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

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

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

extreme points

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

Minimum (0, - 1)

Solution steps

Find the critical points:

(0, - 1)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(0, - 1) :

Minimum

Minimum (0, - 1)

f (x, y) = 3 x + 4 y

f (t) = e^{k t}

x 8 - x^{2}

f (x) = \frac{( x + 4 ) ( x + 1 ) ( x - 1 ) ( x - 3 )}{x}

f (x) = 2 x^{3} - 12 x^{2} + 8

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