Algebra (II) Maximize income

Question

I need help, I’m doing a problem that says that a high school play brought in $5400 last year. The drama teacher intends to raise the $6.00 ticket price. It is estimated that for each $0.25 increase in price, 15 fewer people will come to the play.

The question is, what is the maximum income that could be received from the play and how much should be charged per ticket in order to receive the maximum income.

Answer 1

Last year, since the income was $5400 and the ticket price was $6.00, there were 900 people in attendance (5400/6 = 900).

It might be useful then to make a table of the number of people who would attend vs the ticket price:

Ticket Number

Price of People

(p) (n)

6.00 900

6.25 885

6.50 870

6.75 855

7.00 840

Using this information, it's easy to see that you could create the equation of a line that relates ticket price (p) to number of people attending (n). The slope of the line is (885 - 900)/(6.25 - 6.00) = -60

So the equation of the line (y = mx + b, or this case, n = mp + b) is n = -60p + b. Plugging in one data point, we can solve for "b": 900 = -60(6.00) + b so b = 1260

So the equation relating ticket price and number of people attending is:

n = -60p + 1260

The income is the number of tickets sold multiplied by the price of the ticket or:

Income = n*p but since n = -60p + 1260 then:

Income (I) is (-60p + 1260)*p or:

I(p) = (-60p + 1260)*p

I(p) = -60p² + 1260p

To find the maximum income, we are going to want to find the vertex of this parabola. There are several ways of doing that, one of which is to find the x-intercepts and go halfway between them. To find the x-intercepts, factor the equation and set it equal to zero:

I(p) = -60p² + 1260p

I(p) = -60p(p - 21)

-60p(p - 21) = 0

p = 0 and p - 21

The vertex will be halfway between 0 and 21 so p = 10.5

The ticket price that gives the maximum income is $10.50/ticket. To find the associated income, just plug in p = 10.5:

I(p) = -60p(p - 21)

I(10.50) = -60(10.50)(10.50 - 21) = $6615