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

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

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

extreme points

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

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

Solution steps

Find the critical points:

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

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

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

Check the sign of at each critical point

Check critical point

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

Minimum

Check critical point

(0, - 1) :

Saddle

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

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

f (x) = y^{4} - 4 y^{3} + 2 x^{2} + 8 x y

f (x) = x^{2} e^{- 9 x}

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

y = \frac{ln ( x )}{x}

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