SOLUTION: When a school play charges 2$ for admission, an average of 100 people attend. For each 10 cents increase in admission price, the average number decreases by 1. What charge would ma
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Question 1146167: When a school play charges 2$ for admission, an average of 100 people attend. For each 10 cents increase in admission price, the average number decreases by 1. What charge would make the most money? Found 2 solutions by richwmiller, ikleyn:Answer by richwmiller(17219) (Show Source):
You can put this solution on YOUR website! 5.9 61 359.9
61 people at $5.90 yields $359.90
6.0 60 360
$6 with 60 people yields $360
6.1 59 359.90
but $6.10 with 59 people yields $359.90 same as $5.90 for 61 people
Maximum is $6 and 60 people
The meaning of this problem is that when the admission price varies as p = 2+0.1x dollars,
the number of people who attend varies as n(x) = 100 - x.
The revenue function then is
R(x) = n*p = (100-x)*(2+0.1x) (1)
It is a quadratic function with negative leading coefficient.
They want you find an admission price which provides the maximum value to the revenue.
For it, you need to find the value of "x" which provides the maximum to the quadratic function (1), and then to calculate the price 2 + 0.1x.
This value of x is exactly half way between the roots of linear binomials 100-x and 2+0.1x.
The roots are x= 100 and x= -20; hence, the value of x which provides the maximum value is = = 40.
Then the optimum price is 2 + 40*0.1 = 2 + 4 = 6 dollars for admission.
The attendance then is p(40) = 100 - 40 = 60 and the revenue is 60*6 = 360 dollars.
Compare it with the revenue of 2*100 = 200 dollars before the optimization.
The plot of the revenue as the function (1) is below
Plot y = (100-x)*(2+0.1x)
