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

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

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

extreme points

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

Minimum (3, - \frac{3}{2})

Solution steps

Find the critical points:

(3, - \frac{3}{2})

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(3, - \frac{3}{2}) :

Minimum

Minimum (3, - \frac{3}{2})

f (x) = x^{3} + y^{3} - 33 x y

f (x) = x^{3} - 2 x^{2} - 15 x + 3

f (x) = x^{3} - 2 x^{2} - 15 x + 5

g (x) = 2 x^{3} - 3 x^{2} - 12 x + 13

F (x, y) = \frac{1 + x ^{2} + y ^{2}}{1 - x ^{2} - y ^{2}}

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