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

The extreme f(x,y)=x^2+2y^2-xy+3x+2y is Minimum(-2,-1)

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

extreme points

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

Minimum (- 2, - 1)

Solution steps

Find the critical points:

(- 2, - 1)

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

D (x, y) = 7

Check the sign of at each critical point

Check critical point

(- 2, - 1) :

Minimum

Minimum (- 2, - 1)

f (x) = - 5 x^{3} + 15 x + 6

f (x, y) = (x - 5) (30 - 6 x + 4 y) + (y - 12) (80 + 8 x - 7 y)

f (x) = - 0.001 x^{2} + 3.4 x - 40

\frac{sin ( 5 x )}{sin ^{5} ( x )}

f (x) = \frac{x ^{2} - 2 x + 1}{x - 4}

What is the extreme f(x,y)=x^2+2y^2-xy+3x+2y ?
The extreme f(x,y)=x^2+2y^2-xy+3x+2y is Minimum(-2,-1)