A manufacturer can save X dollars per unit in production

Hi VJesus12,

You can solve this question as follow,

Total Storage Costs for excess units when kept for m days = m * n * y
Total Savings = nX (X dollars for 1 unit, hence for n units n*X).

Question asks when will Total Storage costs > Total Savings
Therfore m*n*y >= nX

Hence m >= nX/ny = X/y. Hence C is the correct answer.

Regards!

VJesus12 wrote:A manufacturer can save x dollars per unit in production costs by overproducing in certain seasons. If storage costs for the excess are y dollars per unit per day ( x > y), which of the following expresses the maximum number of days that n excess units can be stored before the storage costs exceed the savings on the excess units?

A) x−y
B) (x−y)n
C) x/y
D) xn/y
E) x/yn

Let d denote the number of days that n excess units are stored.

Since each unit costs y dollars per day, n units will cost nyd dollars for storage of d days. And, since each unit saves x dollars , then the savings for n units is nx.. The break-even for this scenario, when the storage cost is equal to the savings, is:

nyd - nx = 0

We see, too, that if (nyd - nx) > 0, then the company will lose money (i.e., the cost of storage is greater than the savings). Let’s study this inequality:

nyd - nx > 0

nyd > nx

dy > x

d > x/y

or

x/y < d

As soon as the ratio x/y is less than d, the storage costs are greater than the savings. Therefore, the (maximum) number of days the n units can be stored before the storage costs exceed the savings is x/y.

Answer: C