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

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

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

extreme points

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

Minimum (2, 1)

Solution steps

Find the critical points:

(2, 1)

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

D (x, y) = 3

Check the sign of at each critical point

Check critical point

(2, 1) :

Minimum

Minimum (2, 1)

x - x^{3}

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

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

f (x) = x^{2} + 4 x + 16

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

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