What is the extreme f(x)=(x-4)^2+(y-4)^2-4/(sqrt(5)) ?

The extreme f(x)=(x-4)^2+(y-4)^2-4/(sqrt(5)) is Minimum(4,4)

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extreme f(x)=(x-4)^2+(y-4)^2-4/(sqrt(5))

extreme points

f (x) = (x - 4)^{2} + (y - 4)^{2} - \frac{4}{5}

Minimum (4, 4)

Solution steps

Find the critical points:

(4, 4)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(4, 4) :

Minimum

Minimum (4, 4)

3 x^{4} + 7 x^{3} - 1 8^{2} - 32

f (x) = 0.2 x^{2} - 20 x + 900

14 + 4 x - x^{2}

f (x) = - x (x + 1)

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

What is the extreme f(x)=(x-4)^2+(y-4)^2-4/(sqrt(5)) ?
The extreme f(x)=(x-4)^2+(y-4)^2-4/(sqrt(5)) is Minimum(4,4)