Quadratic Functions

Question

Last year the Sports Time Athletic Club charged $20 to participate in an aerobics class. Seventy people attended the classes. The club wants to increase the class price this year. They expect to lose one customer for each $1 increase in price. What price should the club charge to maximize the income from the aerobics classes? What is the maximum income the Sports Time Athletic Club can expect to make?

Answer 1

let revenue = (price)(#tickets sold)

so f(x)=(20+x)(70-x),

where x represents the number of dollars increased in the price of a ticket, and -x happens to represent how one ticket sale is lost per one dollar increase in ticket price.

multiplying the above function out, the general form of this equation is as follows:

f(x) -x² + 50x + 1400,

where a=-1, b=50, and c=1400

the max value of such a function is expressed like so:

MAX= c - (b²/4a)

So, we have MAX= 1400 - (2500/(-4))

or MAX=2025

This addresses part B of the question.

Similarly, -b/2a gives us the x-value when f(x) is at its maximum value.

we've got x= -50/(-2), or x=25

Therefore, at an optimal price of $25/ticket, we make a maximum revenue of $2,025