What is the extreme f(x,y)=8e^{(((-x^2-y^2))/4)} ?

The extreme f(x,y)=8e^{(((-x^2-y^2))/4)} is Maximum(0,0)

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extreme f(x,y)=8e^{(((-x^2-y^2))/4)}

extreme points

f (x, y) = 8 e^{(\frac{( - x ^{2} - y ^{2} )}{4})}

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 4 (- \frac{e ^{\frac{- x ^{2} - y ^{2}}{4}} y ^{2}}{2} + e^{\frac{- x ^{2} - y ^{2}}{4}})

D (x, y) = - 8 e^{\frac{- x ^{2} - y ^{2}}{2}} x^{2} - 8 e^{\frac{- x ^{2} - y ^{2}}{2}} y^{2} + 16 e^{\frac{- x ^{2} - y ^{2}}{2}}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

f (x, y) = ln (16 - x^{2} - 16 y^{2})

f (x) = \frac{1}{3} x^{3} + \frac{13}{2} x^{2} + 42 x + 3

(x^{2} + 2 x) e^{- x}

- 7 x^{3} + 21 x + 5

5 x^{2} e^{- x}

What is the extreme f(x,y)=8e^{(((-x^2-y^2))/4)} ?
The extreme f(x,y)=8e^{(((-x^2-y^2))/4)} is Maximum(0,0)