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

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

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

extreme points

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

Minimum (- 8, 8)

Solution steps

Find the critical points:

(- 8, 8)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(- 8, 8) :

Minimum

Minimum (- 8, 8)

f (x) = - 0.001 x^{2} + 9 x - 1296

f (x) = \frac{( 500 + 500 \cdot x )}{1 + 0.04 \cdot x ^{2}}

\frac{( x ^{2} + 2 )}{( x ^{2} - 16 )}

y = 2 x^{2} + 5

f (x) = 5 x^{4} - 20 x^{3} + 1

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