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

The extreme f(x)=4x^2+4y^2-xy is Minimum(0,0)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

extreme points

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

Minimum (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 8

D (x, y) = 63

Check the sign of at each critical point

Check critical point

(0, 0) :

Minimum

Minimum (0, 0)

f (x) = 600 + 35 x

f (x) = 3 cos (π x), 0 \leq x \leq \frac{1}{6}

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

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

f (x) = ln (3 x + 1)

What is the extreme f(x)=4x^2+4y^2-xy ?
The extreme f(x)=4x^2+4y^2-xy is Minimum(0,0)