What is the extreme f(x,y)=x^2+y^2-18x+16y-7 ?

The extreme f(x,y)=x^2+y^2-18x+16y-7 is Minimum(9,-8)

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extreme f(x,y)=x^2+y^2-18x+16y-7

extreme points

f (x, y) = x^{2} + y^{2} - 18 x + 16 y - 7

Minimum (9, - 8)

Solution steps

Find the critical points:

(9, - 8)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(9, - 8) :

Minimum

Minimum (9, - 8)

f (x) = \frac{x - 4}{x ^{2} - 7}

f (x, y) = x^{3} - 48 x y + 64 y^{3}

f (x) = 7 x - 14 cos (x), - 2 \leq x \leq 0

y = (x - 9) \cdot 2 x

f (x) = x y^{2} - y^{2} + \frac{1}{2} x^{2} - 5 x + 4

What is the extreme f(x,y)=x^2+y^2-18x+16y-7 ?
The extreme f(x,y)=x^2+y^2-18x+16y-7 is Minimum(9,-8)