Question

Find the number of units x that produce the minimum average cost per unit

Answer 1

C=.001x^3 - 5x + 250

Answer 2

do u know what average cost is?

Answer 3

c/x

Answer 4

exactly. so u have to minimize the function
\[ \large f(x)=\frac{C(x)}{x}=\frac{.001x^3-5x+250}{x} \]

Answer 5

.003x^2- 5 + 250/x ? xD

Answer 6

which becomes
\[ \large f(x)=.001x^2-5+\frac{250}{x} \]

Answer 7

C = 0.001x^3 + 5x + 250
C(avg) = (0.001x^3 + 5x + 250) / x
C(avg) = 0.001x^2 + 5 + 250 / x

Answer 8

C ' (avg) = 0.002x - 250/x^2
2x / 1000 - 250 / x^2 = 0
2x / 1000 = 250 / x^2
2x^3 = 250000
x^3 = 125000
x = 50

Answer 9

shouldn't it be \[.003x ^{2} -5x +\frac{ 250 }{ x }\]

Answer 10

Ok some one was faster... but i also have x=50 as result..

Answer 11

yeah thats the answer but how do u get that

Answer 12

did u understand my last post. how to get f(x)

Answer 13

mayankdevnani did solve this pretty well i guess..
steps are:
-devide by x (if u have only the cost function)
-set derivative equal zero
-solve function

Answer 14

now find the derivative of f(x).
do u know that?

Answer 15

@lilfayfay ?

Answer 16

yeah scroll up :P

Answer 17

.006+250/x^2=0

Answer 18

no.
\[ \large f(x)=.001x^2-5+\frac{250}{x} \]
so
\[ \large f'(x)=.002x-\frac{250}{x^2} \]

Answer 19

then solve
\[ \large 0=f'(x) \]
so
\[ \large .002x=\frac{250}{x^2} \]
\[ \large x^3=\frac{250}{.002}=125000=50^3 \]
so x=50

Answer 20

the second derivative is
\[ \large f''(x)=.002+\frac{500}{x^3} \]
and
\[ \large f''(50)=.002+\frac{500}{50^3}>0 \]
so f has a minimun at x=50.

Answer 21

u can also prove that it's a minimum byusing FDT rite?

Answer 22

what is FDT?

Answer 23

first derivative test

Answer 24

yes. the first derivative has to change from negative to positive at x=50

Answer 25

yes ty i get it now :P
can u help me with my next question plss? ^-^""

Answer 26

sure. i hope my internet line doesn't crash.
not the best service :)

Answer 27

haha, i noe how u feel iposted the question in a new tab <3