An arcade manager finds that revenue, R, based on a per-game fee, f, for unlimited play can be modeled by the function R=−480f2+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 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.

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

1 answer

To find the average rate of change in revenue, \( R \), when the per-game fee, \( f \), increases from $1 to $3, we can evaluate the function \( R = -480f^2 + 3120f \) at these two points.

  1. First, calculate \( R \) at \( f = 1 \):

\[ R(1) = -480(1^2) + 3120(1) = -480 + 3120 = 2640 \]

  1. Next, calculate \( R \) at \( f = 3 \):

\[ R(3) = -480(3^2) + 3120(3) = -480(9) + 9360 = -4320 + 9360 = 5040 \]

  1. Now, calculate the change in revenue as the fee changes from $1 to $3:

\[ \Delta R = R(3) - R(1) = 5040 - 2640 = 2400 \]

  1. The change in the per-game fee is:

\[ \Delta f = 3 - 1 = 2 \]

  1. The average rate of change in revenue per $1 increase in the per-game fee can be found by dividing the change in revenue by the change in the per-game fee. Since the change in the fee is $2, the average increase in revenue per $1 increase in fee is:

\[ \text{Average rate of change} = \frac{\Delta R}{\Delta f} = \frac{2400}{2} = 1200 \]

Thus, the average rate of change in revenue is $1,200 per $1 increase in the per-game fee.

Therefore, the correct interpretation is:

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