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

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

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

extreme points

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

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = - 132

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

f (x, y) = 4 x y^{2} - 2 x^{2} - 16 y^{2}

f (x) = \frac{3 x + 1}{1 - 2 x}

f (x, y) = 16 x + 12 y - 2 x^{2} - 3 y^{2}

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

x^{3} - 12 x + 3

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