Question 730485


A function {{{f}}} is a quadratic function if {{{f(x) = ax^2 + bx + c}}} where {{{a}}}, {{{b}}}, and {{{c}}} are real numbers and not equal to {{{0}}}. 

The graph of a quadratic function is a {{{parabola}}}. The most basic aid in graphing a parabola is knowing whether {{{a > 0}}} (the graph opens {{{upward}}}) or {{{a < 0}}} (the graph opens {{{downward}}}). 

The two simplest quadratic functions are {{{f(x) = x^2}}} and {{{g(x) = - x^2}}} .

 a word problem involving a quadratic function:

An inquiry shows that {{{60000}}} students will attend a theater play in one week if the ticket price is {{{40}}} dollars. Suppose that for every {{{2.50}}} dollars added to the ticket price, {{{2000}}} fewer students will attend the play.

1. What ticket price will give the {{{greatest}}} {{{revenue}}} for the week?
2. what is the approximate {{{maximum}}} {{{possible}}} {{{profit}}} for the week if the theater play spends approximately {{{1.5}}} million dollars to have the play showed for one week?

answer:

1. Let {{{x}}} be the number of times ${{{2.50}}} is added to the ticket price (and {{{2000}}} fewer students attend).

Revenue for the week is {{{(40+2.50x)(60000-2000x)}}}

= {{{2.5(16+x)*2000(30-x)}}}

= {{{5000(16+x)(30-x)}}}

= {{{5000(-x^2+14x+480)}}}

Thus the greatest revenue is given by the maximum of function {{{-x^2+14x+480}}}. It is a quadratic function with a maximum of {{{-b/2a= -14/(-2)=7}}} where {{{7}}} is the value of {{{x}}} for which the revenue will be greatest. 
To answer the question, calculate the corresponding ticket price by adding ${{{2.50*7}}} to the original ${{{40=40+2.5*7=57.50}}}.

2. {{{Profit=revenue - spending}}}

Since spendings are fixed, maximum profit is obtained when maximum revenue is obtained as above, for {{{x=7}}}.

{{{Max _profit=5000(-7^2+14*7+480)-1500000}}}

=${{{1145000}}}