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

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

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

extreme points

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

Minimum (0, 0), Saddle (4, 2), Saddle (4, - 2)

Solution steps

Find the critical points:

(0, 0), (4, 2), (4, - 2)

\frac{\partial ^{2} f}{\partial x ^{2}} = 32 - 8 x

D (x, y) = 4 (32 - 8 x) - 64 y^{2}

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Check critical point

(4, 2) :

Saddle

Check critical point

(4, - 2) :

Saddle

Minimum (0, 0), Saddle (4, 2), Saddle (4, - 2)

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

f (x) = x^{4} - 16 x^{3}

f (x) = \frac{2 ^{2}}{2 ^{4} + 16}

f (x, y) = 2 x^{2} - x y - 3 y^{2} - 3 x + 7 y

f (x, y) = x^{3} + 6 x y + y^{3}

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