What is the extreme f(x,y)=8x^4+y^2-16xy ?

The extreme f(x,y)=8x^4+y^2-16xy is Saddle(0,0),Minimum(-2,-16),Minimum(2,16)

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extreme f(x,y)=8x^4+y^2-16xy

extreme points

f (x, y) = 8 x^{4} + y^{2} - 16 x y

Saddle (0, 0), Minimum (- 2, - 16), Minimum (2, 16)

Solution steps

Find the critical points:

(0, 0), (- 2, - 16), (2, 16)

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

D (x, y) = 192 x^{2} - 256

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(- 2, - 16) :

Minimum

Check critical point

(2, 16) :

Minimum

Saddle (0, 0), Minimum (- 2, - 16), Minimum (2, 16)

f (x) = - 2 x - 1

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

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

f (x, y) = e^{- y} (x^{2} + y^{2}) + 6

f (x) = 5 x^{3} - 60 x + 5

What is the extreme f(x,y)=8x^4+y^2-16xy ?
The extreme f(x,y)=8x^4+y^2-16xy is Saddle(0,0),Minimum(-2,-16),Minimum(2,16)