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

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

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

extreme points

f (a, p) = 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) = (7 x + y)^{2} + 2 y + 900

- \frac{2}{3} x + \frac{1}{6}

f (x) = \frac{1}{2} x + 11

g (x, y) = - x^{2} + x y - \frac{25}{512} y^{4} - 7

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