What is the extreme f(x)=x^2-xy+y^2-9x+6y+8 ?

The extreme f(x)=x^2-xy+y^2-9x+6y+8 is Minimum(4,-1)

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extreme f(x)=x^2-xy+y^2-9x+6y+8

extreme points

f (x) = x^{2} - x y + y^{2} - 9 x + 6 y + 8

Minimum (4, - 1)

Solution steps

Find the critical points:

(4, - 1)

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

D (x, y) = 3

Check the sign of at each critical point

Check critical point

(4, - 1) :

Minimum

Minimum (4, - 1)

x^{3} - 4 x^{2}

f (x) = x^{2} + 2 \cdot y^{2} - 2 \cdot x \cdot y - 2 \cdot y

y = x^{2} - 2 x + 6

f (x y) = x^{3} + 4 y^{2} - 3 x + 1

f (x) = \frac{x \frac{5}{3}}{2 + x}

What is the extreme f(x)=x^2-xy+y^2-9x+6y+8 ?
The extreme f(x)=x^2-xy+y^2-9x+6y+8 is Minimum(4,-1)