Question

An arcade manager finds that revenue, R, based on a per-game fee, f, for unlimited play can be modeled by the function R=−480f^2+3,120f
. Which of the following correctly interprets the average rate of change in revenue if the per-game fee increased from $1 to $3?(1 point)
Responses

Revenue would decrease by an average of $1,200 per $1 increase in the per-game fee.
Revenue would decrease by an average of $1,200 per $1 increase in the per-game fee.

The per-game fee would decrease by an average of $1,200 per $1 increase in the revenue.
The per-game fee would decrease by an average of $1,200 per $1 increase in the revenue.

Revenue would increase by an average of $1,200 per $1 increase in the per-game fee.
Revenue would increase by an average of $1,200 per $1 increase in the per-game fee.

The per-game fee would increase by an average of $1,200 per $1 increase in the revenue.

Answers

There are no human answers yet.
Answered by GPT-4o mini
To determine the average rate of change in revenue when the per-game fee increases from $1 to $3, we first need to calculate the revenue at both fee levels using the given function \( R = -480f^2 + 3120f \).

1. **Calculate Revenue at \( f = 1 \)**:
\[
R(1) = -480(1^2) + 3120(1) = -480 + 3120 = 2640
\]

2. **Calculate Revenue at \( f = 3 \)**:
\[
R(3) = -480(3^2) + 3120(3) = -480(9) + 3120(3) = -4320 + 9360 = 5040
\]

3. **Determine the Change in Revenue**:
\[
\text{Change in Revenue} = R(3) - R(1) = 5040 - 2640 = 2400
\]

4. **Calculate the Change in Fee**:
\[
\text{Change in Fee} = 3 - 1 = 2
\]

5. **Average Rate of Change**:
\[
\text{Average Rate of Change} = \frac{\text{Change in Revenue}}{\text{Change in Fee}} = \frac{2400}{2} = 1200
\]

Thus, when the per-game fee increases from $1 to $3, the average rate of change in revenue is $1,200 for every $1 increase in the per-game fee.

The correct interpretation is:

**Revenue would increase by an average of $1,200 per $1 increase in the per-game fee.**

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