{VERSION 4 0 "IBM INTEL NT" "4.0" }
{USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 
1 0 0 0 0 1 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 
0 0 1 }{CSTYLE "2D Output" 2 20 "" 0 1 0 0 255 1 0 0 0 0 0 0 0 0 0 1 }
{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 
2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Maple Output" -1 
11 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }3 3 0 
0 0 0 1 0 1 0 2 2 0 1 }}
{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 70 "Finding Maximum and Minimu
m of Functions with three or more variables:" }}{PARA 0 "" 0 "" {TEXT 
-1 43 "Step 1: take the first partial derivatives." }}{PARA 0 "" 0 "" 
{TEXT -1 54 "Step 11: set the all the first partials equal to zero." }
}{PARA 0 "" 0 "" {TEXT -1 60 "Step 111. Solve the system of equations \+
for x1, x2, x3...xn." }}{PARA 0 "" 0 "" {TEXT -1 62 "Step IV: take all
 the second partials, and the cross partials." }}{PARA 0 "" 0 "" 
{TEXT -1 122 "Step V: form a matrix by placing the second partials on \+
the principal diagonal and the cross partials on the off  diagonal" }}
{PARA 0 "" 0 "" {TEXT -1 123 "locations. Find the principal minors of \+
the matrix  thus formed. Find the determinants of the minors. Check th
e sign of all" }}{PARA 0 "" 0 "" {TEXT -1 188 "the minors. If the mino
rs are all positive, the function is minimized. If the minors alternat
e in sign, the function has a minimum. Examine the following example: \+
see page 283 of the text." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{MPLTEXT 1 0 13 
"with(linalg):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 71 "A:=hessian(-5*x1^
2+10*x1+x1*x3-2*x2^2+4*x2+2*x2*x3-4*x3^2, [x1,x2,x3]);" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#>%\"AG-%'matrixG6#7%7%!#5\"\"!\"\"\"7%F+!\"%\"\"#7
%F,F/!\")" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "detr:=det(A);
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%detrG!$w#" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 8 "m1:=-10;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%
#m1G!#5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "m2:=minor(A,3,3)
;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#m2G-%'matrixG6#7$7$!#5\"\"!7$F
+!\"%" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "det(m2);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#\"#S" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
108 "Notes: A minor is the DETERMINANT of the submatrix remaining afte
rdeleting the ith row and jth column of the" }}{PARA 0 "" 0 "" {TEXT 
-1 123 "matrix. In the example above, if we delete the third row and t
he third column, and find the determinant of  the submatrix--" }}
{PARA 0 "" 0 "" {TEXT -1 235 "m2, note  that the determinant is the mi
nor for the submatrix formed.  A hessian is a determinant composed of \+
all the second-otder partial derivatives on the principal diagonal and
 the second order cross partials placed as off-diagonal" }}{PARA 0 "" 
0 "" {TEXT -1 9 "elements." }}{PARA 0 "" 0 "" {TEXT -1 73 "Testing for
 Minimum and Maximum Conditions using the Hessian determinant:" }}
{PARA 0 "" 0 "" {TEXT -1 245 "The First minor of the A matrix is -10--
this is the first element on the principal diagnoal. It is obtained by
 deleting all the other elements in the A matrix. The second hessian i
s obtained by deleting the third row and the third col. The third" }}
{PARA 0 "" 0 "" {TEXT -1 449 "hessian is the determinant of the whole \+
matrix.  As shown above,  the firt hessian is equal -10 and is negativ
e. The second hessian is 40 and is positive. The third hessian is -276
 and is negative. Since the principal minors (determinants) alternate \+
in sign, the Hessian is negative definite and the function is maximize
d at x1=1.04, x2=1.22, and x3=0.43.  The hessian method is used to fin
d the min and mx of functions with three or more varioables" }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}}}{MARK "0 \+
7 0" 188 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 
}