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

The extreme f(x,y)=-2x^3+6xy+3y^3 is Saddle(0,0),Minimum(-\sqrt[3]{2/3},(2/3)^{2/3})

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

extreme points

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

Saddle (0, 0), Minimum (- 3 \frac{2}{3}, (\frac{2}{3})^{\frac{2}{3}})

Solution steps

Find the critical points:

(0, 0), (- 3 \frac{2}{3}, (\frac{2}{3})^{\frac{2}{3}})

\frac{\partial ^{2} f}{\partial x ^{2}} = 18 y

D (x, y) = - 216 x y - 36

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(- 3 \frac{2}{3}, (\frac{2}{3})^{\frac{2}{3}}) :

Minimum

Saddle (0, 0), Minimum (- 3 \frac{2}{3}, (\frac{2}{3})^{\frac{2}{3}})

f (x) = 14 x^{2} - x^{3}

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

f (x) = \frac{4 ( x - 10 ) ^{2}}{x - 6}, (0, 6)

f (x) = 15 x^{\frac{2}{3}} - x

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

What is the extreme f(x,y)=-2x^3+6xy+3y^3 ?
The extreme f(x,y)=-2x^3+6xy+3y^3 is Saddle(0,0),Minimum(-\sqrt[3]{2/3},(2/3)^{2/3})