What is the extreme f(xy)=x^3+3y^3+3x^2+3y^2+24 ?

The extreme f(xy)=x^3+3y^3+3x^2+3y^2+24 is Minimum(0,0),Saddle(0,-2/3),Saddle(-2,0),Maximum(-2,-2/3)

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extreme f(xy)=x^3+3y^3+3x^2+3y^2+24

extreme points

f (x y) = x^{3} + 3 y^{3} + 3 x^{2} + 3 y^{2} + 24

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

Solution steps

Find the critical points:

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

\frac{\partial ^{2} f}{\partial x ^{2}} = 18 y + 6

D (x, y) = (6 x + 6) (18 y + 6)

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Check critical point

(0, - \frac{2}{3}) :

Saddle

Check critical point

(- 2, 0) :

Saddle

Check critical point

(- 2, - \frac{2}{3}) :

Maximum

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

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

f (x, y) = ln (1 + x^{2} + y^{2})

f (x) = x - 3 x^{\frac{2}{3}}

f (x) = \frac{x ^{5}}{5} - \frac{8}{3} x^{3}

f (x) = 4 x^{2} - 5 x^{3}

What is the extreme f(xy)=x^3+3y^3+3x^2+3y^2+24 ?
The extreme f(xy)=x^3+3y^3+3x^2+3y^2+24 is Minimum(0,0),Saddle(0,-2/3),Saddle(-2,0),Maximum(-2,-2/3)