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
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(13195) (Show Source):
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,
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.
