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

The extreme f(x,y)=x^2+4y^2 is Minimum(0,0)

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

extreme points

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

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = 16

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

y = x^{3} - 6 x^{2} + 9 x

f (x) = 2 x^{4} - 16 x^{2} + 20 x - 7

f (x) = 4 x^{3} - 4 x

f (x) = x - 6 x - 1

f (x, y) = x^{3} + y^{3} - 15 x y

What is the extreme f(x,y)=x^2+4y^2 ?
The extreme f(x,y)=x^2+4y^2 is Minimum(0,0)