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

The extreme f(x,y)=3x^2+3x-2y+4y^2-12xy is Saddle(0, 1/4)

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

extreme points

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

Saddle (0, \frac{1}{4})

Solution steps

Find the critical points:

(0, \frac{1}{4})

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

D (x, y) = - 96

Check the sign of at each critical point

Check critical point

(0, \frac{1}{4}) :

Saddle

Saddle (0, \frac{1}{4})

f (x, y) = x^{2} - y^{2} - 9 x - 2 y

f (x) = 2 x^{3} - 9 x^{2} + 3, - 2 \leq x \leq 2

f (x, y) = x^{3} + y^{3} + 3 x y^{2} - 18 x - 18 y

f (x) = 2 x^{4} - 36 x^{2}

f (x) = 25 (x^{2} - y^{2}) - 2 (x^{2} + y^{2})^{2}

What is the extreme f(x,y)=3x^2+3x-2y+4y^2-12xy ?
The extreme f(x,y)=3x^2+3x-2y+4y^2-12xy is Saddle(0, 1/4)