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

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

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

extreme points

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

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 4

D (x, y) = 28

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

x^{4} - 8 x^{2} + 9

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

5 x^{2} ln (x)

f (x) = (6 x - x^{2})

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

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