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

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

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

extreme points

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

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}} = 6 y

D (x, y) = 36 x y - 1

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})

f (x) = 2 + x^{\frac{2}{5}}

x^{2} - 4 x + 1

f (x) = x \frac{4}{3} (x - 4)

f (x) = x^{4} - 8 x^{2} - 6

f (x) = x^{4} - 8 x^{2} - 7

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