What is the extreme f(x)=x^3+2y^2+3xy+8 ?

The extreme f(x)=x^3+2y^2+3xy+8 is Saddle(0,0),Minimum(3/4 ,-9/16)

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extreme f(x)=x^3+2y^2+3xy+8

extreme

Saddle (0, 0), Minimum (\frac{3}{4}, - \frac{9}{16})

Solution steps

Find the critical points:

(0, 0), (\frac{3}{4}, - \frac{9}{16})

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

D (x, y) = 24 x - 9

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(\frac{3}{4}, - \frac{9}{16}) :

Minimum

Saddle (0, 0), Minimum (\frac{3}{4}, - \frac{9}{16})

What is the extreme f(x)=x^3+2y^2+3xy+8 ?
The extreme f(x)=x^3+2y^2+3xy+8 is Saddle(0,0),Minimum(3/4 ,-9/16)