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

The extreme f(x,y)=x^2-xy+2y^2 is Minimum(0,0)

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

extreme points

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

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = 7

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

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

f (x) = 2 sin (x) + cos (2 x)

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

f (x) = 2 - 5 x + x^{2}

f (x, y) = 33 x + 66 y + x y - x^{2} - 3 y^{2}

What is the extreme f(x,y)=x^2-xy+2y^2 ?
The extreme f(x,y)=x^2-xy+2y^2 is Minimum(0,0)