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

The extreme f(x,y)=xe^{-x}+ye^{-2y} is Maximum(1, 1/2)

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

extreme points

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

Maximum (1, \frac{1}{2})

Solution steps

Find the critical points:

(1, \frac{1}{2})

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

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

Check the sign of at each critical point

Check critical point

(1, \frac{1}{2}) :

Maximum

Maximum (1, \frac{1}{2})

y = - x^{6} - 3 x^{3} + x + 1, (0, 1)

f (x, y) = x^{2} + 12 \cdot x \cdot y + 8 \cdot y^{2}

f (x) = - 5 x^{2} + 4 x - 6

f (x) = 9 sin (x)

P (q, s) = q + 6 + s

What is the extreme f(x,y)=xe^{-x}+ye^{-2y} ?
The extreme f(x,y)=xe^{-x}+ye^{-2y} is Maximum(1, 1/2)