What is the extreme f(x,y)= 1/3 x^3+y^3+2x^2-12x-3y ?

The extreme f(x,y)= 1/3 x^3+y^3+2x^2-12x-3y is Minimum(2,1),Saddle(2,-1),Saddle(-6,1),Maximum(-6,-1)

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extreme f(x,y)= 1/3 x^3+y^3+2x^2-12x-3y

extreme points

f (x, y) = \frac{1}{3} x^{3} + y^{3} + 2 x^{2} - 12 x - 3 y

Minimum (2, 1), Saddle (2, - 1), Saddle (- 6, 1), Maximum (- 6, - 1)

Solution steps

Find the critical points:

(2, 1), (2, - 1), (- 6, 1), (- 6, - 1)

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

D (x, y) = 6 y (2 x + 4)

Check the sign of at each critical point

Check critical point

(2, 1) :

Minimum

Check critical point

(2, - 1) :

Saddle

Check critical point

(- 6, 1) :

Saddle

Check critical point

(- 6, - 1) :

Maximum

Minimum (2, 1), Saddle (2, - 1), Saddle (- 6, 1), Maximum (- 6, - 1)

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

f (x) e^{2 x} (x^{2} - 3 x + 2)

f (x, y) = 3 x^{3} - 5 y^{2} - 225 x + 70 y + 23

2 x^{3} + 9 x^{2} + 12 x + 1

f (x) = \frac{x ^{3}}{6} - \frac{5 x ^{2}}{4}

What is the extreme f(x,y)= 1/3 x^3+y^3+2x^2-12x-3y ?
The extreme f(x,y)= 1/3 x^3+y^3+2x^2-12x-3y is Minimum(2,1),Saddle(2,-1),Saddle(-6,1),Maximum(-6,-1)