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

The extreme f(x,y)=x^3+y^3-12xy is Saddle(0,0),Minimum(4,4)

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

extreme points

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

Saddle (0, 0), Minimum (4, 4)

Solution steps

Find the critical points:

(0, 0), (4, 4)

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

D (x, y) = 36 x y - 144

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(4, 4) :

Minimum

Saddle (0, 0), Minimum (4, 4)

f (x) = 2 x^{3} - x^{2} + 16

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

x y e^{- x^{2} - y^{2}}

f (x) = x^{4} - 32 x + 4

f (x, y) = 12 x y - x^{2} - 3 y^{2}

What is the extreme f(x,y)=x^3+y^3-12xy ?
The extreme f(x,y)=x^3+y^3-12xy is Saddle(0,0),Minimum(4,4)