What is the extreme f(x,y)=3x^4-2xy^2+4y^3+1 ?

The extreme f(x,y)=3x^4-2xy^2+4y^3+1 is Minimum(1/54 , 1/162)

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extreme f(x,y)=3x^4-2xy^2+4y^3+1

extreme points

f (x, y) = 3 x^{4} - 2 x y^{2} + 4 y^{3} + 1

Minimum (\frac{1}{54}, \frac{1}{162})

Solution steps

Find the critical points:

(\frac{1}{54}, \frac{1}{162})

\frac{\partial ^{2} f}{\partial x ^{2}} = 24 y - 4 x

D (x, y) = 36 x^{2} (24 y - 4 x) - 16 y^{2}

Check the sign of at each critical point

Check critical point

(\frac{1}{54}, \frac{1}{162}) :

Minimum

Minimum (\frac{1}{54}, \frac{1}{162})

f (x) = \frac{1}{3} x^{3} - x

f (x, y) = - 3 x^{2} - 2 x y - 2 y^{2} + 400 x + 200 y

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

f (x, y) = ln (9 - x^{2} - y^{2})

y = 4 - x^{2}

What is the extreme f(x,y)=3x^4-2xy^2+4y^3+1 ?
The extreme f(x,y)=3x^4-2xy^2+4y^3+1 is Minimum(1/54 , 1/162)