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

The extreme f(x,y)=x^2y+2y^2-2xy is Saddle(2,0),Saddle(0,0),Minimum(1, 1/4)

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

extreme points

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

Saddle (2, 0), Saddle (0, 0), Minimum (1, \frac{1}{4})

Solution steps

Find the critical points:

(2, 0), (0, 0), (1, \frac{1}{4})

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

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

Check the sign of at each critical point

Check critical point

(2, 0) :

Saddle

Check critical point

(0, 0) :

Saddle

Check critical point

(1, \frac{1}{4}) :

Minimum

Saddle (2, 0), Saddle (0, 0), Minimum (1, \frac{1}{4})

f (x, y) = x^{3} + y^{2} - 4 x y + 17 x - 10 y + 2021

f (x) = (x - 3) (x - 2)^{2}

y = 7 x + 7 sin (x)

f (x) = 4 x^{2} + 3 x^{2} - 6 x + 1

\frac{4 x + 13}{- 2 x - 6}

What is the extreme f(x,y)=x^2y+2y^2-2xy ?
The extreme f(x,y)=x^2y+2y^2-2xy is Saddle(2,0),Saddle(0,0),Minimum(1, 1/4)