What is the extreme f(x,y)=x^3+y^3+15xy ?

The extreme f(x,y)=x^3+y^3+15xy is Saddle(0,0),Maximum(-5,-5)

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extreme f(x,y)=x^3+y^3+15xy

extreme points

f (x, y) = x^{3} + y^{3} + 15 x y

Saddle (0, 0), Maximum (- 5, - 5)

Solution steps

Find the critical points:

(0, 0), (- 5, - 5)

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

D (x, y) = 36 x y - 225

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(- 5, - 5) :

Maximum

Saddle (0, 0), Maximum (- 5, - 5)

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

f (x) = x ln (\frac{x}{3}), 0.1 \leq x \leq 3

f (x) = 2 x^{3} - 6 x^{2} + 6 x - 8

x^{3} - 27 x + 51

x^{3} - 27 x + 57

What is the extreme f(x,y)=x^3+y^3+15xy ?
The extreme f(x,y)=x^3+y^3+15xy is Saddle(0,0),Maximum(-5,-5)