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

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

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extreme f(x,y)=x^3+3xy+y^3+k

extreme points

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

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) = 5 + 5 x - 5 x^{2}, 0 \leq x \leq 3

f (x) = - 2 x^{2} - x + 4

y = 1250 sin (2 x)

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

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

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