What is the extreme f(x,y)=x^3y+12x^2-8y ?

The extreme f(x,y)=x^3y+12x^2-8y is Saddle(2,-4)

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extreme f(x,y)=x^3y+12x^2-8y

extreme points

f (x, y) = x^{3} y + 12 x^{2} - 8 y

Saddle (2, - 4)

Solution steps

Find the critical points:

(2, - 4)

\frac{\partial ^{2} f}{\partial x ^{2}} = 0

D (x, y) = - 9 x^{4}

Check the sign of at each critical point

Check critical point

(2, - 4) :

Saddle

Saddle (2, - 4)

What is the extreme f(x,y)=x^3y+12x^2-8y ?
The extreme f(x,y)=x^3y+12x^2-8y is Saddle(2,-4)