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Stupid problem I can't solve

  1. Jun 16, 2009 #1
    Hi all,

    I am experiencing some problems in solving this simple problem regarding optimization. Here it is:

    A fitness center charges $500 for each registration per group of 20 people. The cost will diminish by $5 every time a new member registers. For example, if the group is now formed of 21 people, the cost of each registration will then be $495.

    I know it is not that hard, but I don't get the idea behind the group of 20 people where each additional member diminishes the cost of each registration. I need to find the maximum revenue, but how can you write this equation in the first place?

    Thanks!

    Jeremy
     
  2. jcsd
  3. Jun 17, 2009 #2
    Hi there,

    You need to lay down some equation that explains the problem, and then it is quite simple to solve. From 0 to 20, you charge [tex]500\$ \times x[/tex]. For a group bigger than 20, you charge [tex][500\$ - 5(x-20)]x[/tex].

    I believe I help you enough here. You should be able to do the rest. Cheers
     
    Last edited by a moderator: Jun 26, 2009
  4. Jun 17, 2009 #3
    There seems to be a prob with the latex....

    Anyways you could differentiate the following and then equal it to zero to get the max revenue....

    Number of people = 20 + x

    Wage for each guy if there are 20+X peolpe = 500 - 50x

    Hence diff (20+x)(500-5x)
     
  5. Jun 17, 2009 #4

    Borek

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    I don't get it, perhaps my English fails me.

    The more people registers, the higher the revenue, as long as they pay (at some moment registration becomes $0). So to calculate maximum revenue you just need to calculate which is the last person paying.

    If you need max revenue per person - as long as there is no more than 20 people registered, it doesn't matter how many people there is.

    Both cases are completely unrealistic, but we have to deal with the data as given.

    Something doesn't click.
     
  6. Jun 17, 2009 #5
    Precisely. There comes a point in this problem, where a maximum is reached. Make is simple for example: if 20 people subscribe, then you (as the owner) charge them [tex]500\$ \times 20 = 10'000\$[/tex], if 30 people subscribe, the you charge each membership 50$ less, therefore [tex]450\$ \times 30 = 13'500$[/tex], if 50 people subscribe, you get [tex]350\$ \times 50 = 17'500\$[/tex], if 60 people subscribe, you get [tex]200\$ \times 60 = 13'000\$[/tex].

    As you can see the amount reaches a maximum, and then starts going down, because you are charging to little per person.

    Cheers
     
    Last edited by a moderator: Jun 26, 2009
  7. Jun 17, 2009 #6

    Borek

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    Please, can't you just write 200*60=1300 to make you posts readable? At the moment they don't seem to make any sense.

    But I think I know where the problem lies, I misunderstood the question. If there is more than 20 people each one pays less, not only those over the 20 limit. Somehow missed the each part.
     
  8. Jun 17, 2009 #7
    Yeah, I am sorry. I forgot to preview my posts before sending them out.

    What I was saying is the up to a groupd of 20, the progression is linear. After that, you reach a max amount of money (not per person, but in total). The equation is not terribly complicated to put and to solve.

    Hope this is better. I'll pay attention to my future post. Sorry again. Cheers
     
  9. Jun 25, 2009 #8
    hi,
    actual i m also not good in english
    i would like to ask you .. how about if register people is over 100?
     
  10. Jun 26, 2009 #9
    x<20, y=500

    x>20, y=500-5*(x-20)=600-5x=5(120-x)

    So for one person y there is sum of 5(120-x) fee for the fitness center, depending from the total number of clients x (if x>20)

    If there are x number of clients the total sum is x*(5(120-x))=5x(120-x)

    Am I right?
     
    Last edited: Jun 26, 2009
  11. Jun 26, 2009 #10

    HallsofIvy

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    Suppose there are x people. If x> 20, then there are x- 20 "additional members" so the cost is reduced by $5- the cost per person would be 50- 5(x- 20)= 50 -5x+ 100= 150- 5x. The total cost would then be x(100- 5x)= 150x- 5x2. The graph of that would be a parabola opening upward. You could complete the square to find the vertex, giving the lowest total cost.

    Why 100? If x= 30, there would be 10 "additional members" and the cost per person would be 50- 5(10)= $0. Presumably this problem is predicated on there being at most a few more than 20 people in the group.
     
  12. Jun 26, 2009 #11

    cristo

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    $ is a special latex character, hence to print '$' one needs to write '\$'
     
  13. Jun 26, 2009 #12
    Halls of Ivy, I think you got some error in the equation. Check my post above yours.

    Jeremy111 said that the cost for each client is 500$ (for <20), and y=500-5*(x-20)=600-5x=5(120-x) for x>20

    The cost will diminish by $5 every time a new member registers.

    Plug x=21 you will get y=495$, x=22, y=490$... etc...

    So the total sum of registrations is the number of members x * the number of individual cost

    or

    x*(5(120-x))=5x(120-x)

    Regards.
     
  14. Jun 26, 2009 #13

    Borek

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    I think didihurt refers to the fact that we are not told what happens when number of people registering is high - are they really get paid then? Seem like registration fee is negative.
     
  15. Jun 26, 2009 #14
    Yes, that's true. The maximum number of members is 120 until the fee reaches 0$.
     
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