fozzzy 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) \(\frac{x}{y}\)
D) \(\frac{xn}{y}\)
E) \(\frac{x}{yn}\)
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
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