What is the extreme f(x,y)=x^3+y^3-24xy ?

The extreme f(x,y)=x^3+y^3-24xy is Saddle(0,0),Minimum(8,8)

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extreme f(x,y)=x^3+y^3-24xy

extreme points

f (x, y) = x^{3} + y^{3} - 24 x y

Saddle (0, 0), Minimum (8, 8)

Solution steps

Find the critical points:

(0, 0), (8, 8)

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

D (x, y) = 36 x y - 576

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(8, 8) :

Minimum

Saddle (0, 0), Minimum (8, 8)

f (x, y) = - 2 x^{2} - 2 y^{2} - 4 x y + 5 x + 4 y

f (x) = (x^{2} + x)^{\frac{2}{3}}

(x^{2} + 12) (4 - x^{2})

f (x y) = x^{3} + y^{3} + 2 x^{2} + 4 y^{2} + 6

f (x) = 2 x - 4 x

What is the extreme f(x,y)=x^3+y^3-24xy ?
The extreme f(x,y)=x^3+y^3-24xy is Saddle(0,0),Minimum(8,8)