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

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

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

extreme points

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

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = 8

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

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

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

y = 2 x^{3} - 9 x^{2} + 12 x + 2

f (x) = x^{4} - 50 x^{2} + 9

f (x) = - 4 x^{2} + 8 x - 2, 0 \leq x \leq 3

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