junhoma said:
Please help me to find and classify all relatives extrema and saddle points of \(\displaystyle f(x,y)=x^{3}+3xy-y^{3}\)
Since \(\displaystyle f_{x}(x,y)=3x^{2}+3y\) and \(\displaystyle f_{y}(x,y)=3x-3y^{2}\), the critical points of f satisfy the equations
\(\displaystyle 3x^{2}+3y=0\Rightarrow x^{2}+y=0\).........[1]
\(\displaystyle 3x-3y^{2}=0\Rightarrow x-y^{2}=0\)..........[2]
From [2], \(\displaystyle x=y^{2}\)...........[3]
Sub into [1]: \(\displaystyle y^{4}+y=0\)
\(\displaystyle y=0 \;\ or \;\ y=-1\)
Sub back into [3] and we see that \(\displaystyle x=0 \;\ or \;\ x=1\)
So, (0,0) and (1,-1) are critical points.
We also need the second order partials to test what sort of extrema occurs at those points.
\(\displaystyle f_{xx}=6x\)
\(\displaystyle f_{yy}=-6y\)
\(\displaystyle f_{xy}^{2}=9\)
By the second partials test:
\(\displaystyle D=f_{xx}(x_{0},y_{0})f_{yy}(x_{0},y_{0})-f_{xy}^{2}(x_{0},y_{0})\)
\(\displaystyle D=6x(-6y)-9=-36xy-9\)
\(\displaystyle At \;\ (0,0), \;\ D=-36(0)(0)-9=-9\)
\(\displaystyle At \;\ (1,-1), \;\ D=-36(1)(-1)-9=27\)
Since \(\displaystyle D<0\) at \(\displaystyle (0,0)\), there is a saddle point.
Since \(\displaystyle D>0\) and \(\displaystyle f_{xx}(1,-1)>0\) at \(\displaystyle (1,-1)\), there is a relative minimum.
