What is the extreme f(x)=x^2+2y^2-4x+8y ?

The extreme f(x)=x^2+2y^2-4x+8y is Minimum(2,-2)

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extreme f(x)=x^2+2y^2-4x+8y

extreme points

f (x) = x^{2} + 2 y^{2} - 4 x + 8 y

Minimum (2, - 2)

Solution steps

Find the critical points:

(2, - 2)

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

D (x, y) = 8

Check the sign of at each critical point

Check critical point

(2, - 2) :

Minimum

Minimum (2, - 2)

f (x) = \frac{2 - x}{5 + x}

f (x) = x^{(\frac{2}{5})} (\frac{7}{2 + x})

Q (x, y) = 5 x + 3 x y

f (x, y) = x^{2} + y^{2} - 10 x + 26

f (x) = x^{3} - 9 x^{2} + 10

What is the extreme f(x)=x^2+2y^2-4x+8y ?
The extreme f(x)=x^2+2y^2-4x+8y is Minimum(2,-2)