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

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

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extreme f(x,y)=(-x^2-y^2+2y-1)

extreme points

f (x, y) = (- x^{2} - y^{2} + 2 y - 1)

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) = 7 (\frac{1}{5} (y + 3) e^{- x y})

y = 5 x^{2} - 15 x + 3

f (x, y) = x (e^{x} - e^{y})

f (x) = 9 x^{2} - 27 x + 6

f (x) = 2 x^{3} - 24 x + 6

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