What is the extreme f(x,y)=4^2x^4+4^2y^4-4xy+19 ?

The extreme f(x,y)=4^2x^4+4^2y^4-4xy+19 is Saddle(0,0),Minimum(-1/4 ,-1/4),Minimum(1/4 , 1/4)

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extreme f(x,y)=4^2x^4+4^2y^4-4xy+19

extreme points

f (x, y) = 4^{2} x^{4} + 4^{2} y^{4} - 4 x y + 19

Saddle (0, 0), Minimum (- \frac{1}{4}, - \frac{1}{4}), Minimum (\frac{1}{4}, \frac{1}{4})

Solution steps

Find the critical points:

(0, 0), (- \frac{1}{4}, - \frac{1}{4}), (\frac{1}{4}, \frac{1}{4})

\frac{\partial ^{2} f}{\partial x ^{2}} = 192 y^{2}

D (x, y) = 36864 x^{2} y^{2} - 16

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(- \frac{1}{4}, - \frac{1}{4}) :

Minimum

Check critical point

(\frac{1}{4}, \frac{1}{4}) :

Minimum

Saddle (0, 0), Minimum (- \frac{1}{4}, - \frac{1}{4}), Minimum (\frac{1}{4}, \frac{1}{4})

f (x) = x^{3} - \frac{3}{2} x^{2} - 6 x + 8

f (x) = x^{3} - 6 x + 5

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

y = \frac{1}{3} x^{3} + 4

f (x, y) = \frac{1}{x + y}

What is the extreme f(x,y)=4^2x^4+4^2y^4-4xy+19 ?
The extreme f(x,y)=4^2x^4+4^2y^4-4xy+19 is Saddle(0,0),Minimum(-1/4 ,-1/4),Minimum(1/4 , 1/4)