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A manufacturer can save x dollars per unit in production cos
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Updated on: 17 Sep 2022, 12:56
Updated on: 17 Sep 2022, 12:56
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Question Stats:
48% (02:22) correct
51% (02:13) wrong based on 168 sessions
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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) \(\frac{x}{yn}\)
A) \(x - y\)
B) \((x - y)n\)
C) \(\frac{x}{y}\)
D) \(\frac{xn}{y}\)
E) \(\frac{x}{yn}\)
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Re: A manufacturer can save x dollars per unit in production cos
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03 Feb 2020, 09:46
03 Feb 2020, 09:46
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Asif1234 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 ) ( y x > , 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) n(x-y) (C) x/y (D) xn/y (E) x/yn
(A) x − y (B) n(x-y) (C) x/y (D) xn/y (E) x/yn
Question:
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 \((y > x)\), 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?
Amount saved per unit = \($x\)
=> Amount saved for \(n\) units = \($(nx)\)
Extra cost per unit per day = \($y\)
=> Extra cost for \(n\) units per day = \($(ny)\)
Let the required number of days in storage be \(d\)
=> Extra cost for \(n\) units in \(d\) days = \($(nyd)\)
Thus, for storage costs to remain less than or be equal to the savings on the excess units, we have:
\(nyd ≤ nx\)
\(=> d ≤ x/y\)
Thus, maximum number of days = \(x/y\)
Answer C
General Discussion
Re: A manufacturer can save x dollars per unit in production cos
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28 Aug 2022, 18:01
28 Aug 2022, 18:01
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The problem states y>x. If the storage cost per unit per day (y) exceeds the savings per unit (x), then I don't understand how a unit could be stored even one day without losing money. Shouldn't the answer just be 0?
Example: x=5, y=6 (since y>x).
I produce a widget for $10. But if I produce extra, I can do it for $5, thus saving $5.
It will cost me $6 per day to store the widget. This already exceeds the $5 in savings. (If I store the widget for 1 day, it costs me $5 to produce it plus $6 to store it = $11 now instead of $10 if I had just not overproduced it.)
Example: x=5, y=6 (since y>x).
I produce a widget for $10. But if I produce extra, I can do it for $5, thus saving $5.
It will cost me $6 per day to store the widget. This already exceeds the $5 in savings. (If I store the widget for 1 day, it costs me $5 to produce it plus $6 to store it = $11 now instead of $10 if I had just not overproduced it.)
