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, y) = x^{2} + y^{2} - 6 x + 8 y + 35

f (x) = \frac{1}{3} x^{3} - 15 x^{2} + 78.2 x + 3630

x^{3} + 3 x^{2} - 9 x + 7

f (x) = 100 - \frac{x}{100} - \frac{y}{400}

f (x, y) = x^{3} + y^{3} + 2 x^{2} + 4 y^{2} + 6

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)