What is the extreme f(x,y)=x^2+5y^2-4xy+2x-6y+6 ?

The extreme f(x,y)=x^2+5y^2-4xy+2x-6y+6 is Minimum(1,1)

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extreme f(x,y)=x^2+5y^2-4xy+2x-6y+6

extreme points

f (x, y) = x^{2} + 5 y^{2} - 4 x y + 2 x - 6 y + 6

Minimum (1, 1)

Solution steps

Find the critical points:

(1, 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = 10

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(1, 1) :

Minimum

Minimum (1, 1)

f (x) = 3 x + ln (x)

f (x) = 1.7 (x + 4)^{2} (x + 0.5) (x + 4)

f (x, y) = 2 x^{2} - 6 x y + 5 y^{2} - x + 3 y + 2

f (x) = 2 x^{3} - x^{2} - 4 x + 2, - 1 \leq x \leq 0

(x - 1)^{3}

What is the extreme f(x,y)=x^2+5y^2-4xy+2x-6y+6 ?
The extreme f(x,y)=x^2+5y^2-4xy+2x-6y+6 is Minimum(1,1)