What is the extreme f(x,y)=11x^2-2xy+2y^2+3 ?

The extreme f(x,y)=11x^2-2xy+2y^2+3 is Minimum(0,0)

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extreme f(x,y)=11x^2-2xy+2y^2+3

extreme points

f (x, y) = 11 x^{2} - 2 x y + 2 y^{2} + 3

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 4

D (x, y) = 84

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

y = \frac{2 e ^{3 x}}{- 3 x - 1}

f (x) = 6 x^{4} + 11 x^{3} - x^{2} + x

f (x) = - 25 x^{2} - 20 y^{2} + 300 x + 480 y + 200

24 x^{3} - 3 x^{4} - x^{3} (24 - 3 x)

y = 9 x + 9 sin (x), 0 \leq x \leq 2 π

What is the extreme f(x,y)=11x^2-2xy+2y^2+3 ?
The extreme f(x,y)=11x^2-2xy+2y^2+3 is Minimum(0,0)