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
The OA is the option C .
What are the equations that I should use here? Experts, could you show me how to solve this PS question? Please.
A manufacturer can save X dollars per unit in production
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- Jay@ManhattanReview
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Say the number of days that n excess units can be stored = pVJesus12 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
The OA is the option C .
What are the equations that I should use here? Experts, could you show me how to solve this PS question? Please.
Total saving out of overproduction of n excess units = xn
Total storage cost for p days of n excess units = ypn
Since the total saving must be greater than or equal to the total storage cost, we have xn ≥ ypn
Cancelling, n, we get x ≥ yp
Thus, p ≤ x/y. The maximum value of p (the number of days that n excess units can be stored) = x/y.
The correct answer: C
Hope this helps!
-Jay
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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!
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!
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- Jeff@TargetTestPrep
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Let d denote the number of days that n excess units are stored.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
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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