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

The extreme f(x,y)=3x^3y-2x^2y^2+y^3 is Saddle(81/32 , 729/128)

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

extreme points

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

Saddle (\frac{81}{32}, \frac{729}{128})

Solution steps

Find the critical points:

(\frac{81}{32}, \frac{729}{128})

\frac{\partial ^{2} f}{\partial x ^{2}} = 6 y - 4 x^{2}

D (x, y) = - 81 x^{4} + 72 x^{3} y - 48 x^{2} y^{2} + 108 x y^{2} - 24 y^{3}

Check the sign of at each critical point

Check critical point

(\frac{81}{32}, \frac{729}{128}) :

Saddle

Saddle (\frac{81}{32}, \frac{729}{128})

What is the extreme f(x,y)=3x^3y-2x^2y^2+y^3 ?
The extreme f(x,y)=3x^3y-2x^2y^2+y^3 is Saddle(81/32 , 729/128)