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

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

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

extreme points

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

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

Solution steps

Find the critical points:

(2, - \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

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

Minimum

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

f (x, y) = 2^{3} - y^{2} + 2 x y + 1

y = x^{2} + 2

f (x) = 2^{(x + 0.528)} - 10 (x + 0.528)

y = x^{2} - 5

- x^{2} - 4 x + 1

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