What is the extreme f(x,y)=3x^2+6y^2 ?

The extreme f(x,y)=3x^2+6y^2 is Minimum(0,0)

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extreme f(x,y)=3x^2+6y^2

extreme points

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

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 12

D (x, y) = 72

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

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

f (x, y) = ln (36 - 4 x^{2} - 9 y^{2})

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

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

(x + 2)^{2} (x - 3)

What is the extreme f(x,y)=3x^2+6y^2 ?
The extreme f(x,y)=3x^2+6y^2 is Minimum(0,0)