What is the extreme f(x,y)=-x^2-y^2-2y+5 ?

The extreme f(x,y)=-x^2-y^2-2y+5 is Maximum(0,-1)

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extreme f(x,y)=-x^2-y^2-2y+5

extreme points

f (x, y) = - x^{2} - y^{2} - 2 y + 5

Maximum (0, - 1)

Solution steps

Find the critical points:

(0, - 1)

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

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(0, - 1) :

Maximum

Maximum (0, - 1)

f (x, y) = x^{3} + y^{2} - 2 x y + 7 x - 8 y + 4

f (x) = x^{2} y + y^{3} - 12 y

\frac{e ^{x}}{x + 1}

f (x) = x^{2} - 2 x - 3, 0 \leq x \leq 2

f (x) = x (14 - 2 x) (15 - x)

What is the extreme f(x,y)=-x^2-y^2-2y+5 ?
The extreme f(x,y)=-x^2-y^2-2y+5 is Maximum(0,-1)