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

The extreme f(x)=e^{-2x^2}-3y^2+2y is Maximum(0, 1/3)

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

extreme

Maximum (0, \frac{1}{3})

Solution steps

Find the critical points:

(0, \frac{1}{3})

\frac{\partial ^{2} f}{\partial x ^{2}} = - 6

D (x, y) = 24 (- 4 e^{- 2 x^{2}} x^{2} + e^{- 2 x^{2}})

Check the sign of at each critical point

Check critical point

(0, \frac{1}{3}) :

Maximum

Maximum (0, \frac{1}{3})

What is the extreme f(x)=e^{-2x^2}-3y^2+2y ?
The extreme f(x)=e^{-2x^2}-3y^2+2y is Maximum(0, 1/3)