What is the extreme f(x,y)=2^2x^4+2^2y^4-4xy+16 ?

The extreme f(x,y)=2^2x^4+2^2y^4-4xy+16 is Saddle(0,0),Minimum(-1/2 ,-1/2),Minimum(1/2 , 1/2)

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extreme f(x,y)=2^2x^4+2^2y^4-4xy+16

extreme points

f (x, y) = 2^{2} x^{4} + 2^{2} y^{4} - 4 x y + 16

Saddle (0, 0), Minimum (- \frac{1}{2}, - \frac{1}{2}), Minimum (\frac{1}{2}, \frac{1}{2})

Solution steps

Find the critical points:

(0, 0), (- \frac{1}{2}, - \frac{1}{2}), (\frac{1}{2}, \frac{1}{2})

\frac{\partial ^{2} f}{\partial x ^{2}} = 48 y^{2}

D (x, y) = 2304 x^{2} y^{2} - 16

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(- \frac{1}{2}, - \frac{1}{2}) :

Minimum

Check critical point

(\frac{1}{2}, \frac{1}{2}) :

Minimum

Saddle (0, 0), Minimum (- \frac{1}{2}, - \frac{1}{2}), Minimum (\frac{1}{2}, \frac{1}{2})

What is the extreme f(x,y)=2^2x^4+2^2y^4-4xy+16 ?
The extreme f(x,y)=2^2x^4+2^2y^4-4xy+16 is Saddle(0,0),Minimum(-1/2 ,-1/2),Minimum(1/2 , 1/2)