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Response Details:
f( x, y ) = x4 + y4 - 4xy + 2
fx = 4x3 - 4y
fxx = 12x2
fxy = -4
fy = 4y3 - 4x
fyy = 12y2
D
= ( fxx )( fyy ) - ( fxy )2
= ( 12x2 )( 12y2 ) - ( -4 )2
= 144x2y2 - 16
Find the critical points of f:
fx = 0 =>
4x3 - 4y = 0 =>
4y = 4x3 =>
y = x3
fy = 0 =>
4y3 - 4x = 0 =>
4( x3 )3 - 4x = 0 =>
4x9 - 4x = 0 =>
x( 4x8 - 4 ) = 0 =>
x = 0 or x = -1 or x = 1
x = 0 =>
y = ( 0 )3 = 0
x = -1 =>
y = ( -1 )3 = -1
x = 1 =>
y = ( 1 )3 = 1
The critical points of f are:
( 0, 0 )
( -1, -1 )
( 1, 1 )
Classify the critical points of f:
f( 0, 0 ) = ( 0 )4 + ( 0 )4 - 4( 0 )( 0 ) + 2 = 2
D( 0, 0 ) = 144( 0 )2( 0 )2 - 16 = -16
D( 0, 0 ) < 0 => f( 0, 0 ) = 2 is a saddle point
f( -1, -1 ) = ( -1 )4 + ( -1 )4 - 4( -1 )( -1 ) + 2 = 0
D( -1, -1 ) = 144( -1 )2( -1 )2 - 16 = 128
fxx( -1, -1 ) = 12( -1 )2 = 12
D( -1, -1 ) > 0 and fxx( -1, -1 ) > 0 => f( -1, -1 ) = 0 is a local minimum
f( 1, 1 ) = ( 1 )4 + ( 1 )4 - 4( 1 )( 1 ) + 2 = 0
D( 1, 1 ) = 144( 1 )2( 1 )2 - 16 = 128
fxx( 1, 1 ) = 12( 1 )2 = 12
D( 1, 1 ) > 0 and fxx( 1, 1 ) > 0 => f( 1, 1 ) = 0 is a local minimum
Final answer:
f has a local minimum at the points ( -1, -1, 0 ) and ( 1, 1, 0 ), and a saddle point at the point ( 0, 0, 2 ).
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