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

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

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

extreme points

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

Saddle (0, 0), Minimum (1, 1)

Solution steps

Find the critical points:

(0, 0), (1, 1)

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

D (x, y) = 36 x y - 9

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(1, 1) :

Minimum

Saddle (0, 0), Minimum (1, 1)

p (x) = 3 x^{3} - 13 x^{2} + a x - 8

f (x) = 2 x^{2} - 7 x - 9

f (x) = (x^{2} - 16)^{7}

f (x) = 10 x + 11 x^{\frac{10}{11}}

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