What is the extreme f(x,y)=(7x+9y+8)e^{(-(x^2+y^2))} ?

The extreme f(x,y)=(7x+9y+8)e^{(-(x^2+y^2))} is Minimum(-7/10 ,-9/10),Maximum(7/26 , 9/26)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x,y)=(7x+9y+8)e^{(-(x^2+y^2))}

extreme points

f (x, y) = (7 x + 9 y + 8) e^{(- (x^{2} + y^{2}))}

Minimum (- \frac{7}{10}, - \frac{9}{10}), Maximum (\frac{7}{26}, \frac{9}{26})

Solution steps

Find the critical points:

(- \frac{7}{10}, - \frac{9}{10}), (\frac{7}{26}, \frac{9}{26})

\frac{\partial ^{2} f}{\partial x ^{2}} = - 2 e^{- x^{2} - y^{2}} (- 18 y^{3} - 14 x y^{2} - 16 y^{2} + 27 y + 7 x + 8)

D (x, y) = 4 e^{- 2 x^{2} - 2 y^{2}} (- 14 x^{3} - 18 x^{2} y - 16 x^{2} + 21 x + 9 y + 8) (- 18 y^{3} - 14 x y^{2} - 16 y^{2} + 27 y + 7 x + 8) - 4 e^{2 (- x^{2} - y^{2})} (y (7 - 2 x (8 + 7 x + 9 y)) + 9 x)^{2}

Check the sign of at each critical point

Check critical point

(- \frac{7}{10}, - \frac{9}{10}) :

Minimum

Check critical point

(\frac{7}{26}, \frac{9}{26}) :

Maximum

Minimum (- \frac{7}{10}, - \frac{9}{10}), Maximum (\frac{7}{26}, \frac{9}{26})

x^{3} - 3 x + 3 x y^{2}

f (x) = \frac{x}{( x + 3 ) ^{2}}

y = u e^{x}

f (x, y) = x^{3} - y^{2} - 3 x + 8 y

f (x, y) = 2 x^{2} + 5 y^{2} - x y

What is the extreme f(x,y)=(7x+9y+8)e^{(-(x^2+y^2))} ?
The extreme f(x,y)=(7x+9y+8)e^{(-(x^2+y^2))} is Minimum(-7/10 ,-9/10),Maximum(7/26 , 9/26)