What is the extreme f(x)=5000-15x+0.08x^2 ?

The extreme f(x)=5000-15x+0.08x^2 is Minimum(375/4 ,4296.875)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=5000-15x+0.08x^2

extreme points

f (x) = 5000 - 15 x + 0.08 x^{2}

Minimum (\frac{375}{4}, 4296.875)

Solution steps

f^{'} (x) = 0.16 x - 15

Find intervals:

Decreasing

: - \infty < x < \frac{375}{4},

Increasing

: \frac{375}{4} < x < \infty

Plug

x = \frac{375}{4}

into

5000 - 15 x + 0.08 x^{2} : 4296.875

Minimum (\frac{375}{4}, 4296.875)

2 y^{2} + 2 x y + x^{2} - 16 x - 20 y

f (x) = - 2 x^{2} + 200 x

y = x^{2} - 5 x - 14

x \cdot 7 - x

f (x) = - 5 x^{2} - 8 y^{2} - 2 x y + 102 y

What is the extreme f(x)=5000-15x+0.08x^2 ?
The extreme f(x)=5000-15x+0.08x^2 is Minimum(375/4 ,4296.875)