Get reliable answers to your questions at Westonci.ca, where our knowledgeable community is always ready to help. Get quick and reliable solutions to your questions from a community of experienced experts on our platform. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.

A store uses the expression [tex]\(-2p + 50\)[/tex] to model the number of backpacks it sells per day, where the price [tex]\(p\)[/tex] can be anywhere from \[tex]$9 to \$[/tex]15. Which price gives the store the maximum amount of revenue, and what is the maximum revenue? (Revenue = price [tex]\(\times\)[/tex] number of backpacks.)

A. \[tex]$9.00 per backpack; maximum revenue is \$[/tex]32.00.
B. \[tex]$12.00 per backpack; maximum revenue is \$[/tex]312.00.
C. \[tex]$12.50 per backpack; maximum revenue is \$[/tex]312.50.
D. \[tex]$15.00 per backpack; maximum revenue is \$[/tex]20.00.

Sagot :

To determine the price that gives the store the maximum revenue and what that maximum revenue is, we need to:

1. Define the revenue function in terms of price [tex]\( p \)[/tex].
2. Find the price [tex]\( p \)[/tex] that maximizes this revenue function.

Here's the detailed step-by-step solution:

1. Defining the Revenue Function:

The store uses the expression [tex]\(-2p + 50\)[/tex] to model the number of backpacks sold per day, where [tex]\( p \)[/tex] is the price of each backpack. The revenue [tex]\( R \)[/tex] is given by the formula:
[tex]\[ R = p \times (\text{number of backpacks sold}) \][/tex]
Substituting the number of backpacks sold expression:
[tex]\[ R = p \times (-2p + 50) \][/tex]
Thus, the revenue function becomes:
[tex]\[ R(p) = p(-2p + 50) = -2p^2 + 50p \][/tex]

2. Finding the Maximum Revenue:

To find the maximum revenue, we need to find the critical points of the function [tex]\( R(p) \)[/tex]. We do this by taking the derivative of [tex]\( R(p) \)[/tex] with respect to [tex]\( p \)[/tex] and setting it equal to zero to solve for [tex]\( p \)[/tex].

[tex]\[ \frac{dR}{dp} = -4p + 50 \][/tex]
Set the derivative equal to zero:
[tex]\[ -4p + 50 = 0 \][/tex]
Solving for [tex]\( p \)[/tex]:
[tex]\[ -4p = -50 \implies p = 12.5 \][/tex]

So, the price that maximizes revenue is [tex]\( p = 12.5 \)[/tex].

3. Calculating the Maximum Revenue:

To find the maximum revenue, substitute [tex]\( p = 12.5 \)[/tex] back into the revenue function [tex]\( R(p) \)[/tex]:
[tex]\[ R(12.5) = -2(12.5)^2 + 50(12.5) \][/tex]
[tex]\[ R(12.5) = -2(156.25) + 625 \][/tex]
[tex]\[ R(12.5) = -312.5 + 625 \][/tex]
[tex]\[ R(12.5) = 312.5 \][/tex]

So, the price that gives the store the maximum revenue is \[tex]$12.50 per backpack, and the maximum revenue is \$[/tex]312.50.

Therefore, the correct answer is:
[tex]\[ \$ 12.50 \text{ per backpack gives the maximum revenue; the maximum revenue is } \$ 312.50. \][/tex]