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

The extreme f(x,y)=2xy-2x^2-y^3+3 is Saddle(0,0),Maximum(1/6 , 1/3)

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

extreme points

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

Saddle (0, 0), Maximum (\frac{1}{6}, \frac{1}{3})

Solution steps

Find the critical points:

(0, 0), (\frac{1}{6}, \frac{1}{3})

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

D (x, y) = 24 y - 4

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(\frac{1}{6}, \frac{1}{3}) :

Maximum

Saddle (0, 0), Maximum (\frac{1}{6}, \frac{1}{3})

What is the extreme f(x,y)=2xy-2x^2-y^3+3 ?
The extreme f(x,y)=2xy-2x^2-y^3+3 is Saddle(0,0),Maximum(1/6 , 1/3)