Answer to Question #350679 in Calculus for Dudu

Question #350679

Consider the R²-R function f defined by f(x,y)=x²+2y²-x²y.


Show that f has two saddle points

1
Expert's answer
2022-06-15T16:42:17-0400
"f_x=2x-2xy"

"f_y=4y-x^2"

"f_x=0""f_y=0"

"2x-2xy=0""4y-x^2=0"

If "x=0," then "y=0."

If "y=1," then "x=-2" or "x=2."

Point (0,0), Point(-2, 1), Point(2, 1).



"f_{xx}=2-2y, f_{yy}=4, f_{xy}=f_{yx}=-2x"

"D=\\begin{vmatrix}\n 2-2y & -2x \\\\\n -2x & 4\n\\end{vmatrix}=8-8y-4x^2"

Point (0,0)


"D=8-0-0=8>0"

"f_{xx}(0,0)=2>0"

Then "f(0,0)" is a local minimum.


Point (-2,1)


"D=8-8-16=-16<0"

Then "f(-2,1)" is a saddle point.


Point (2,1)


"D=8-8-16=-16<0"

Then "f(2,1)" is a saddle point.


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