3. SB 7 Quadratic Equations

Question

3. The Valley High School student council is sponsoring a dance. They anticipate that 500 students will attend. The ticket price is $3.00, and they estimate that for every $0.20 increase in ticket price, 25 fewer students will attend. What price will maximize the student council’s profit?
? $3.80
? $3.40
? $3.50
? $3.25

Answer 1

There are a few steps here.  First, we must develop an equation to model the problem.  Then (since it's blatantly going to be a parabola opening downwards), we have to find the vertex.  Scary stuff!  But not impossible.

 

First, the profit is equal to the number of students attending, multiplied by the ticket price.  For example, if 2 students attend and each buys a ticket for $3, the profit is $6.00 (2 students times $3).  So this looks like a 2-variable problem.  Letting P = profit, s = number of students attending, and t = ticket price, we have

 

P = s*t. 

 

Our goal is to make this into a one-variable problem.  Which variable?  Since we're trying to find the ticket price which maximizes profit, we want our equation to only include t.  So, we'll solve for s in terms of t, and plug the solution into our profit formula. This means that, in an equation describing s and t, t will be chosen as the independent variable and s the dependent variable.

 

How can we do this?  By looking at the relation between the number of students, and the ticket price!  There is a linear relationship:  For every $.2 the ticket price goes up, the number of students attending drops by 25.  When the ticket price is $3, 500 will attend.  When the ticket price is $3.2, 475 (500-25) will attend. We need two pieces of information to write the equation of a line in point-slope form:

 

1) a point on the line (t,s).  This is a freebie:  The problem statement gives us (3, 500).

2) the slope of the line.  This is also given in the problem statement.  Since s is the dependent variable, the slope is (change in s)/(change in t) = -25/.2.  This comes out to -125.

 

slope = -125.

 

Now we use point slope form:  (s - 500) = -125(t - 3).  Solving for s, we have

 

s = 875 - 125t. 

 

Great!  We can plug this into our profit equation:

 

P = s*t

P = (875 - 125t)*t

P = -125t² + 875 t

 

As expected, this is a parabola opening downwards, so we have a maximum.  The vertex of a parabola (at²+bt+c) is located at -b/2a.  Here, this is (-875)/(2*-125) = 3.5

 

So, selling the tickets for $3.50 each maximizes profit.

Answer 2

Let x = number of $0.20 increases

 

P(x) = profit if there are x $0.20 increases

 

       = (price per ticket)(number of tickets)

 

       = (3 + 0.20x)(500 - 25x)

 

P(x) is a quadratic function.  The graph is a parabola opening downward whose vertex (highest point) has x-coordinate that is halfway between the x-intercepts.  

 

To find the x-intercepts, set P(x) = 0 and get x = -15 or 20.  So, the x-coordinate of the vertex is (-15+20)/2 = 2.5.  

 

So, the profit is maximized when there are 2.5 $0.20 increases from the original $3.00 price.  The profit is maximized if the ticket price is $3.50.