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

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

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

extreme points

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

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

Solution steps

Find the critical points:

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

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

D (x, y) = 12 x

Check the sign of at each critical point

Check critical point

(- 2, 1) :

Saddle

Check critical point

(2, 1) :

Minimum

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

f (x) = x^{2} + 81

f (x, y) = \frac{414}{23} x^{\frac{2}{3}} y^{\frac{1}{3}}

f (x) = 5 x - 4 ln (4 x)

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

g (x, y) = x^{3} + y^{3} - 3 x - 3 y

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