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{SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT 256 25 "Several Variables, Part
 2" }}{PARA 256 "" 0 "" {TEXT -1 10 "March 2010" }}{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 261 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 27 " we obtain jut one solution" }}{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 alwa
ys return just this answer, but we have options to find other solution
s.  For one, we can we can specify a starting point for the search of \+
a stationary point:" }}{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 example 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 -1 105 "If we pick a rectangle without a stationary poin
t, maple cannot create one for us and 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 st
ationary points 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);\n
C := D[2,2](f)(0,0);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 23 "and then \+
we calculate 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 "As 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 obtain" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 7 "d(0,0);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 44 "while at the other stationary poin
t 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 have a relative minimum at (1,1)." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 142 "We conclude this porti
on with a contour plot and with a graph of the function.  The relative
 minimum and the saddle 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,vi
ew=[-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 262 8 "Problem:" }{TEXT -1 41 " Find the minimum
 value 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-" }{TEXT -1 30 "   subject to the constraint  " }{XPPEDIT 
18 0 "2*x-3*y-4*z = 49;" "6#/,(*&\"\"#\"\"\"%\"xGF'F'*&\"\"$F'%\"yGF'!
\"\"*&\"\"%F'%\"zGF'F,\"#\\" }{TEXT -1 33 ".\nThis is Example 2 on pag
e 723.\n" }{TEXT 263 9 "Solution:" }{TEXT -1 43 " First we define the \+
repective functions   " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "f := (x,y
,z) -> 2*x^2 +y^2 +3*z^2;\ng := (x,y,z) -> 2*x -3*y - 4*z;" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 133 "The Lagrange multiplier methods leads to
 the following system of equations.  We abbreviate it as LMS  (Lagrang
e Multiplier System).  " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 136 
"LMS := \nD[1](f)(x,y,z)=lambda*D[1](g)(x,y,z),\nD[2](f)(x,y,z)=lambda
*D[2](g)(x,y,z),\nD[3](f)(x,y,z)=lambda*D[3](g)(x,y,z),\ng(x,y,z) = 49
;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 123 "This is the resulting syste
m given in the book.  We solve the problem with the fsolve command.  N
otice the braces on LMS,  " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "fsolv
e(\{LMS\});" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 39 "This is the soluti
on found in the book." }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 16 "# --------------" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
264 8 "Problem:" }{TEXT -1 94 "  Find the highest and lowest point on \+
the curve defined by the intersection of the surfaces  " }{XPPEDIT 18 
0 "x^2+y^2 = z^2;" "6#/,&*$%\"xG\"\"#\"\"\"*$%\"yGF'F(*$%\"zGF'" }
{TEXT -1 7 "  and  " }{XPPEDIT 18 0 "x+2*z = 4;" "6#/,&%\"xG\"\"\"*&\"
\"#F&%\"zGF&F&\"\"%" }{TEXT -1 4 "    " }}{PARA 0 "" 0 "" {TEXT 265 
11 "Solution:  " }{TEXT -1 214 "This leads to a problem with two const
raint equations, and thus two multipliers.  First wedefine the pertine
nt functions.  The height is given by the z-coordinate, which leads 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-" }{TEXT -1 27 " subject to the constraint " }
{XPPEDIT 18 0 "2*x-3*y-4*z = 49;" "6#/,(*&\"\"#\"\"\"%\"xGF'F'*&\"\"$F
'%\"yGF'!\"\"*&\"\"%F'%\"zGF'F,\"#\\" }{TEXT -1 56 ".  As before, we n
eed to define functions f and g:      " }}{PARA 0 "> " 0 "" {MPLTEXT 
1 0 66 "f := (x,y,z) -> 2*x^2 +y^2 +3*z^2;\ng := (x,y,z) -> 2*x -3*y -
 4*z;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 77 "Minimization is straight
forward (notice that we need braces on the constaint)" }}{PARA 0 "> " 
0 "" {MPLTEXT 1 0 35 "Minimize( f(x,y,z), \{g(x,y,z)=49\});" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 76 "Reworking the second example (highest and lowest point on
 a curve) leads to " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 73 "f := (x,y,z)
 -> z;\ng := (x,y,z) -> x^2+y^2-z^2;\nh := (x,y,z) -> x + 2*z; " }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 14 "Highest point:" }}{PARA 0 "> " 0 "
" {MPLTEXT 1 0 48 "Maximize( f(x,y,z), \{g(x,y,z)=0, h(x,y,z) = 4\});
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 13 "Lowest point:" }}{PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 48 "Minimize( f(x,y,z), \{g(x,y,z)=0, h(x,y,z) = 4
\});" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 
"" 0 "" {TEXT -1 165 "The optimization package is convenient to find e
xtrema on bounded regions (extrame on boundary or in the inside of the
 region).  Here we have inequality constraints." }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 38 "Example:  Find the extreme values of  " }{XPPEDIT 
18 0 "f(x,y) = x^2+2*y^2-2*x+3;" "6#/-%\"fG6$%\"xG%\"yG,**$F'\"\"#\"\"
\"*&F+F,*$F(F+F,F,*&F+F,F'F,!\"\"\"\"$F," }{TEXT -1 29 "  subject to t
he constraint  " }{XPPEDIT 18 0 "x^2+y^2 <= 10;" "6#1,&*$%\"xG\"\"#\"
\"\"*$%\"yGF'F(\"#5" }{TEXT -1 109 " .  This is Example 3, page 723.\n
We applied the maple statements directly (without defining functions f
irst)." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "Maximize( x^2+2*y^2-2*x+3
 , \{x^2 + y^2 <= 10\});" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 
"Minimize( x^2+2*y^2-2*x+3 , \{x^2 + y^2 <= 10\});" }}}{EXCHG {PARA 0 
"" 0 "" {TEXT -1 88 "The maximizer (-1,3) lies on the boundary, and th
e minimizer (1,0) is inside the region." }}{PARA 0 "" 0 "" {TEXT -1 1 
" " }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }
}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 83 "Occasionally we need to select i
nitial approximations.  Our example is the function" }}{PARA 0 "> " 0 
"" {MPLTEXT 1 0 28 "f := (x,y) -> cos(x)*sin(y);" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 34 "plot3d(f,-5..5,-5..5,axes=normal);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 67 "The graph shows that f attains its
 maximum at several points.  The " }{TEXT 266 8 "Maximize" }{TEXT -1 
21 " command takes us to " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "Maximi
ze(f(x,y));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 22 "This translates in
to  " }{XPPEDIT 18 0 "x = 0;" "6#/%\"xG\"\"!" }{TEXT -1 7 "  and  " }
{XPPEDIT 18 0 "y = Pi/2;" "6#/%\"yG*&%#PiG\"\"\"\"\"#!\"\"" }{TEXT -1 
36 "   .   But the function equas 1 at  " }{XPPEDIT 18 0 "x = Pi;" "6#
/%\"xG%#PiG" }{TEXT -1 6 " and  " }{XPPEDIT 18 0 "y = -Pi/2;" "6#/%\"y
G,$*&%#PiG\"\"\"\"\"#!\"\"F*" }{TEXT -1 45 "  as well.  By selecting x
=3 and y=-1.5, the " }{TEXT 267 12 "Maximization" }{TEXT -1 41 " routi
ne takes us to the desired point.  " }{MPLTEXT 1 0 1 " " }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "Maximize(f(x,y), initialpoint=\{x=3
,y=-1.5\});" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}}{MARK 
"0 1 0" 10 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 
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