What is the extreme f(x,y)=(2e^{-x^2}+e^{-3y^2}) ?

The extreme f(x,y)=(2e^{-x^2}+e^{-3y^2}) is Maximum(0,0)

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

extreme points

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

Maximum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = - 6 (- 6 e^{- 3 y^{2}} y^{2} + e^{- 3 y^{2}})

D (x, y) = 24 (- 2 e^{- x^{2}} x^{2} + e^{- x^{2}}) (- 6 e^{- 3 y^{2}} y^{2} + e^{- 3 y^{2}})

Check the sign of at each critical point

Check critical point

(0, 0) :

Maximum

Maximum (0, 0)

f (x, y) = \frac{x ^{4} + 2 x ^{3} y ^{2} + 2}{3 ( x - 1 ) y}

f (x) = (x - 1)^{2} + 4 - 4 x^{2}

x^{2} - 5 x y + 4 y^{2} - 2 x + 5 y

f (x) = x^{2} + 4 x + 11

f (x) = x^{2} + 4 x + 10

What is the extreme f(x,y)=(2e^{-x^2}+e^{-3y^2}) ?
The extreme f(x,y)=(2e^{-x^2}+e^{-3y^2}) is Maximum(0,0)