(x,y,z) + mu*D[2](h)(x,y,z),\nD[3](f)(x,y,z) = lambda
*D[3](g)(x,y,z) + mu*D[3](h)(x,y,z),\ng(x,y,z) = 0,\nh(x,y,z) = 4; " }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "fsolve(\{ LMS \},\{x,y,z,l
ambda,mu\});" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 195 "This is a first \+
solution, and judging by the graph (z=4), we have identified the highe
st point.  We include estimates for x and y in the fsolve statement be
low, in order to find the lowest point." }}{PARA 0 "> " 0 "" {MPLTEXT 
1 0 38 "fsolve(\{ LMS \},\{x=2,y=0,z,lambda,mu\});" }}}{EXCHG {PARA 0 
"" 0 "" {TEXT -1 89 "Result.  The highst point is located at (-4,0,4) \+
and the lowest point is at (4/3, 0,4/3)." }}{PARA 0 "" 0 "" {TEXT -1 
101 "Quick checks: (substitute the points into the constaint functions
, g must be zero and h must be four)" }}{PARA 0 "> " 0 "" {MPLTEXT 1 
0 50 "g(-4,0,4); g(4/3,0,4/3);\nh(-4,0,4); h(4/3,0,4/3); " }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}
{SECT 1 {PARA 3 "" 0 "" {TEXT -1 24 "The Optimization Package" }}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 63 "Here we rework the problems using \+
maple's optimization package." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "re
start;\nwith(Optimization);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 38 "We
 find the minimum of the function   " }{XPPEDIT 18 0 "f(x,y,z) = 2*x^2
+y^2+3*z^2;" "6#/-%\"fG6%%\"xG%\"yG%\"zG,(*&\"\"#\"\"\"*$)F'F,F-F-F-*$
)F(F,F-F-*&\"\"$F-*$)F)F,F-F-F-" }{TEXT -1 27 " subject to the constra
int " }{XPPEDIT 18 0 "2*x-3*y-4*z = 49;" "6#/,(*&\"\"#\"\"\"%\"xGF'F'*
&\"\"$F'%\"yGF'!\"\"*&\"\"%F'%\"zGF'F,\"#\\" }{TEXT -1 56 ".  As befor
e, we need to define functions f and g:      " }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 66 "f := (x,y,z) -> 2*x^2 +y^2 +3*z^2;\ng := (x,y,z) -> 2
*x -3*y - 4*z;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 77 "Minimization is
 straightforward (notice that we need braces on the constaint)" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "Minimize( f(x,y,z), \{g(x,y,z)=49\}
);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 76 "Reworking the second example (highest and lowest p
oint on a curve) leads to " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 73 "f := \+
(x,y,z) -> z;\ng := (x,y,z) -> x^2+y^2-z^2;\nh := (x,y,z) -> x + 2*z; \+
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 14 "Highest point:" }}{PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 48 "Maximize( f(x,y,z), \{g(x,y,z)=0, h(x,y,z) = 4
\});" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 13 "Lowest point:" }}{PARA 0 
"> " 0 "" {MPLTEXT 1 0 48 "Minimize( f(x,y,z), \{g(x,y,z)=0, h(x,y,z) \+
= 4\});" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 165 "The optimization package is convenient t
o find extrema on bounded regions (extrame on boundary or in the insid
e of the region).  Here we have inequality constraints." }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 38 "Example:  Find the extreme values of  " }
{XPPEDIT 18 0 "f(x,y) = x^2+2*y^2-2*x+3;" "6#/-%\"fG6$%\"xG%\"yG,**$)F
'\"\"#\"\"\"F-*&F,F-*$)F(F,F-F-F-*&F,F-F'F-!\"\"\"\"$F-" }{TEXT -1 29 
"  subject to the constraint  " }{XPPEDIT 18 0 "x^2+y^2 <= 10;" "6#1,&
*$)%\"xG\"\"#\"\"\"F)*$)%\"yGF(F)F)\"#5" }{TEXT -1 109 " .  This is Ex
ample 3, page 723.\nWe applied the maple statements directly (without \+
defining functions first)." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "Maxim
ize( x^2+2*y^2-2*x+3 , \{x^2 + y^2 <= 10\});" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 47 "Minimize( x^2+2*y^2-2*x+3 , \{x^2 + y^2 <= 10\})
;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 88 "The maximizer (-1,3) lies on
 the boundary, and the minimizer (1,0) is inside the region." }}{PARA 
0 "" 0 "" {TEXT -1 1 " " }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}
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