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

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

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

extreme points

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

Minimum (0, - \frac{1}{2})

Solution steps

Find the critical points:

(0, - \frac{1}{2})

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

D (x, y) = 8

Check the sign of at each critical point

Check critical point

(0, - \frac{1}{2}) :

Minimum

Minimum (0, - \frac{1}{2})

f (x) = 3 sin (2 x)

f (x, y) = 11 x^{2} + 4 x y + 5 y^{2} + 5 x + 5 y

f (x) = ln (\frac{1}{x ^{2} + 2 x + 10})

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

f (x, y) = 72 x + 30 y - 2 x^{2} - 3 y^{2}

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