What is the extreme f(x,y)=x^4+3xy^3-xy ?

The extreme f(x,y)=x^4+3xy^3-xy is

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extreme f(x,y)=x^4+3xy^3-xy

extreme points

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

Solution steps

Find the critical points:

\frac{\partial ^{2} f}{\partial x ^{2}} = 18 x y

D (x, y) = 216 x^{3} y - 81 y^{4} + 18 y^{2} - 1

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

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

Saddle

Check critical point

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

Saddle

Check critical point

Minimum

Check critical point

Minimum

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

f (x, y) = 1 - x - y

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

y = 3 x^{4} + 4 x^{3}

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