What is the extreme f(x,y)=6y^3-5x^3y+3x^2 ?

The extreme f(x,y)=6y^3-5x^3y+3x^2 is Saddle(\sqrt[5]{72/125},(2\sqrt[5]{125})/(5\sqrt[5]{72)})

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extreme f(x,y)=6y^3-5x^3y+3x^2

extreme points

f (x, y) = 6 y^{3} - 5 x^{3} y + 3 x^{2}

Saddle (5 \frac{72}{125}, \frac{2 5 125}{5 5 72})

Solution steps

Find the critical points:

(5 \frac{72}{125}, \frac{2 5 125}{5 5 72})

\frac{\partial ^{2} f}{\partial x ^{2}} = 36 y

D (x, y) = 36 y (- 30 x y + 6) - 225 x^{4}

Check the sign of at each critical point

Check critical point

(5 \frac{72}{125}, \frac{2 5 125}{5 5 72}) :

Saddle

Saddle (5 \frac{72}{125}, \frac{2 5 125}{5 5 72})

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

f (x) = (x - 1)^{2} \cdot (x - 3) + 2

f (x) = 14 x + 19 x^{\frac{14}{19}}

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

f (x) = 3 cos (5 x) + sin (5 x)

What is the extreme f(x,y)=6y^3-5x^3y+3x^2 ?
The extreme f(x,y)=6y^3-5x^3y+3x^2 is Saddle(\sqrt[5]{72/125},(2\sqrt[5]{125})/(5\sqrt[5]{72)})