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The condition says that in the domain n >= 100 the following formula works for the single ticket price P:
P = 800 - 5*(n-100) dollars.
The revenue R is the product R = n*P, or
R = n*(800-5*(n-100)) = 800n - 5n^2 + 500n = 1300n - 5n^2 = 5n*(260-n).
The function R(x) = 5x*(260-x) is the parabola, and you need to find its maximum.
This parabola has the roots x = 0 and x = 260.
Therefore, its maximum is exactly mid-point between the roots, i.e. x = 130.
Thus, the optimal number of people in the tour is 130: this number gives the maximum to Revenue.
To calculate the value of this maximum, substitute x = n = 130 into the parabola equation
R(n) = 5n*(260-n) = 5*130*130 and calculate.
On finding the maximum of a quadratic function see the lessons
- HOW TO complete the square to find the minimum/maximum of a quadratic function
- Briefly on finding the minimum/maximum of a quadratic function
- HOW TO complete the square to find the vertex of a parabola
- Briefly on finding the vertex of a parabola
in this site.
On solving similar problems see the lesson
- Using quadratic functions to solve problems on maximizing revenue/profit
in this site.
Also, you have this free of charge online textbook in ALGEBRA-I in this site
- ALGEBRA-I - YOUR ONLINE TEXTBOOK.
The referred lessons are the part of this textbook under the topic "Finding minimum/maximum of quadratic functions".
Save the link to this online textbook together with its description
Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson
to your archive and use it when it is needed.