Find the extreme values of
x^4+y^4-2(x-y)^2
The matrix "H(0,0)" is negative semidefinite sine
"\\lang\n\\begin{bmatrix}\n -4 & 4 \\\\\n 4 & -4\n\\end{bmatrix}\n\\begin{bmatrix}\n x \\\\\n y \n\\end{bmatrix},\n\\begin{bmatrix}\n x \\\\\n y \n\\end{bmatrix} \n \\rang =-4(x-y)^2\\le0"
with equality if x=y
This is a necessary condition for a local maximum, but not sufficient. Therefore, the test is inconclusive.
Indeed, (0,0) is a saddle point. If we approach (0,0) along x=y then we have
"f(x,x)=2x^4>0"
However, if we approach along "x=-y" , then
"f(x,-x)=2x^4-8x^2<0"
near (0,0).
extrema at "x=(-\\sqrt2,\\sqrt2)" and "y=(\\sqrt2,-\\sqrt2)"
Comments
Leave a comment