What is the extreme f(x,y)=2x^2+y^2-4x+6y ?

The extreme f(x,y)=2x^2+y^2-4x+6y is Minimum(1,-3)

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extreme f(x,y)=2x^2+y^2-4x+6y

extreme points

f (x, y) = 2 x^{2} + y^{2} - 4 x + 6 y

Minimum (1, - 3)

Solution steps

Find the critical points:

(1, - 3)

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

D (x, y) = 8

Check the sign of at each critical point

Check critical point

(1, - 3) :

Minimum

Minimum (1, - 3)

y = x e^{- 2 x}

f (x) = x + 3 y^{2} = 25

f (x) = 110.2 x - 0.4 x^{2} - 20.2 - 2000

f (x) = \frac{x ^{3}}{3} - 16 x^{2} + 240 x - 36

y = 4 x^{3} - 8 x

What is the extreme f(x,y)=2x^2+y^2-4x+6y ?
The extreme f(x,y)=2x^2+y^2-4x+6y is Minimum(1,-3)