What is the extreme f(x,y)=(x+y)^3-12xy ?

The extreme f(x,y)=(x+y)^3-12xy is Minimum(1,1),Minimum(1,1),Saddle(0,0)

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extreme f(x,y)=(x+y)^3-12xy

extreme points

f (x, y) = (x + y)^{3} - 12 x y

Minimum (1, 1), Minimum (1, 1), Saddle (0, 0)

Solution steps

Find the critical points:

(1, 1), (1, 1), (0, 0)

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

D (x, y) = 144 x + 144 y - 144

Check the sign of at each critical point

Check critical point

(1, 1) :

Minimum

Check critical point

(1, 1) :

Minimum

Check critical point

(0, 0) :

Saddle

Minimum (1, 1), Minimum (1, 1), Saddle (0, 0)

f (x_{1}, x_{2}) = x_{1} + x_{1} x_{2} + x_{2}

f (x) = - x^{2} + 2 x + 6

f (x) = (x - 2)^{2} - 1

f (x) = 395 x^{2} - 2920 x^{3}

f (x) = 4 x e^{- 0.2 x}

What is the extreme f(x,y)=(x+y)^3-12xy ?
The extreme f(x,y)=(x+y)^3-12xy is Minimum(1,1),Minimum(1,1),Saddle(0,0)