every year danielle sells 29720 cases of her cookie mix. it costs $1 per year to store a case plus annual warehouse fees of $2 per case for the maximum number of cases she will store. if it costs her $743 to set up a production run, plus $10 per case to manufacture a single case, how many production runs will minimize cost?

To minimize a cost function (or any other function that deals with minimums or maximums) we need to find the first derivative of the cost function and set it equal to zero (no slope):

First to translate the cost function described in the question into an equation:

x = 29,720 per case of cookie mix

n = number of runs

holding costs = $1x (to store case) + $2(x/n) (to store maximum number of cases)

production costs = $743*n (for setup) + $10*x (to manufacture case)

cost function = C = $743n + $10x + $1x + $2(x/n) = **$743n + $11x + $2(x/n)**

For a single run the cost is $387,103 (743 + 13*29,720). However, this cost will decrease as we increase the number of runs to a certain point. This is primarily due to the lowered cost of holding less at any given time. Our costs will increase after a certain point due to the $743 charge to set up a production run each time.

If the run is split into two orders (n = 2):

x = 29,720

C = 2*743 + 11*29,720 + 2*(29,720/2) = 1,486 + 326,920 + 29,720 = **358,125**

The cost to produce in two separate production runs decreased $28,977. This should continue to the point where we can minimize costs. Instead of trial and error, we can determine the first derivative of our cost function in terms of n and set it equal to zero.

C = 743n + 11x + 2(x/n) (rewrite to: 743n + 11x + 2xn^(-1)) then derive:

C prime = 743 + 0 + (-2xn^(-2)) = 0 (x is treated as a constant when deriving in terms of n)

Rewrite to: **C prime = 743 – 2(x/n^2) = 0** (now solve for n)

743 = 2x/n^2 ⇒ 743*n^2 = 2*29,720 (insert value for x)

n^2 = 59,440/743 = 80 (square both sides)

n = +/- 8.9443 = **+9**

Since you can’t have negative production runs or partial runs, choose the positive number and round up.
**So, the answer is 9 runs. This will minimize the cost of production and storage.**

At this level of production runs it would cost: C = 743*9 + 11*29,720 + 2*(29,720/9) = 6,687 + 326,920 + 6,604.44 =
**340,211.44**

To check your answer, we can see what it would cost to complete 8 runs and 10 runs.

n = 8

C = 743*8 + 326,920 + 2*(29,720/8) = 5,944 + 326,920 + 7,430 = **340,294**

So, the cost is roughly $83 more when completing only 8 runs.

n = 10

C = 743*10 + 326,920 + 2*(29,720/10) = 7,430 + 326,920 + 5,944 = **340,294**

Again, it will cost roughly $83 more to complete 10 production runs.