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

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

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

extreme points

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

Saddle (0, 0), Saddle (0, - \frac{3}{2})

Solution steps

Find the critical points:

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

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

D (x, y) = - 16 y^{2} - 24 y - 9

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(0, - \frac{3}{2}) :

Saddle

Saddle (0, 0), Saddle (0, - \frac{3}{2})

\frac{1}{3} (x - 2.1)^{2} + 7.9

f (x, y) = 120 x + 120 y - x y - x^{2} - y^{2}

f (x) = (x + 7)^{\frac{2}{3}}

f (x) = - x + 2 cos (x)

\frac{x + 4}{x ^{2} - 5 x - 36}

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