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

The extreme f(x,y)=e^{-(x^2+y^2-6y)} is Maximum(0,3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x,y)=e^{-(x^2+y^2-6y)}

extreme

Maximum (0, 3)

Solution steps

Find the critical points:

(0, 3)

\frac{\partial ^{2} f}{\partial x ^{2}} = 2 e^{6 y - x^{2} - y^{2}} (2 y^{2} - 12 y + 17)

D (x, y) = - 4 e^{6 y - x^{2} - y^{2}} (- 2 e^{- x^{2} - y^{2} + 6 y} x^{2} + e^{- x^{2} - y^{2} + 6 y}) (2 y^{2} - 12 y + 17) - 4 e^{2 (- x^{2} - y^{2} + 6 y)} x^{2} (- 2 y + 6)^{2}

Check the sign of at each critical point

Check critical point

(0, 3) :

Maximum

Maximum (0, 3)

What is the extreme f(x,y)=e^{-(x^2+y^2-6y)} ?
The extreme f(x,y)=e^{-(x^2+y^2-6y)} is Maximum(0,3)