What is the extreme f(x,y)=x^2+3y^2-2xy+10x-2y+4 ?

The extreme f(x,y)=x^2+3y^2-2xy+10x-2y+4 is Minimum(-7,-2)

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extreme f(x,y)=x^2+3y^2-2xy+10x-2y+4

extreme points

f (x, y) = x^{2} + 3 y^{2} - 2 x y + 10 x - 2 y + 4

Minimum (- 7, - 2)

Solution steps

Find the critical points:

(- 7, - 2)

\frac{\partial ^{2} f}{\partial x ^{2}} = 6

D (x, y) = 8

Check the sign of at each critical point

Check critical point

(- 7, - 2) :

Minimum

Minimum (- 7, - 2)

f (x) = x^{2} \cdot ln (x)

3 x^{\frac{2}{3}} - 2 x, - 1 \leq x \leq 1

f (x) = - 25 (x - 9)^{2} + 200

f (x) = - x^{3} + 3 x - 10

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What is the extreme f(x,y)=x^2+3y^2-2xy+10x-2y+4 ?
The extreme f(x,y)=x^2+3y^2-2xy+10x-2y+4 is Minimum(-7,-2)