What is the extreme f(x,y)=x^3+y^3-6xy+27 ?

The extreme f(x,y)=x^3+y^3-6xy+27 is Saddle(0,0),Minimum(2,2)

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extreme f(x,y)=x^3+y^3-6xy+27

extreme points

f (x, y) = x^{3} + y^{3} - 6 x y + 27

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

Solution steps

Find the critical points:

(0, 0), (2, 2)

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

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

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(2, 2) :

Minimum

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

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

f (x) = \frac{100 ln ( x + 2 )}{x + 2}

3 (e^{- 0.8 t} - e^{- 1.2 t})

f (x) = 6 x^{2} - 8 x + 2

f (x) = ∣ 2 x + 1 ∣

What is the extreme f(x,y)=x^3+y^3-6xy+27 ?
The extreme f(x,y)=x^3+y^3-6xy+27 is Saddle(0,0),Minimum(2,2)