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

The extreme f(x,y)=4x^2+4y^2-4x-8y is Minimum(1/2 ,1)

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

extreme points

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

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

Solution steps

Find the critical points:

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

\frac{\partial ^{2} f}{\partial x ^{2}} = 8

D (x, y) = 64

Check the sign of at each critical point

Check critical point

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

Minimum

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

f (x, y) = 10 x y - 5 x^{2} - 7 y^{2} + 40 x

f (x) = ln (1 + 8 x^{3})

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

f (x) = 2 x ln (x)

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

What is the extreme f(x,y)=4x^2+4y^2-4x-8y ?
The extreme f(x,y)=4x^2+4y^2-4x-8y is Minimum(1/2 ,1)