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

The extreme f(x,y)=2x^4+y^2-16xy is Saddle(0,0),Minimum(-4,-32),Minimum(4,32)

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

extreme points

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

Saddle (0, 0), Minimum (- 4, - 32), Minimum (4, 32)

Solution steps

Find the critical points:

(0, 0), (- 4, - 32), (4, 32)

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

D (x, y) = 48 x^{2} - 256

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(- 4, - 32) :

Minimum

Check critical point

(4, 32) :

Minimum

Saddle (0, 0), Minimum (- 4, - 32), Minimum (4, 32)

f (x, y) = 2 x^{3} + x y^{2} + 5 x^{2} + y^{2} + 4

f (x, y) = x^{3} - 12 x + y^{3} + 3 y^{2} - 9 y

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

f (x) = a e^{2 x} - e^{3 x}

What is the extreme f(x,y)=2x^4+y^2-16xy ?
The extreme f(x,y)=2x^4+y^2-16xy is Saddle(0,0),Minimum(-4,-32),Minimum(4,32)