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

The extreme f(x,y)=e^{-x^2-3y^2-x+y} is Maximum(-1/2 , 1/6)

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

extreme

Maximum (- \frac{1}{2}, \frac{1}{6})

Solution steps

Find the critical points:

(- \frac{1}{2}, \frac{1}{6})

\frac{\partial ^{2} f}{\partial x ^{2}} = e^{y - x^{2} - 3 y^{2} - x} (1 - 6 y + 6) (1 - 6 y - 6)

D (x, y) = e^{- 2 x^{2} - 2 x - 6 y^{2} + 2 y} (- 1 - 2 x + 2) (- 1 - 2 x - 2) (1 - 6 y + 6) (1 - 6 y - 6) - e^{2 (- x^{2} - 3 y^{2} - x + y)} (- 2 x - 1)^{2} (- 6 y + 1)^{2}

Check the sign of at each critical point

Check critical point

(- \frac{1}{2}, \frac{1}{6}) :

Maximum

Maximum (- \frac{1}{2}, \frac{1}{6})

What is the extreme f(x,y)=e^{-x^2-3y^2-x+y} ?
The extreme f(x,y)=e^{-x^2-3y^2-x+y} is Maximum(-1/2 , 1/6)