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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

extreme points

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

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

Solution steps

Find the critical points:

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

\frac{\partial ^{2} f}{\partial x ^{2}} = - 4 e^{- x^{2} - y^{2}} y^{3} + 4 e^{- x^{2} - y^{2}} x y^{2} + 6 e^{- x^{2} - y^{2}} y - 2 e^{- x^{2} - y^{2}} x

D (x, y) = - 2 e^{- x^{2} - y^{2}} (- 2 x^{3} + 2 x^{2} y + 3 x - y) (- 4 e^{- x^{2} - y^{2}} y^{3} + 4 e^{- x^{2} - y^{2}} x y^{2} + 6 e^{- x^{2} - y^{2}} y - 2 e^{- x^{2} - y^{2}} x) - 4 e^{2 (- x^{2} - y^{2})} (- y (1 - 2 x (x - y)) + x)^{2}

Check the sign of at each critical point

Check critical point

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

Minimum

Check critical point

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

Maximum

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

\frac{d}{d x} (5 x^{2} y^{4} - 23 x y^{3} + 4 y^{5})

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

f (x) = x^{3} - x^{2}, 0 \leq x \leq 5

f (x, y) = 72 x^{4} + y^{2} - 24 x y

f (x) = x^{3} + y^{3} - 18 x y

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