What is the extreme f(x,y)=-2x^3-2y^3+6xy+10 ?

The extreme f(x,y)=-2x^3-2y^3+6xy+10 is Saddle(0,0),Maximum(1,1)

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extreme f(x,y)=-2x^3-2y^3+6xy+10

extreme points

f (x, y) = - 2 x^{3} - 2 y^{3} + 6 x y + 10

Saddle (0, 0), Maximum (1, 1)

Solution steps

Find the critical points:

(0, 0), (1, 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 12 y

D (x, y) = 144 x y - 36

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(1, 1) :

Maximum

Saddle (0, 0), Maximum (1, 1)

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

f (x) = 12 x - x^{3}

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

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

y = x + z

What is the extreme f(x,y)=-2x^3-2y^3+6xy+10 ?
The extreme f(x,y)=-2x^3-2y^3+6xy+10 is Saddle(0,0),Maximum(1,1)