SOLUTION: A community group holds a dance every few months. It costs $5 to get into the dance and usually 300 people attend. According to the surveys, the group has conducted, for every $0.5

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Question 1146702: A community group holds a dance every few months. It costs $5 to get into the dance and usually 300 people attend. According to the surveys, the group has conducted, for every $0.50 increase in the price of admission, 15 fewer people will go to the dance. The revenue for holding a dance is modeled by the function R(x)= (300-15x)(5+0.50x), where the x is the number for $0.50 increases in price Determine the maximum revenue and the admission price that must be charged to earn it.
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13198)   (Show Source): You can put this solution on YOUR website!




The maximum value of the quadratic function y = ax^2+bx+c is when . Use that to find the desired value of x.

Then remember that "x" is not the answer to the question; the answers to the two questions are
(1) the maximum revenue, , and
(2) the admission price that will yield that maximum revenue,


Answer by ikleyn(52776)   (Show Source): You can put this solution on YOUR website!
.
When the problems comes in such a formulation, presenting the revenue function as th product of two linear binomials,

the problem can be easily solved mentally.


The roots of the quadratic function  R(x) =  (300-15x)(5+0.50x)  are   = 20  and   = -10, OBVIOUSLY.


The maximum value of the parabola is achieved exactly half way between the roots, at  x =  = 5.


So, maximum revenue is achieved at 5 increments by  $0.50  from  $5,  i.e. at  the ticket price of  5 + 5*0.5 = 7.5 dollars,


and the value of the maximum revenue is then


     = (300 - 15*5)*(5 + 5*0.5) = 1687.50 dollars,


comparing with 300*5 = 1500 dollars without optimization.



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