A community theater sells about 150 tickets to play each week when it charges $20 per ticket. For each $1 decrease in price, about 10 more tickets per week are sold. The theater has fixed expenses of $1500 per week. Write a quadratic function to represent the theater's weekly profit.
@whpalmer4 Thank you!

Question

A community theater sells about 150 tickets to play each week when it charges $20 per ticket.  For each $1 decrease in price, about 10 more tickets per week are sold.  The theater has fixed expenses of $1500 per week.  Write a quadratic function to represent the theater's weekly profit.
@whpalmer4 Thank you!

anonymous · Answer

f(n) = (20-n) (150 + 10n) - 1500

where n is an integer.

anonymous · Answer

make sense?

anonymous · Answer

@sourwing yes it does thanks :)

whpalmer4 · Answer

Sorry, I got distracted.  I'll try to make amends by showing you the result graphically:

whpalmer4 · Answer

the x-axis is the value of $n$ in that function.  If $n=0$, the theater sells 150 tickets @ $20, for $3000, less $1500 in expenses for a profit of $1500.  We see that the y-intercept is indeed 1500 on the graph.  Looking at the graph, we also see that if we keep raising prices (moving to negative values of $n$), we sell fewer tickets and the revenue drops off faster than the number of tickets sold.  If we sell cheaper tickets, we sell more of them, but after a while, we don't sell enough additional tickets to make up for the smaller amount we make on each ticket.  The vertex of the parabola gives us the best ticket price, and by a bit of algebra you can find that $n = 2.5$ is the vertex, so either $17 or $18 would be the best ticket price (assuming we had to stay with round numbers).

whpalmer4 · Answer

In real life, we might have other constraints, like the total number of seats in the theater.  There's a class of problems called optimization problems where you try to solve much more complicated versions of this.  We will not be covering that today :-)