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

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

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

extreme points

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

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

Solution steps

Find the critical points:

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

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

D (x, y) = 3

Check the sign of at each critical point

Check critical point

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

Minimum

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

f (x) = \frac{x ^{3}}{4 - x ^{2}}

f (x) = sin^{2} (x), 0 \leq x \leq \frac{2 π}{3}

3 x^{2} e^{x}

f (x) = 4 x^{\frac{1}{3}} + x^{\frac{4}{3}}

f (x) = \frac{1}{6} x^{3} - x^{2} - 6 x + 11

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