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

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

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

extreme points

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

Minimum (0, 0), Saddle (2, 1), Saddle (- 2, 1)

Solution steps

Find the critical points:

(0, 0), (2, 1), (- 2, 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = 4

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

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Check critical point

(2, 1) :

Saddle

Check critical point

(- 2, 1) :

Saddle

Minimum (0, 0), Saddle (2, 1), Saddle (- 2, 1)

x + \frac{32}{x ^{2}}

f (x, y) = 1 - x^{2} - 4 - y^{2}

f (x) = x - 2 cos (x), 0 \leq x \leq 2 π

f (x) = x^{4} - 20 x^{3} + 28 x^{2} - 4

1448.2 x^{2} - 81667 x + 1000000

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