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

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

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

extreme points

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

Saddle (0, 0), Minimum (- 16, - 4), Minimum (16, 4)

Solution steps

Find the critical points:

(0, 0), (- 16, - 4), (16, 4)

\frac{\partial ^{2} f}{\partial x ^{2}} = 12 y^{2}

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

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(- 16, - 4) :

Minimum

Check critical point

(16, 4) :

Minimum

Saddle (0, 0), Minimum (- 16, - 4), Minimum (16, 4)

f (x, y) = (7 x + 9 y + 8) e^{(- (x^{2} + y^{2}))}

x^{3} - 3 x + 3 x y^{2}

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

y = u e^{x}

f (x, y) = x^{3} - y^{2} - 3 x + 8 y

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