What is the extreme f(x,y)=e^{-x^2-y^2} ?

The extreme f(x,y)=e^{-x^2-y^2} is Maximum(0,0)

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extreme f(x,y)=e^{-x^2-y^2}

extreme points

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

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 2 (- 2 e^{- x^{2} - y^{2}} y^{2} + e^{- x^{2} - y^{2}})

D (x, y) = 4 (- 2 e^{- x^{2} - y^{2}} x^{2} + e^{- x^{2} - y^{2}}) (- 2 e^{- x^{2} - y^{2}} y^{2} + e^{- x^{2} - y^{2}}) - 16 e^{2 (- x^{2} - y^{2})} x^{2} y^{2}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

f (x, y) = 72 y^{2} + x^{2} - x^{2} y

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

f (x) = 3 x^{\frac{2}{3}}

f (x) = \frac{1}{x} \cdot ln^{2} (3 x)

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

What is the extreme f(x,y)=e^{-x^2-y^2} ?
The extreme f(x,y)=e^{-x^2-y^2} is Maximum(0,0)