What is the extreme q(x,y)=x^2-xy+5y^2 ?

The extreme q(x,y)=x^2-xy+5y^2 is Minimum(0,0)

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extreme q(x,y)=x^2-xy+5y^2

extreme points

q (x, y) = x^{2} - x y + 5 y^{2}

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 10

D (x, y) = 19

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

f (x, y) = 6 - x^{2} - 4 x - y^{2}

f (x) = 5 x^{3} - 4

f (x) = - x - \frac{144}{x} [9.13]

f (x) = - x^{3} + 12 x - 24

f (x) = x^{2} - 4 x + 16

What is the extreme q(x,y)=x^2-xy+5y^2 ?
The extreme q(x,y)=x^2-xy+5y^2 is Minimum(0,0)