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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

extreme points

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

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

Solution steps

Find the critical points:

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

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

D (x, y) = 24

Check the sign of at each critical point

Check critical point

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

Maximum

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

P (x, y) = 36 x^{2} - 9 y^{2}

f (x, y) = \frac{500}{4 + x ^{2} + y ^{2}}

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

f (x, y) = 2 y^{2} + 2 x y + x^{2} - 16 x - 20 y

f (x, y) = \frac{1}{x} - \frac{64}{y} + x y

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