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

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

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

extreme points

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

Minimum (1, 1)

Solution steps

Find the critical points:

(1, 1)

\frac{\partial ^{2} f}{\partial x ^{2}} = 4

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(1, 1) :

Minimum

Minimum (1, 1)

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}

f (x) = - 14 + 7 x - x^{2}

2 x 4 - x^{2}

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