What is the extreme f(x,y)=x^2+y^2-6x+8y+35 ?

The extreme f(x,y)=x^2+y^2-6x+8y+35 is Minimum(3,-4)

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extreme f(x,y)=x^2+y^2-6x+8y+35

extreme points

f (x, y) = x^{2} + y^{2} - 6 x + 8 y + 35

Minimum (3, - 4)

Solution steps

Find the critical points:

(3, - 4)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(3, - 4) :

Minimum

Minimum (3, - 4)

f (x) = \frac{1}{3} x^{3} - 15 x^{2} + 78.2 x + 3630

x^{3} + 3 x^{2} - 9 x + 7

f (x) = 100 - \frac{x}{100} - \frac{y}{400}

f (x, y) = x^{3} + y^{3} + 2 x^{2} + 4 y^{2} + 6

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

What is the extreme f(x,y)=x^2+y^2-6x+8y+35 ?
The extreme f(x,y)=x^2+y^2-6x+8y+35 is Minimum(3,-4)