What is the extreme f(x,y)=x^2+2xy+2y^2-6x+10y+2 ?

The extreme f(x,y)=x^2+2xy+2y^2-6x+10y+2 is Minimum(11,-8)

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extreme f(x,y)=x^2+2xy+2y^2-6x+10y+2

extreme points

f (x, y) = x^{2} + 2 x y + 2 y^{2} - 6 x + 10 y + 2

Minimum (11, - 8)

Solution steps

Find the critical points:

(11, - 8)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(11, - 8) :

Minimum

Minimum (11, - 8)

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

f (x) = 4 - x^{2}, - 1 \leq x \leq 2

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

f (x) = \frac{3 x}{x ^{2} + 9}

f (x, y) = 10 x y - x^{3} - 5 y^{2}

What is the extreme f(x,y)=x^2+2xy+2y^2-6x+10y+2 ?
The extreme f(x,y)=x^2+2xy+2y^2-6x+10y+2 is Minimum(11,-8)