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

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

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

extreme points

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

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) = \frac{2 x ^{\frac{5}{2}}}{5} - \frac{2 x ^{\frac{3}{2}}}{3} - 6

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

f (x) = 16 x + \frac{9}{x}

f (x) = x + \frac{22}{x}

y = 6 x - 7 x^{\frac{6}{7}}

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