What is the extreme f(x,y)=x^2+y^2-12x+16y ?

The extreme f(x,y)=x^2+y^2-12x+16y is Minimum(6,-8)

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extreme f(x,y)=x^2+y^2-12x+16y

extreme points

f (x, y) = x^{2} + y^{2} - 12 x + 16 y

Minimum (6, - 8)

Solution steps

Find the critical points:

(6, - 8)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(6, - 8) :

Minimum

Minimum (6, - 8)

f (x) = 14 x + 15 x^{\frac{14}{15}}

f (x) = \frac{x ^{3}}{3} + \frac{15 x ^{2}}{2} + 50 x - 2

y = x^{\frac{6}{7}} (x^{2} - 4)

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

f (x) = 3 x^{3} - 3 x^{2} - 3 x + 8

What is the extreme f(x,y)=x^2+y^2-12x+16y ?
The extreme f(x,y)=x^2+y^2-12x+16y is Minimum(6,-8)