What is the extreme f(x,y)=8x^3-12xy+y^3+11 ?

The extreme f(x,y)=8x^3-12xy+y^3+11 is Saddle(0,0),Minimum(1,2)

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extreme f(x,y)=8x^3-12xy+y^3+11

extreme points

f (x, y) = 8 x^{3} - 12 x y + y^{3} + 11

Saddle (0, 0), Minimum (1, 2)

Solution steps

Find the critical points:

(0, 0), (1, 2)

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

D (x, y) = 288 x y - 144

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(1, 2) :

Minimum

Saddle (0, 0), Minimum (1, 2)

f (x) = I x - 2 I + 1

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

f (x, y) = x^{2} - 4 x y + 2 y^{2} + 4 x + 8 y + 7

3 - \frac{x ^{2}}{2}

P (y, x) = - 2 y^{3} - 7 y^{2} + 5 x y

What is the extreme f(x,y)=8x^3-12xy+y^3+11 ?
The extreme f(x,y)=8x^3-12xy+y^3+11 is Saddle(0,0),Minimum(1,2)