What is the extreme f(x,y)=2xy-x^2-2y^2+3x+4 ?

The extreme f(x,y)=2xy-x^2-2y^2+3x+4 is Maximum(3, 3/2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x,y)=2xy-x^2-2y^2+3x+4

extreme points

f (x, y) = 2 x y - x^{2} - 2 y^{2} + 3 x + 4

Maximum (3, \frac{3}{2})

Solution steps

Find the critical points:

(3, \frac{3}{2})

\frac{\partial ^{2} f}{\partial x ^{2}} = - 4

D (x, y) = 4

Check the sign of at each critical point

Check critical point

(3, \frac{3}{2}) :

Maximum

Maximum (3, \frac{3}{2})

f (x) = 2 x^{3} - 6 x + 1

f (x) = 2 x^{3} - 6 x + 5

\frac{7 x}{x ^{2} - 9}

(\frac{1}{2} - x^{2} + y^{2}) e^{1 - x^{2} - y^{2}}

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

What is the extreme f(x,y)=2xy-x^2-2y^2+3x+4 ?
The extreme f(x,y)=2xy-x^2-2y^2+3x+4 is Maximum(3, 3/2)