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

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

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

extreme points

f (x, y) = 2 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) = 8

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

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

x x^{2} + 25

f (x, y) = 2 x^{2} + 3 x y + 4 y^{2} + 2 x - 10 y

\frac{1}{2 x + 3}

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

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