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

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

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

extreme points

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

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

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

D (x, y) = 32 (- 4 e^{- 2 x^{2} - 4 y^{2}} x^{2} + e^{- 2 x^{2} - 4 y^{2}}) (- 8 e^{- 2 x^{2} - 4 y^{2}} y^{2} + e^{- 2 x^{2} - 4 y^{2}}) - 1024 e^{2 (- 2 x^{2} - 4 y^{2})} x^{2} y^{2}

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

6 x^{2} - 48 x - 190

f (x) = \frac{e ^{x}}{( 4 + e ^{x} )}

f (x) = 4 x^{3} - 2 x^{2} - 5

T (x, y) = ln (3 x y + 2 x^{2} - y)

f (x) = 5000 - 15 x + 0.08 x^{2}

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