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

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

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

extreme points

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

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 12

D (x, y) = 47

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

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

f (x) = x - sin (- x), 0 \leq x \leq 2 π

f (x) = \frac{( x ^{4} + 1 )}{x ^{2}}

f (x) = x^{\frac{1}{5}} (x + 1)

P (x, y) = 0.4 x + 0.6 y

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