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

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

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

extreme points

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

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = - 52

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

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

f (t) = (4 - t) 4^{t}

f (x) = 2 x^{3} - 15 x^{2} - 30

f (x) = 15 x^{2} - 15 x^{4}

6 x^{3} - 9 x^{2} - 216 x + 9

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