What is the extreme x^3+y^3-4xy+1 ?

The extreme x^3+y^3-4xy+1 is Saddle(0,0),Minimum(4/3 , 4/3)

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extreme x^3+y^3-4xy+1

extreme points

x^{3} + y^{3} - 4 x y + 1

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})

x^{\frac{1}{3}} (1 - x)

x^{2} - 2 x y + 3 y^{2} - 12 y

f (x) = 2 x^{3} + 4 x^{2} + \frac{2}{1} x

f (x) = 2 x + \frac{32}{x}, x \neq = 0

f (\frac{24}{13}, \frac{88}{39}) = 3 x y - 4 x^{2} - 3 y^{2} + 8 x + 8 y - 4

What is the extreme x^3+y^3-4xy+1 ?
The extreme x^3+y^3-4xy+1 is Saddle(0,0),Minimum(4/3 , 4/3)