What is the extreme f(x,y)=x^3-4xy+y^3 ?

The extreme f(x,y)=x^3-4xy+y^3 is Saddle(0,0),Minimum(4/3 , 4/3)

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extreme f(x,y)=x^3-4xy+y^3

extreme points

f (x, y) = x^{3} - 4 x y + y^{3}

Saddle (0, 0), Minimum (\frac{4}{3}, \frac{4}{3})

Solution steps

Find the critical points:

(0, 0), (\frac{4}{3}, \frac{4}{3})

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

D (x, y) = 36 x y - 16

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(\frac{4}{3}, \frac{4}{3}) :

Minimum

Saddle (0, 0), Minimum (\frac{4}{3}, \frac{4}{3})

f (x) = 2 sin^{2} (x), 0 \leq x \leq π

f (x) = 4 cos (3 x) + 5, (0, 2 π)

f (x, y) = 14 x y - x^{3} - 7 y^{2}

f (x, y) = x^{2} + x y + y^{2} - 25 y + 208

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

What is the extreme f(x,y)=x^3-4xy+y^3 ?
The extreme f(x,y)=x^3-4xy+y^3 is Saddle(0,0),Minimum(4/3 , 4/3)