What is the extreme f(x,y)=4x^2-xy+4y^2 ?

The extreme f(x,y)=4x^2-xy+4y^2 is Minimum(0,0)

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extreme f(x,y)=4x^2-xy+4y^2

extreme points

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

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 8

D (x, y) = 63

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

2 x^{3} + 3 x^{2} - 12 x - 18

f (x) = 4.5 x^{2} - 7.1 x + 3.9

y = 6 - 2 x^{2} - z^{2}

x^{3} - 9 x^{2} + 24 x - 12

f (x) = (1 - 5 x)^{3} x - 3

What is the extreme f(x,y)=4x^2-xy+4y^2 ?
The extreme f(x,y)=4x^2-xy+4y^2 is Minimum(0,0)