What is the extreme f(x,y)=4x^2-xy+9y^2 ?

The extreme f(x,y)=4x^2-xy+9y^2 is Minimum(0,0)

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extreme f(x,y)=4x^2-xy+9y^2

extreme points

f (x, y) = 4 x^{2} - x y + 9 y^{2}

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 18

D (x, y) = 143

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

f (x) = 200 y^{2} + x^{2} - x^{2} y

f (x) = x^{4} - 4 x^{3} - x^{2} + 12 x - 2

f (x) = 5 x^{5} - 3 x^{2} + x - 10

f (x) = ln (x^{2} + x + 1), - 1 \leq x \leq 1

f (x) = x^{2} + x y + y^{2} - 28 y + 261

What is the extreme f(x,y)=4x^2-xy+9y^2 ?
The extreme f(x,y)=4x^2-xy+9y^2 is Minimum(0,0)