What is the extreme f(x,y)=x^2+4y^2+6 ?

The extreme f(x,y)=x^2+4y^2+6 is Minimum(0,0)

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extreme f(x,y)=x^2+4y^2+6

extreme points

f (x, y) = x^{2} + 4 y^{2} + 6

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 8

D (x, y) = 16

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

f (x) = - (x^{2} + 4 x - 4) (x + 2) (x - 1)

f (x, y) = (x + 1)^{2} + (y - 3)^{2}

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

f (x) = 8 x + 11 x^{\frac{8}{11}}

f (x, y) = x - 2 y g (x, y) = x^{2} + y^{2} = 25

What is the extreme f(x,y)=x^2+4y^2+6 ?
The extreme f(x,y)=x^2+4y^2+6 is Minimum(0,0)