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

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

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

extreme points

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

Minimum (- 2, 1)

Solution steps

Find the critical points:

(- 2, 1)

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

D (x, y) = 8

Check the sign of at each critical point

Check critical point

(- 2, 1) :

Minimum

Minimum (- 2, 1)

f (x) = x^{4} - 8 x^{2} + 18

f (x) = x^{3} + 29 x + 128

x^{2} + \frac{4}{x}

f (x) = x e^{- x} [0.2]

f (x) = - 2 x^{2} - 12 x - 23

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