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

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

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

extreme points

f (x, y) = 4 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) = 16

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

f (x, y) = 3 x^{2} y - y^{3} + 36 x - 15 y + 18

f (x, y) = x y (1 - x - y)

f (x) = x^{3} + x^{2} - 8 x + 5

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

f (x, y) = 2 x^{2} + x y^{2} - 6 x y + 5 x + 2

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