What is the extreme f(x,y)=6x^2-xy+4y^2 ?

The extreme f(x,y)=6x^2-xy+4y^2 is Minimum(0,0)

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extreme f(x,y)=6x^2-xy+4y^2

extreme points

f (x, y) = 6 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) = 95

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

- x^{4} + 4 x^{3} + 8 x^{2}

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

f (x) = x^{\frac{7}{5}} - 12 x^{\frac{1}{5}}

f (x) = x^{2} - 12

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

What is the extreme f(x,y)=6x^2-xy+4y^2 ?
The extreme f(x,y)=6x^2-xy+4y^2 is Minimum(0,0)