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

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

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

extreme points

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

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = 1

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

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

x (r, t) = 2 r (2 t)

\frac{1}{4 x ^{2}} - \frac{1}{x ^{3}}

- x^{2} + 4 x y - 5 y^{2} - 6 x + 22 y + 9

y = \frac{5000}{1 + 0.01 e ^{z x}}

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