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

f (x) = 2 x + \frac{6}{x}

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

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

f (x, y) = 3 x - x^{3} - 2 y^{2} + 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)