Question

C) what is the maximum revenue that the magazine can expect for each issue.D) what price per copy wi maximize revenue?

Answer 1

Explanation

We are given the following information:

• A local magazine charges $5 per copy and sells approximately 1000 copies.

,

• The magazine expects to sell 100 copies more for every $0.25 reduction in price.

We are required to determine:

• The maximum revenue that the magazine can expect for each issue.

,

• The price per copy that will maximize revenue.

First, we need to determine the number of copies and the price per copy as follows:

`\begin{gathered} Price\text{ }per\text{ }copy=5-0.25x \\ Number\text{ }of\text{ }copies=1000+100x \end{gathered}`

Next, we determine the revenue as:

`\begin{gathered} Revenue=(5-0.25x)*(1000+100x) \\ Revenue=-25x^2+250x+5000 \end{gathered}`

Showing this function in a graph, we have:

The maximum revenue that the magazine can expect for each issue is represented in the graph as:

`\begin{gathered} (5,5625) \\ i.e.\text{ }Revenue=\text{ \$}5625 \end{gathered}`

The price per copy that will maximize revenue is:

`\begin{gathered} From\text{ }the\text{ }graph,we\text{ }have:(5,5625) \\ Substituting\text{ }the\text{ }value\text{ }of\text{ }x\text{ }in\text{ }the\text{ }formula:(5-0.25x) \\ (5-0.25x)\text{ }when\text{ }x=5 \\ \Rightarrow5-0.25(5) \\ =5-1.25 \\ =\text{ \$}3.75 \end{gathered}`

Hence, the answers are:

The maximum revenue that the magazine can expect for each issue is: $5,625

The price per copy that will maximize revenue is: $3.75