What is the extreme f(x,y)=x^2+y^2-18x+12y-7 ?

The extreme f(x,y)=x^2+y^2-18x+12y-7 is Minimum(9,-6)

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extreme f(x,y)=x^2+y^2-18x+12y-7

extreme points

f (x, y) = x^{2} + y^{2} - 18 x + 12 y - 7

Minimum (9, - 6)

Solution steps

Find the critical points:

(9, - 6)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(9, - 6) :

Minimum

Minimum (9, - 6)

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

f (x) = 2 x^{2} - 8 x + 11, 0 \leq x \leq 6

- 4 + 4 x + 2 x^{2}

f (x) = x^{\frac{3}{2}} (x - 5)

f (x, y) = x^{2} - y^{2} - 10 x + 2 y + 8

What is the extreme f(x,y)=x^2+y^2-18x+12y-7 ?
The extreme f(x,y)=x^2+y^2-18x+12y-7 is Minimum(9,-6)