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{SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 257 "" 
0 "" {TEXT -1 37 "Extrema of Functions of Two Variables" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 166 "In this assignment we will be solving si
multaneous equations and using matrices in order to locate possible re
lative maxima and minima.  After defining the function  " }{XPPEDIT 
18 0 "f" "6#%\"fG" }{TEXT -1 46 "  and determining the partial derivat
ives of  " }{TEXT 256 1 "f" }{TEXT -1 90 " , we set those partials equ
al to zero and solve the equations.  We obtain an unexpected \223" }
{TEXT 257 6 "RootOf" }{TEXT -1 83 " \224 in our first solution, so the
 example shows how to resolve that.  The notation \223" }{TEXT 258 4 "
s[1]" }{TEXT -1 54 "\224refers to the first set of brackets in the sol
ution \223" }{TEXT 259 1 "s" }{TEXT -1 6 "\224and \223" }{TEXT 260 9 "
allvalues" }{TEXT -1 33 "\224yields the two solutions where \223" }
{TEXT 261 6 "RootOf" }{TEXT -1 39 "\224 has occurred.  At a point, suc
h as  (" }{XPPEDIT 18 0 "(x[0],y[0])" "6$&%\"xG6#\"\"!&%\"yG6#F&" }
{TEXT -1 94 ") , where both partial derivatives are zero we test to se
e if the determinant of the Hessian, " }{XPPEDIT 18 0 "H(x,y)" "6#-%\"
HG6$%\"xG%\"yG" }{TEXT -1 26 " , a matrix with top row  " }{XPPEDIT 
18 0 "f[xx]" "6#&%\"fG6#%#xxG" }{TEXT -1 3 "   " }{XPPEDIT 18 0 "f[xy]
" "6#&%\"fG6#%#xyG" }{TEXT -1 20 "   and bottom row   " }{XPPEDIT 18 
0 "f[yx]" "6#&%\"fG6#%#yxG" }{TEXT -1 3 "   " }{XPPEDIT 18 0 "f[yy]" "
6#&%\"fG6#%#yyG" }{TEXT -1 17 "  evaluated at  (" }{XPPEDIT 18 0 "(x[0
],y[0])" "6$&%\"xG6#\"\"!&%\"yG6#F&" }{TEXT -1 127 ")   is positive or
 negative.  The positive case yields a relative maximum or minimum, wh
ile the negative case indicates that  (" }{XPPEDIT 18 0 "(x[0],y[0])" 
"6$&%\"xG6#\"\"!&%\"yG6#F&" }{TEXT -1 150 ")   is a saddle point.  We \+
test further if the determinant of the Hessian is positive because thi
s can only occur if the second partial derivatives,  " }{XPPEDIT 18 0 
"f[xx]" "6#&%\"fG6#%#xxG" }{TEXT -1 8 "  and   " }{XPPEDIT 18 0 "f[yy]
" "6#&%\"fG6#%#yyG" }{TEXT -1 150 " , have the same sign and may be in
terpreted as indicating concave \223up\224 or \223down\224.  Because t
he entry of the Hessian, H , in the  (1,1)  position is  " }{XPPEDIT 
18 0 "f[xx]" "6#&%\"fG6#%#xxG" }{TEXT -1 147 " , we simply evaluate  H
[1,1]  at the point in question and we have our answer.  We will need \+
three packages for Maple: student, linalg, and plots." }}{PARA 0 "" 0 
"" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 262 8 "Example:" }
{TEXT -1 45 "  Find all critical points of the function   " }{XPPEDIT 
18 0 "f(x,y)=x^3+3*x*y^2-4*y^3-15*x" "6#/-%\"fG6$%\"xG%\"yG,**$F'\"\"$
\"\"\"*(F+F,F'F,F(\"\"#F,*&\"\"%F,*$F(F+F,!\"\"*&\"#:F,F'F,F2" }{TEXT 
-1 112 "  and determine which, if any, are relative maxima or minima o
r saddle points.  For simplicity, we will define  " }{TEXT 263 1 "f" }
{TEXT -1 19 "  as an expression." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 50 "restart: with(student): with(plots): with(linalg):" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "f:=x^3+3*x*y^2-4*y^3-15*x;
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "fx:=diff(f,x);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "fy:=diff(f,y);" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 38 "Solve the simultaneous equations for  " }
{TEXT 264 1 "x" }{TEXT -1 7 "  and  " }{TEXT 265 1 "y" }{TEXT -1 2 " .
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "s:=solve(\{fx=0,fy=0\},
\{x,y\});" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 130 "### WARNING: \+
allvalues now returns a list of symbolic values instead of a sequence \+
of lists of numeric values\ns1:=allvalues(s[1]);" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 20 "H:=hessian(f,[x,y]);" }}}{EXCHG {PARA 0 "" 0 
"" {TEXT -1 45 "Insert the first of the solutions listed in \223" }
{TEXT 266 2 "s1" }{TEXT -1 107 "\224 into H .  The use of  \223op\224 \+
reminds Maple that  H  is a matrix and allows access to the components
 of  H ." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "H1:=subs(s1[1],
op(H));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "a1:=det(H1);" }}
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 21 "Testing the sign of  " }{XPPEDIT 
18 0 "f[xx]" "6#&%\"fG6#%#xxG" }{TEXT -1 22 "  or fxx at the point." }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "b1:=subs(s1[1],H[1,1]);" }
}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 128 "Because the determinant of the \+
Hessian is positive and b1 is positive (indicating \221concave up\222)
, there is a relative minimum at\n" }{TEXT 267 2 "s1" }{TEXT -1 5 "[1]
=(" }{XPPEDIT 18 0 "sqrt(5)" "6#-%%sqrtG6#\"\"&" }{TEXT -1 6 " , 0)." 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "v1:=subs(s1[1],f);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "s2:=s1[2];" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 57 "This line was to demonstrate what  \221s1[2]
\222 would produce." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "H2:=
subs(s2,op(H));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "a2:=det(
H2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "b2:=subs(s2,H[1,1])
;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 36 "Because the Hessian is posit
ive at  " }{TEXT 268 2 "s2" }{TEXT -1 7 "  and  " }{TEXT 269 2 "b2" }
{TEXT -1 75 "  is negative (indicating \221concave down\222), there is
 a relative maximum at  " }{TEXT 270 3 "s2 " }{TEXT -1 3 "= (" }
{XPPEDIT 18 0 "sqrt(5)" "6#-%%sqrtG6#\"\"&" }{TEXT -1 36 " ,0) . Now w
e compute the value of  " }{TEXT 272 1 "f" }{TEXT -1 5 "  at " }{TEXT 
271 2 "s2" }{TEXT -1 2 " ." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
15 "v2:=subs(s2,f);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "H3:=
subs(s[2],op(H));\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 1 "\221" }
{TEXT 274 4 "s[2]" }{TEXT -1 49 "\222 is the second set listed of the \+
original set, \221" }{TEXT 273 1 "s" }{TEXT -1 2 "\222." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "a3:=det(H3);" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 77 "Because the determinant of the Hessian is negative,w
e have a saddle point at " }{TEXT 275 4 "s[2]" }{TEXT -1 10 " = (2 ,1)
." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "v3:=subs(s[2],f);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "H4:=subs(s[3],op(H));" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "a4:=det(H4);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "v4:=subs(s[3],f);" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 64 "Because the determinant of the Hessian is negat
ive at s [3] = ( " }{XPPEDIT 18 0 "-2,-1" "6$,$\"\"#!\"\",$\"\"\"F%" }
{TEXT -1 70 " ), there is a saddle point there.  Now let \222s see wha
t the graph of  " }{TEXT 276 1 "f" }{TEXT -1 13 "  looks like." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "plot3d(f,x=-6..6,y=-5..5,axe
s=BOXED);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 102 "contourplot(f
,x=-6..6,y=-4..4,contours=[-28,-24,-20,-16,-12,-8,-4,0,4,8,12,16,20,24
,28],color=black);\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 174 "Check out the location on the co
ntour plot of the four critical points we located above.  See if they \+
make sense as far as being relative maxima, minima, or as saddle point
s." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 167 "Solve for relative maxima, minima, and saddle points by using \+
MAPLE and by pencil and paper.  Include a MAPLE contour plot, but scal
e it down so as to not waste paper." }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 5 "  1. " }{XPPEDIT 18 0 "f(x,y)=2*x^3+6*x*y^2+3*y^3-150*x+4" "6#/-
%\"fG6$%\"xG%\"yG,,*&\"\"#\"\"\"*$F'\"\"$F,F,*(\"\"'F,F'F,F(F+F,*&F.F,
*$F(F.F,F,*&\"$]\"F,F'F,!\"\"\"\"%F," }{TEXT -1 13 "          2. " }
{XPPEDIT 18 0 "f(x,y)=4*x^3+3*x*y^2-y^3-24*x+4" "6#/-%\"fG6$%\"xG%\"yG
,,*&\"\"%\"\"\"*$F'\"\"$F,F,*(F.F,F'F,F(\"\"#F,*$F(F.!\"\"*&\"#CF,F'F,
F2F+F," }{TEXT -1 3 "   " }}}}{MARK "41" 0 }{VIEWOPTS 1 1 0 3 2 1804 
1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }