What is the extreme f(x,y)=x^3-6xy+8y^3 ?

The extreme f(x,y)=x^3-6xy+8y^3 is Saddle(0,0),Minimum(1, 1/2)

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extreme f(x,y)=x^3-6xy+8y^3

extreme points

f (x, y) = x^{3} - 6 x y + 8 y^{3}

Saddle (0, 0), Minimum (1, \frac{1}{2})

Solution steps

Find the critical points:

(0, 0), (1, \frac{1}{2})

\frac{\partial ^{2} f}{\partial x ^{2}} = 48 y

D (x, y) = 288 x y - 36

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(1, \frac{1}{2}) :

Minimum

Saddle (0, 0), Minimum (1, \frac{1}{2})

f (x, y) = x^{2} + x^{2} y + y^{2} + 1

f (x) = 2 x + cos (x)

f (x) = (x^{2} + 1)^{3}

f (x) = x^{2} - 12 x

f (x, y) = \frac{1}{3} x^{3} + 2 y^{2} - 3 x^{2} + 5 x - 8 y + 4

What is the extreme f(x,y)=x^3-6xy+8y^3 ?
The extreme f(x,y)=x^3-6xy+8y^3 is Saddle(0,0),Minimum(1, 1/2)