What is the extreme f(x,y)=x^2y-xy+3y^2 ?

The extreme f(x,y)=x^2y-xy+3y^2 is Saddle(1,0),Saddle(0,0),Minimum(1/2 , 1/24)

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extreme f(x,y)=x^2y-xy+3y^2

extreme points

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

Saddle (1, 0), Saddle (0, 0), Minimum (\frac{1}{2}, \frac{1}{24})

Solution steps

Find the critical points:

(1, 0), (0, 0), (\frac{1}{2}, \frac{1}{24})

\frac{\partial ^{2} f}{\partial x ^{2}} = 6

D (x, y) = 12 y - 4 x^{2} + 4 x - 1

Check the sign of at each critical point

Check critical point

(1, 0) :

Saddle

Check critical point

(0, 0) :

Saddle

Check critical point

(\frac{1}{2}, \frac{1}{24}) :

Minimum

Saddle (1, 0), Saddle (0, 0), Minimum (\frac{1}{2}, \frac{1}{24})

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

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

f (x) = x^{2} - 5 x + 1

f (x, y) = e^{4 x y} + 5 y

f (x) = \frac{x}{2} - cos (x)

What is the extreme f(x,y)=x^2y-xy+3y^2 ?
The extreme f(x,y)=x^2y-xy+3y^2 is Saddle(1,0),Saddle(0,0),Minimum(1/2 , 1/24)