A baseball team plays in a stadium that holds 56,000 spectators. With the ticket price at \$10, the average attendance at recent games has been 30,000. A market survey indicates that for every dollar the ticket price is lowered, attendance increases by 3000.

Let t be the ticket price and a be the attendance.

We can relate the attendance to the ticket price as a=mt+c where m and c are to be found.

When t=10 a=30000 and when t=9 a=33000.

Plugging these values in we get 30000=10t+c and 33000=9t+c, so if we subtract these two equations from one another we eliminate c: t=-3000. Now we can calculate c: 30000=-30000+c, so c=60000 and a=60000-3000t. We have related attendance to the ticket price.

Let r be the revenue from ticket sales, so r=at.

Now we can relate the revenue to the ticket price: r=60000t-3000t².

The maximum value of a is 56000 so we can determine the ticket price for maximum attendance:

56000=60000-3000t, 3000t=4000, so t=4/3=\$1.33. We have to choose t>1.33 because at \$1 per ticket we exceed the capacity, but at \$2 per ticket we increase the attendance to 54,000 using the formula for a.

But we decrease the revenue because r=at=54000×2=\$108,000.

We get the same answer if we put t=2 into the r equation: 120000-12000=\$108,000.

This is assuming I correctly anticipated your question.

answered Oct 12, 2017 by Top Rated User (569,800 points)