Solve Calculus Problem: Find Max Profit of The Sound of Music Tickets

  • Thread starter Thread starter the_morbidus
  • Start date Start date
  • Tags Tags
    Calculus
the_morbidus
Messages
18
Reaction score
0

Homework Statement


Currently 2000 people attend performances of The Sound of Music if tickets cost $40. Expenses are $8 per person in attendance for each performance. For each $2 decrease in the ticket price, 200 more people attend. Calculate the ticket price that produces maximum profit.



Homework Equations



don't really know of relevant equations besides derivatives.

The Attempt at a Solution



So I'm struggling to solve this one, I've been trying to use another similar problem to help me solve this and so i have come up with 2 functions for it.

2000+200x for how much people will come in
40-2x for the minimum price for the fare
now i don't know where to put $8 and well i just can't figure out the main function so i can come up with a derivative and solve for x.
 
Physics news on Phys.org
What are expenses when n people attend?

What is profit?
 
well i think the expences will be 8 dollars per person so 8(2000+200x)
the profit before expenses will be... (40-2x)(2000+200x) ??
so overall profit will be mmm P = (40-2x)(2000+200x)-8(2000+200x) ?

man, optimization problems are so tricky...but i can't give up, worth a lot of marks hahaha.
 
I can be missing something, but so far looks OK to me.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top