What is the extreme f(x,y)=12x-x^3-4y^2 ?

The extreme f(x,y)=12x-x^3-4y^2 is Saddle(-2,0),Maximum(2,0)

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extreme f(x,y)=12x-x^3-4y^2

extreme points

f (x, y) = 12 x - x^{3} - 4 y^{2}

Saddle (- 2, 0), Maximum (2, 0)

Solution steps

Find the critical points:

(- 2, 0), (2, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 8

D (x, y) = 48 x

Check the sign of at each critical point

Check critical point

(- 2, 0) :

Saddle

Check critical point

(2, 0) :

Maximum

Saddle (- 2, 0), Maximum (2, 0)

What is the extreme f(x,y)=12x-x^3-4y^2 ?
The extreme f(x,y)=12x-x^3-4y^2 is Saddle(-2,0),Maximum(2,0)