What is the extreme f(x,y)=2x^3+xy^2+5x^2+y^2+4 ?

The extreme f(x,y)=2x^3+xy^2+5x^2+y^2+4 is Minimum(0,0),Maximum(-5/3 ,0),Saddle(-1,2),Saddle(-1,-2)

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extreme f(x,y)=2x^3+xy^2+5x^2+y^2+4

extreme points

f (x, y) = 2 x^{3} + x y^{2} + 5 x^{2} + y^{2} + 4

Minimum (0, 0), Maximum (- \frac{5}{3}, 0), Saddle (- 1, 2), Saddle (- 1, - 2)

Solution steps

Find the critical points:

(0, 0), (- \frac{5}{3}, 0), (- 1, 2), (- 1, - 2)

\frac{\partial ^{2} f}{\partial x ^{2}} = 2 x + 2

D (x, y) = (12 x + 10) (2 x + 2) - 4 y^{2}

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Check critical point

(- \frac{5}{3}, 0) :

Maximum

Check critical point

(- 1, 2) :

Saddle

Check critical point

(- 1, - 2) :

Saddle

Minimum (0, 0), Maximum (- \frac{5}{3}, 0), Saddle (- 1, 2), Saddle (- 1, - 2)

What is the extreme f(x,y)=2x^3+xy^2+5x^2+y^2+4 ?
The extreme f(x,y)=2x^3+xy^2+5x^2+y^2+4 is Minimum(0,0),Maximum(-5/3 ,0),Saddle(-1,2),Saddle(-1,-2)