this is a problem dealing with inventory and orders, Please help!!

Jun 2010
4
0
This is the problem im stuck on, any help is appreciated!

A wine warehouse expects to sell 30,000 bottles of wine in a year. Each bottle
costs $9, and there is a fixed charge of $200 per order. If it costs $3 to store a
bottle for a year, how many bottles should be ordered at a time and how many
orders should the warehouse place in a year to minimize inventory costs.
 
Jun 2010
4
0
I found something similar to what your post was about in my book and course notes, these are the answers I came up with, please let me know if they are correct:

C=200y + 1.5x

xy= 30,000, thus y=30000/x

So, C(x)= 200(30000/x)+1.5x

Then, C(x)= 6,000,000/x + 1.5x

C '(x)= -6,000,000/x^2 + 1.5

-6,000,000/x^2 + 1.5 = 0

x^2= 6,000,000/1.5

Divide, then square root to cancel out
x^2= 4,000,000, the square root of 4,000,000= 2,000
Thus, x=2,000, so y= 30,000/2,000 = 15

The final answer:
The company will minimize its inventory cost by ordering 2,000 bottles, 15 times during the year at a time.
 
May 2010
1,034
272
its the right answer, some minor comments in red

you have not defined any of your variables so i have to guess what you mean...

i assume:
C = total annual cost
y = number of orders per year
x = number of bottles per order


In that case:

C=200y+ 1.5x
You have not included the purchase price of the bottles in your cost function. This is actually ok, because the annual cost is a constant (30000*9) that is not affected by your choice of x and y. Since it is a constant, it will disappear when you differenciate anyway. Different professors will have their own preferences about whether this fixed cost should be included or not


you have to do quite a bit of simplification to get the cost of storage, 1.5x

  • if there are y orders per year, then the time between orders is 1/y.
  • So the average time each bottle spends in storage is 0.5(1/y) = 1/2y
  • So the cost of storage between orders is x * 3 * (1/2y) = 3x/2y
  • Multiply by the number of orders: y * (3x/2y) = 3x/2 = 1.5x



xy = 30000, so y=30000/x

\(\displaystyle C(x) = 200(30000/x) + 1.5x = 6000000x^{-1} + 1.5x\)
\(\displaystyle C'(x) = -6000000x^{-2} + 1.5\)

Solve C'(x) =0
\(\displaystyle 6000000 = 1.5x^2\)
\(\displaystyle = +/-2000\)
Discard negative result, so x=2000

y=30000/2000=15
 
Last edited: