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

The extreme f(x,y)=x^2y+2xy^2-2xy is Minimum(2/3 , 1/3),Saddle(0,1)

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

extreme points

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

Minimum (\frac{2}{3}, \frac{1}{3}), Saddle (0, 1)

Solution steps

Find the critical points:

(\frac{2}{3}, \frac{1}{3}), (0, 1)

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

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

Check the sign of at each critical point

Check critical point

(\frac{2}{3}, \frac{1}{3}) :

Minimum

Check critical point

(0, 1) :

Saddle

Minimum (\frac{2}{3}, \frac{1}{3}), Saddle (0, 1)

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

f (x) = 6 x^{4} + 2 x^{3} - 12 x^{2} + 3

f (x, y) = x y + \frac{64}{x} + \frac{64}{y}

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

f (x, y) = 5 (x - 3)^{2} - 6 y^{2} + 4 x y

What is the extreme f(x,y)=x^2y+2xy^2-2xy ?
The extreme f(x,y)=x^2y+2xy^2-2xy is Minimum(2/3 , 1/3),Saddle(0,1)