f(x,y)=4+x3+y3−3xy
fx=3x2−3y
fy=3y2−3x
Find the critical point(s)
fx=0fy=0 Then
3x2−3y=03y2−3x=0
y=x2
x4−x=0
Critical points are (0,0),(1,1)
fxx=6x
fxy=fyx=−3
fyy=6y
D=∣∣fxxfyxfxyfyy∣∣=∣∣6x−3−36y∣∣=36xy−9 Point (0,0)
D(0,0)=36(0)(0)−9=−9<0 Then f(0,0) is not a local maximum or minimum. Point (0,0) is a saddle point of f.
Point (1,1)
D(1,1)=36(1)(1)−9=27>0
fxx(1,1)=6(1)=6>0
Then f(1,1)=3 is a local minimum.
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