What is the extreme f(x,y)=x^3+3xy^2+xy^3-5y-4 ?

The extreme f(x,y)=x^3+3xy^2+xy^3-5y-4 is Saddle(0.50134…,-3.07951…)

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extreme f(x,y)=x^3+3xy^2+xy^3-5y-4

extreme points

f (x, y) = x^{3} + 3 x y^{2} + x y^{3} - 5 y - 4

Saddle (0.50134 \dots, - 3.07951 \dots)

Solution steps

Find the critical points:

(0.50134 \dots, - 3.07951 \dots)

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

D (x, y) = 36 x^{2} y + 36 x^{2} - 9 y^{4} - 36 y^{3} - 36 y^{2}

Check the sign of at each critical point

Check critical point

(0.50134 \dots, - 3.07951 \dots) :

Saddle

Saddle (0.50134 \dots, - 3.07951 \dots)

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

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

f (x) = 3 x + \frac{9}{2} x^{2} - 54 x + 1

Z (x) = 2 x_{1} + 5 x_{2}

f (x, y) = 2 x^{3} + 2 y^{3} - 9 x^{2} + 3 y^{2} - 13 y

What is the extreme f(x,y)=x^3+3xy^2+xy^3-5y-4 ?
The extreme f(x,y)=x^3+3xy^2+xy^3-5y-4 is Saddle(0.50134…,-3.07951…)