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

The extreme f(x,y)=2x^2+2xy+2y^2-6x is Minimum(2,-1)

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

extreme points

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

Minimum (2, - 1)

Solution steps

Find the critical points:

(2, - 1)

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

D (x, y) = 12

Check the sign of at each critical point

Check critical point

(2, - 1) :

Minimum

Minimum (2, - 1)

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

f (x, y) = x y (1 + 2 x - 3 y)

f (x) = \frac{x}{2} - sin (x)

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

f (x) = x^{2} - 4 x + 1, 0 \leq x \leq 3

What is the extreme f(x,y)=2x^2+2xy+2y^2-6x ?
The extreme f(x,y)=2x^2+2xy+2y^2-6x is Minimum(2,-1)