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

The extreme f(x,y)=(x^3)/3-3y^2-(x^2)/2+3xy is Saddle(0,0),Maximum(-1/2 ,-1/4)

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

extreme

Saddle (0, 0), Maximum (- \frac{1}{2}, - \frac{1}{4})

Solution steps

Find the critical points:

(0, 0), (- \frac{1}{2}, - \frac{1}{4})

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

D (x, y) = - 12 x - 3

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(- \frac{1}{2}, - \frac{1}{4}) :

Maximum

Saddle (0, 0), Maximum (- \frac{1}{2}, - \frac{1}{4})

What is the extreme f(x,y)=(x^3)/3-3y^2-(x^2)/2+3xy ?
The extreme f(x,y)=(x^3)/3-3y^2-(x^2)/2+3xy is Saddle(0,0),Maximum(-1/2 ,-1/4)