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f(x,y)=x^4+2y^2-4xy How do you find the critical points?
The answer is:
Saddle point at (0,0)
local min at (1,1) and (-1,-1)
1 Answer
- WenTianQianLv 59 years agoFavorite Answer
Hello !
fx(x , y) = 4x³ - 4y = 0
fy(x , y) = 4y - 4x = 0
Solve this set of equations you get the stationary points (0 , 0), (1 , 1), (-1 , -1).
Now for the second derivative :
A = fxx(x , y) = 12x², B = fxy(x , y) = - 4, C = fyy(x , y) = 4
For (0 , 0), clearly fxx(0 , 0) = 0 ==> saddle point
For (1 , 1), AC - B² ＞ 0 ==> Local min
For (-1 , -1), AC - B² ＞ 0 ==> Local min
