What is the extreme f(x,y)=x^2+3y^2-4x+6y ?

The extreme f(x,y)=x^2+3y^2-4x+6y is Minimum(2,-1)

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extreme f(x,y)=x^2+3y^2-4x+6y

extreme points

f (x, y) = x^{2} + 3 y^{2} - 4 x + 6 y

Minimum (2, - 1)

Solution steps

Find the critical points:

(2, - 1)

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

D (x, y) = 12

Check the sign of at each critical point

Check critical point

(2, - 1) :

Minimum

Minimum (2, - 1)

g (x) = \frac{1}{9} x^{9} - x

f (x, y) = x^{2} + y^{2} - 18 x + 12 y - 7

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

f (x) = 2 x^{2} - 8 x + 11, 0 \leq x \leq 6

- 4 + 4 x + 2 x^{2}

What is the extreme f(x,y)=x^2+3y^2-4x+6y ?
The extreme f(x,y)=x^2+3y^2-4x+6y is Minimum(2,-1)