What is the extreme f(x)=x^2+y^2+8x-12y-7 ?

The extreme f(x)=x^2+y^2+8x-12y-7 is Minimum(-4,6)

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extreme f(x)=x^2+y^2+8x-12y-7

extreme points

f (x) = x^{2} + y^{2} + 8 x - 12 y - 7

Minimum (- 4, 6)

Solution steps

Find the critical points:

(- 4, 6)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(- 4, 6) :

Minimum

Minimum (- 4, 6)

4 x^{3} - 48 x - 6

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

300 x - 4 x^{3}

x^{3} + 12 x^{2} - 27 x + 2

f (x) = x^{3} - 9 x^{2} + 15 x - 4

What is the extreme f(x)=x^2+y^2+8x-12y-7 ?
The extreme f(x)=x^2+y^2+8x-12y-7 is Minimum(-4,6)