What is the extreme f(x,y)=x^3+y^3+4xy ?

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

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extreme f(x,y)=x^3+y^3+4xy

extreme points

f (x, y) = x^{3} + y^{3} + 4 x y

Saddle (0, 0), Maximum (- \frac{4}{3}, - \frac{4}{3})

Solution steps

Find the critical points:

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

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

D (x, y) = 36 x y - 16

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

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

Maximum

Saddle (0, 0), Maximum (- \frac{4}{3}, - \frac{4}{3})

f (x) = 6 x^{5} - 40 x^{3} + 45

f (t) = t^{2} u (t) - (3 t - 2) u (t - 2)

f (x) = x^{\frac{4}{5}} (x + 1)

f (x) = x^{3} - 3 x, - 2 \leq x \leq 2

f (x, y) = - x^{2} - 3 y^{2} - 2 x y + 4 x - 3

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