Question #305438

Classify the critical points of f(x,y)= 4+x^3 + y^3 -3xy


1
Expert's answer
2022-03-07T15:38:01-0500
f(x,y)=4+x3+y33xyf(x,y)= 4+x^3 + y^3 -3xy

fx=3x23yf_x=3x^2-3y

fy=3y23xf_y=3y^2-3x

Find the critical point(s)


fx=0f_x=0fy=0f_y=0

Then


3x23y=03x^2-3y=03y23x=03y^2-3x=0

y=x2y=x^2

x4x=0x^4-x=0

Critical points are (0,0),(1,1)(0, 0), (1, 1)


fxx=6xf_{xx}=6x

fxy=fyx=3f_{xy}=f_{yx}=-3

fyy=6yf_{yy}=6y


D=fxxfxyfyxfyy=6x336y=36xy9D=\begin{vmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{vmatrix}=\begin{vmatrix} 6x & -3 \\ -3 & 6y \end{vmatrix}=36xy-9

Point (0,0)(0,0)


D(0,0)=36(0)(0)9=9<0D(0,0)=36(0)(0)-9=-9<0

Then f(0,0)f(0,0) is not a local maximum or minimum. Point (0,0)(0,0) is a saddle point of f.f.


Point (1,1)(1,1)

D(1,1)=36(1)(1)9=27>0D(1,1)=36(1)(1)-9=27>0

fxx(1,1)=6(1)=6>0f_{xx}(1,1)=6(1)=6>0

Then f(1,1)=3f(1,1)=3 is a local minimum.



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