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)

f (x) = x^{3} + y^{3} - 24 x y

f (x) = x^{4} - 50 x^{2} + 6

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

\frac{3 x ^{2}}{x ^{2} - 25}

f (x) = - x^{3} - x^{2} + 5 x, - 2 \leq x \leq 2

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)