What is the extreme ((pi+4)x^2-640x+25600)/pi ?

The extreme ((pi+4)x^2-640x+25600)/pi is Minimum((320)/(pi+4),(25600)/(pi+4))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme ((pi+4)x^2-640x+25600)/pi

extreme

Minimum (\frac{320}{π + 4}, \frac{25600}{π + 4})

Solution steps

f^{'} (x) = 2 x + \frac{8 x - 640}{π}

Find intervals:

Decreasing

: - \infty < x < \frac{320}{π + 4},

Increasing

: \frac{320}{π + 4} < x < \infty

Plug

x = \frac{320}{π + 4}

into

\frac{( π + 4 ) x ^{2} - 640 x + 25600}{π} : \frac{25600}{π + 4}

Minimum (\frac{320}{π + 4}, \frac{25600}{π + 4})

What is the extreme ((pi+4)x^2-640x+25600)/pi ?
The extreme ((pi+4)x^2-640x+25600)/pi is Minimum((320)/(pi+4),(25600)/(pi+4))