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

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

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

extreme points

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

Saddle (0, 0), Maximum (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) :

Maximum

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

f (x) = x^{3} + 3 x^{2} - 45 x + 100

f (x) = \frac{1}{x ( x - 6 )}

f (x, y) = x + y + \frac{1}{x y}

\frac{1}{3} x^{3} - 2 x^{2} + 3 x

f (x) = x^{\frac{4}{5}} (x - 4)^{2}

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