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

The extreme f(x,y)=x^2+2xy+2y^2-x+2y is Minimum(2,-3/2)

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

extreme points

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

Minimum (2, - \frac{3}{2})

Solution steps

Find the critical points:

(2, - \frac{3}{2})

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(2, - \frac{3}{2}) :

Minimum

Minimum (2, - \frac{3}{2})

f (x) = x^{3} + 9 x^{2} + 24 x + 20

f (x) = x - \frac{1}{6} x^{2} - \frac{2}{3} ln (x)

f (x) = e^{x y}

f (x) = \frac{4}{9 - x ^{2} - y ^{2}}

x^{2} + 1

What is the extreme f(x,y)=x^2+2xy+2y^2-x+2y ?
The extreme f(x,y)=x^2+2xy+2y^2-x+2y is Minimum(2,-3/2)