Maximizing Revenue for Lakeside Amusement Park

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The discussion focuses on maximizing revenue for Lakeside Amusement Park, where the daily revenue function is defined as D(n) = (25.00 + 1.00n)(20,000 - 750n), with n representing the number of price increases. Participants suggest methods for finding the maximum revenue, including using calculus, graphing the function, and testing various values. A key insight shared is that the maximum revenue in a quadratic equation typically occurs halfway between its two zeroes. There is uncertainty about whether lowering the ticket price could also maximize revenue, indicating a need for clarity on the relationship between price increases and customer retention. Overall, the conversation emphasizes understanding the revenue function and exploring different techniques to identify its maximum.
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Homework Statement


An average of 20 000 people visit the Lakeside Amusement Park each day in the summer. The admission fee is $25.00. Consultants predict that, for each $1.00 increase in the admission fee, the park will lose an average of 750 customers a day. Determine the function that represents the projected daily revenue if the admission fee is increased AND estimate the ticket price that will maximize function.


Homework Equations


I figured out the equation of the revenue, which I'm sure is correct. It is:

D(n) = (25.00 + 1.00n)(20 000-750n)

let n represent the number of times the price increases.



The Attempt at a Solution


However, I did not understand the second question, do you have to lower the ticket price to maximize revenue? Also, can anyone explain an fast way to determine a maximum revenue, because my method involves choosing random numbers and plopping them into the equation until I find a good value?

Thanks everyone! :)
 
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There are two ways to do this: one involves calculus, so I won't cover that, and the other involves the graph.

Just take a look at the graph and ask yourself, when does the graph appear to max out?

Also, a little personal tip: in a quadratic equation, the maximum is usually halfway between the two zeroes, at least in my experience.

Sorry if I spoke too much.
 
Congrats, you have the right revenue function, just expand it so it is quadratic. There are three methods:

1. Calculus: This is the easiest method(this question seems to be a preview of differential calculus).

2. Graph it.

3. Plug in different numbers and "zoom-in" into the maximum.

4. Your teacher may have told you the formula for finding the maximum of a quadratic function, I remember our Algebra II teacher gave us the formula.
 
I'm not learning Calc until next semester, but I just graphed it and plugged in some values.
Thanks guys!
 
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