What is the extreme f(x,y)=x^2+5y^2-4xy+2 ?

The extreme f(x,y)=x^2+5y^2-4xy+2 is Minimum(0,0)

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extreme f(x,y)=x^2+5y^2-4xy+2

extreme points

f (x, y) = x^{2} + 5 y^{2} - 4 x y + 2

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 10

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

x^{2} + 8 x - 9

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

5 x ln (x)

f (x) = 2 e^{- x} (x^{2} - x + 1)

\frac{x}{x ^{2} - 49}

What is the extreme f(x,y)=x^2+5y^2-4xy+2 ?
The extreme f(x,y)=x^2+5y^2-4xy+2 is Minimum(0,0)