A publishing company sells 100,000 copies of certain books each year. It costs the company $1 to store each book for a year. Each time it must print additional copies, it costs the company $500 to set up the presses. How many books should the company produce during each printing in order to minimize its total storage and setup costs?

10,000 books

Step-by-step explanation:

Let x be the number of print runs per year and let y the number of books per print run.

Thus, xy = 100,000.

Now from the question, we only start a new print run when we have sold all books in the storage. Thus;

Per print run we now have a cost of;

(x * 1)/(y * 2)

This is because right after the print run, we have y books that last 1/n years (until the next print run). Now, if we plot number of books in storage vs time, we will see a sawtooth pattern where the spikes begin at each print run and will linearly decrease to 0 until the next sprint run which implies constant demand. The area of each triangle will be how many book⋅years we have to pay the storage for. This area is;

(y * (1/x))/2

We'll have to multiply this number by 1 so we can then we get the storage cost per printrun:

(y * (1/x))/2 * 1 = y/2x

Since we do x print runs, the total storage costs is; y/2x * x = y/2

The total print run cost is (500 * x). Therefore, the total cost is;

C_total = (500x) + (y/2)

From initially, we saw that;

xy = 100000

So,x = 100,000/y

C_total = (500*100,000/y) + (y/2)

C_total = 50000000/y + y/2

To minimize its total storage and setup costs, we will find the derivative of the total cost and equate to zero.

So;

dC/dx = -50000000/y² + 1/2

At dC/dx = 0,we have;

0 = -50000000/y² + 1/2

50000000/y² = 1/2

2 × 50000000 = y²

y = √2 × 50000000

y = 10,000 books

step-by-step explanation:

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step-by-step explanation:

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