Following are given:

number of nits overproduced: n

dollars saved for each unit over produced: x

storage cost for each unit excess produced per day : y

Let us assume that in "D" days storage costs exceed the savings on the excess units

total saving on excess units : nx

total expense per day on excess storage : ny

total expense for D days on excess storage: nyD

we want to find D when

nx=nyD

D=x/y

IMO C

With such questions, I generally prefer plugging in values & check with the answer choices. This approach saves times & avoids any algerbriac errors you might make when running against time.

Lets us assume that:
1. saving on each excess unit is $10 (x=10)
2. 10 excess units (n=10)
3. storage costs per day per unit is $1 (y=1).

Savings for the excess units by means of storage is: $100 ($10 * 10units)

For the excess units, the storage cost is: $10 per day (10 units * $1).

So, for the storage cost to exceed the savings, the number of days required is: 100/10 = 10days. We need to look for this answer in the choices.

A. 9
B. 90
C. 10 (our answer)
D. 100
E. 1

Hope this helps.

I think if you step back a second and think about this, you would be able to find the answer quickly. If we save $x per unit and lose $y/day per unit, then you need x/y days to breakeven.
Remark: as a double check you can also exclude all the options with n, since both x and y are in dollars per unit so this variable should not show in the answer, so now you are only left with a & c.

jaimito01 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 express 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

How do you get C?