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

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

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

extreme points

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

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = 12

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

f (x) = - x^{2} - 2 x + 1

3 x^{2} + y^{2} - x y - 11 x

x^{2} - 8 x

f (x) = - x^{3} - 6 x^{2} - 9 x + 3

f (x) = x^{3} - 6 x^{2} + 3

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