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

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

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

extreme points

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

Minimum (- 2, 1)

Solution steps

Find the critical points:

(- 2, 1)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(- 2, 1) :

Minimum

Minimum (- 2, 1)

x (r, t) = 2 r (2 t)

\frac{1}{4 x ^{2}} - \frac{1}{x ^{3}}

- x^{2} + 4 x y - 5 y^{2} - 6 x + 22 y + 9

y = \frac{5000}{1 + 0.01 e ^{z x}}

f (x) = 1 x - \frac{4}{3} x^{\frac{3}{4}}, (0, \infty)

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