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

The extreme f(x,y)=3x^3y-2x^2y^2+3y is Saddle(1/(\sqrt[3]{2)}, 9/(4\sqrt[3]{2)})

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

extreme points

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

Solution steps

Find the critical points:

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

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

Check the sign of at each critical point

Check critical point

Saddle

f (x, y) = 3 x^{2} + 2 y^{2} + 6 x + 8 y

f (x) = - 5 x^{7} ln (2 x), (0, 6)

- x^{3} + 12 x

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

y = x^{2} - 12 x + 1

What is the extreme f(x,y)=3x^3y-2x^2y^2+3y ?
The extreme f(x,y)=3x^3y-2x^2y^2+3y is Saddle(1/(\sqrt[3]{2)}, 9/(4\sqrt[3]{2)})