What is the extreme f(x,y)=x^2+y^2+12x-14y ?

The extreme f(x,y)=x^2+y^2+12x-14y is Minimum(-6,7)

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extreme f(x,y)=x^2+y^2+12x-14y

extreme points

f (x, y) = x^{2} + y^{2} + 12 x - 14 y

Minimum (- 6, 7)

Solution steps

Find the critical points:

(- 6, 7)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(- 6, 7) :

Minimum

Minimum (- 6, 7)

f (x) = - 12 x^{5} + 135 x^{4} - 400 x^{3}

f (x) = 3 x^{5} - 45 x^{3}

x (x + 1) (x + 2) (x + 3)

y = x + 2 cos (x), 0 \leq x \leq 2 π

f (x, y) = y^{3} - x^{3} - 2 x y + 6

What is the extreme f(x,y)=x^2+y^2+12x-14y ?
The extreme f(x,y)=x^2+y^2+12x-14y is Minimum(-6,7)