What is the extreme f(x,y)=x^2+xy+y^2-3x-6y ?

The extreme f(x,y)=x^2+xy+y^2-3x-6y is Minimum(0,3)

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extreme f(x,y)=x^2+xy+y^2-3x-6y

extreme points

f (x, y) = x^{2} + x y + y^{2} - 3 x - 6 y

Minimum (0, 3)

Solution steps

Find the critical points:

(0, 3)

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

D (x, y) = 3

Check the sign of at each critical point

Check critical point

(0, 3) :

Minimum

Minimum (0, 3)

f (x, y) = - x^{2} + y^{2} + 6 x y + 4 x - 4 y + 7

f (x) = \frac{x ^{2}}{x ^{2} + 7}

f (x) = \frac{x ^{2} - 10 x + 21}{2 x - 15}

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

f (x, y) = x^{2} - 6 x y + y^{3} - 3

What is the extreme f(x,y)=x^2+xy+y^2-3x-6y ?
The extreme f(x,y)=x^2+xy+y^2-3x-6y is Minimum(0,3)