What is the extreme f(x,y)=sqrt(x^2+y^2-10x+26) ?

The extreme f(x,y)=sqrt(x^2+y^2-10x+26) is Minimum(5,0)

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extreme f(x,y)=sqrt(x^2+y^2-10x+26)

extreme points

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

Minimum (5, 0)

Solution steps

Find the critical points:

(5, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{x ^{2} - 10 x + 26}{( x ^{2} + y ^{2} - 10 x + 26 ) x ^{2} + y ^{2} - 10 x + 26}

D (x, y) = \frac{1}{( x ^{2} + y ^{2} + 26 - 10 x ) ^{2}}

Check the sign of at each critical point

Check critical point

(5, 0) :

Minimum

Minimum (5, 0)

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

f (x) = \frac{2}{3} x^{3} + 2 x^{2} + 6 x - 2

f (x) = 5 - 2 x^{2}, - 4 \leq x \leq 2

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

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

What is the extreme f(x,y)=sqrt(x^2+y^2-10x+26) ?
The extreme f(x,y)=sqrt(x^2+y^2-10x+26) is Minimum(5,0)