Question

A bike plant assembles 2000 bikes per month. Each production run cost $1200, and it cost $20to store a bike for a month. How many production runs should the plant use to min. inventory costs? How many bikes are assembled in each production run?
(Calculus)

Answer 1

ANYONE??

Answer 2

I feel like we are missing information. You sure you got everything in there?

Answer 3

Yup I have everything

Answer 4

let x for $1200
let y for $20
than you have eqation:
1200x+20y=2000

Answer 5

x+y>0
subtitution x=0 , find for y
subtitution y=0 , find for x

Answer 6

Okay, so I am going to assume it is possible for the plant to do one production run a year and produce 12*2000 sets of bike parts. Then the total cost of production and inventory with only 1 run is:
\[$1200+$20*(12*2000+11*2000+10*2000+...+1*2000)\]

Answer 7

Two production runs a year would look like:
$1200*2+$20(6*2000+5*2000+4*2000+...+1*2000)
And three production runs (every four months):
$1200*3+$20(4*2000+3*2000+2*2000+1*2000)

Answer 8

wait

Answer 9

not sure

Answer 10

So if n is the number of production runs and T is total cost we have:
\[T=$1200n+$20*2000*\frac{n(n+1)}{2}\]

Answer 11

Oops I mean:
\[T=$1200n+$20*2000*\frac{\frac{12}{n}(\frac{12}{n}+1)}{2}\]

Answer 12

So going with those assumptions:
\[T=$1200n+$40,000\frac{6}{n}(\frac{12}{n}+1)=$1200n+$40,000*72n^{-2}+$40,000*6n^{-1}\]

Answer 13

I don't really like these assumptions, how do you interpret the meaning of a production run and storage and assembly?