What is the extreme f(x,y)=2x^2+2xy+5y^2+2x-2y+1 ?

The extreme f(x,y)=2x^2+2xy+5y^2+2x-2y+1 is Minimum(-2/3 , 1/3)

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extreme f(x,y)=2x^2+2xy+5y^2+2x-2y+1

extreme points

f (x, y) = 2 x^{2} + 2 x y + 5 y^{2} + 2 x - 2 y + 1

Minimum (- \frac{2}{3}, \frac{1}{3})

Solution steps

Find the critical points:

(- \frac{2}{3}, \frac{1}{3})

\frac{\partial ^{2} f}{\partial x ^{2}} = 10

D (x, y) = 36

Check the sign of at each critical point

Check critical point

(- \frac{2}{3}, \frac{1}{3}) :

Minimum

Minimum (- \frac{2}{3}, \frac{1}{3})

f (x) = \frac{ln ∣ x ∣}{x}, (- 3, 0)

g (x, y) = - \frac{5}{3} x^{3} - 2 x y - \frac{1}{2} y^{2} + 3 x + 7

f (x) = x^{\frac{1}{3}} (x + 4), - 27 \leq x \leq 27

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

What is the extreme f(x,y)=2x^2+2xy+5y^2+2x-2y+1 ?
The extreme f(x,y)=2x^2+2xy+5y^2+2x-2y+1 is Minimum(-2/3 , 1/3)