What is the extreme f(x,y)=8x^3+y^3+6xy ?

The extreme f(x,y)=8x^3+y^3+6xy is Saddle(0,0),Maximum(-1/2 ,-1)

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extreme f(x,y)=8x^3+y^3+6xy

extreme points

f (x, y) = 8 x^{3} + y^{3} + 6 x y

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

Solution steps

Find the critical points:

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

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

D (x, y) = 288 x y - 36

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

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

Maximum

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

What is the extreme f(x,y)=8x^3+y^3+6xy ?
The extreme f(x,y)=8x^3+y^3+6xy is Saddle(0,0),Maximum(-1/2 ,-1)