What is the extreme f(x,y)=3-x^2-y^2+6y ?

The extreme f(x,y)=3-x^2-y^2+6y is Maximum(0,3)

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

extreme points

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

Maximum (0, 3)

Solution steps

Find the critical points:

(0, 3)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(0, 3) :

Maximum

Maximum (0, 3)

f (x) = 1 - 1

(x + 1)^{2} - 7

f (x, y) = x y (1 - 5 x - 10 y)

f (x) = x^{2} + y^{2} + 18 x - 6 y

f (x) = 100 x + 100 y - x^{2} - y^{2}

What is the extreme f(x,y)=3-x^2-y^2+6y ?
The extreme f(x,y)=3-x^2-y^2+6y is Maximum(0,3)