What is the extreme f(x,y)=x^2+4y^2+1 ?

The extreme f(x,y)=x^2+4y^2+1 is Minimum(0,0)

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extreme f(x,y)=x^2+4y^2+1

extreme points

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

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} + y^{2} - x y + x + y

f (x) = \frac{x}{1 - x}

f (x) = 2 x + 4 y

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

f (x) = x e^{- 7 x^{2}}

What is the extreme f(x,y)=x^2+4y^2+1 ?
The extreme f(x,y)=x^2+4y^2+1 is Minimum(0,0)