What is the extreme f(x,y)=5+4x-2x^2+3y-y^2 ?

The extreme f(x,y)=5+4x-2x^2+3y-y^2 is Maximum(1, 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)=5+4x-2x^2+3y-y^2

extreme points

f (x, y) = 5 + 4 x - 2 x^{2} + 3 y - y^{2}

Maximum (1, \frac{3}{2})

Solution steps

Find the critical points:

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

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

D (x, y) = 8

Check the sign of at each critical point

Check critical point

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

Maximum

Maximum (1, \frac{3}{2})

f (x, y) = x^{3} + y^{3} - 243 x - 27 y - 2

F (x, y) = 3 y^{2} - 2 y^{2} - 3 x^{2} + 6 x y

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

14 + 14 x^{- \frac{3}{17}}

f (x, y) = x^{4} + 2 y^{2} - 8 x y

What is the extreme f(x,y)=5+4x-2x^2+3y-y^2 ?
The extreme f(x,y)=5+4x-2x^2+3y-y^2 is Maximum(1, 3/2)