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

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

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

extreme points

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

Saddle (0, - 3), Maximum (- 1, - 1)

Solution steps

Find the critical points:

(0, - 3), (- 1, - 1)

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

D (x, y) = - 4 x y - 4 x^{2} - 12 x - 4 y^{2} - 12 y - 9

Check the sign of at each critical point

Check critical point

(0, - 3) :

Saddle

Check critical point

(- 1, - 1) :

Maximum

Saddle (0, - 3), Maximum (- 1, - 1)

\frac{x ^{2} - 4}{x ^{2} + 4}

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

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

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

\frac{ln ( x )}{x}

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