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) = \frac{8}{x} + 2 π x^{2}

f (x, y) = 400 - 9 x^{2} - 36 y^{2}

f (x, y) = \frac{1}{3} x^{3} - x

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

f (x) = (x^{2} - 1)^{3}, - 1 \leq x \leq 6

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)