What is the extreme f(x,y)=2x^3-y^2+2xy+1 ?

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

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extreme f(x,y)=2x^3-y^2+2xy+1

extreme

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

Solution steps

Find the critical points:

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

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

D (x, y) = - 24 x - 4

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

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

Maximum

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

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