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:
[tex]\begin{gathered} Price\text{ }per\text{ }copy=5-0.25x \\ Number\text{ }of\text{ }copies=1000+100x \end{gathered}[/tex]
Next, we determine the revenue as:
[tex]\begin{gathered} Revenue=(5-0.25x)\times(1000+100x) \\ Revenue=-25x^2+250x+5000 \end{gathered}[/tex]
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:
[tex]\begin{gathered} (5,5625) \\ i.e.\text{ }Revenue=\text{ \$}5625 \end{gathered}[/tex]
The price per copy that will maximize revenue is:
[tex]\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}[/tex]
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
