The Student Union of Taperoá is going to have a party, selling tickets for R$6.00 each. To encourage the early purchase of tickets, the directors of the Union decided that:

- the tickets will be numbered starting from number 1 and sold in ascending order of their numbering;
- at the end of the party, each participant will receive R$0.01 for each ticket sold that has a number greater than the number of their ticket.

a) If 100 tickets are sold, how much will the person who bought the ticket with number 1 receive at the end of the party? And the one who bought the ticket with number 70?

b) What will be the profit of the Union if 100 tickets are sold?

c) How many tickets should the Union sell to achieve the maximum possible profit?
Solution

a) After the sale of ticket number 1, 100-1 = 99 tickets were sold. Therefore, the person who bought the first ticket will receive $99 \times 0.01 = 0.99$ reais. Similarly, after the sale of ticket number 70, $100-70=30$ tickets were sold, so the person who bought this ticket will receive $30 \times 0.01 = 0.30$ reais.

b) 1st solution: The value of selling 100 tickets is R$600.00. The Grêmio will have to return R$0.01 to the person who bought the 99th ticket, 2 cents to the person who bought the 98th ticket, and so on, up to R$0.99 to the person who bought the 1st ticket. In total, the Grêmio will have to return, in reais,

$$
\frac{1}{100} + \frac{2}{100} + \frac{3}{100} + \cdots + \frac{99}{100} = \frac{1 + 2 + 3 + \cdots + 99}{100} = \frac{\frac{99 \times 100}{2}}{100} = 49.50
$$

and the total profit will be R$600.00 - R$49.50 = R$550.50.

Observation: We notice that this solution is based on the idea used to prove the well-known formula for the sum of consecutive terms of an arithmetic progression.

2nd solution: With tickets numbered 1 and 100, the Grêmio has a profit, in reais, of

$$
(6 - 99 \times 0.01) + (6 - 0 \times 0.01) = 11.01
$$

With tickets numbered 2 and 99, the profit will also be

$$
(6 - 98 \times 0.01) + (6 - 1 \times 0.01) = 11.01
$$

Applying the same argument to the pairs of tickets numbered 3 and 98, 4 and 97, ..., 50 and 51, we get a total of 50 pairs, each giving the Grêmio a profit of R$11.01. Therefore, the Grêmio's profit will be 50 \times R$11.01 = R$550.50.

![](https://cdn.mathpix.com/cropped/2024_05_01_c051a4aaadd76ab6970bg-20.jpg?height=45&width=1528&top_left_y=1551&top_left_x=338)

$$
\frac{1}{100} + \frac{2}{100} + \frac{3}{100} + \cdots + \frac{x-1}{100} = \frac{1 + 2 + 3 + \cdots + (x-1)}{100} = \frac{\frac{(x-1) x}{2}}{100} = \frac{x^2 - x}{200}
$$

Thus, denoting by \( L(x) \) the profit of the Grêmio from selling \( x \) tickets, we have:

$$
L(x) = 6x - \frac{x^2 - x}{200} = \frac{1201x - x^2}{200} = \frac{x(1201 - x)}{200}
$$

The graph of \( L(x) \) is a parabola, and the maximum value of \( L(x) \) occurs when \( x = 600.5 \). To see this, it is not necessary to use the formula for the maximum or minimum points; it is enough to observe the symmetry of the graph of this parabola, drawn below:

![](https://cdn.mathpix.com/cropped/2024_05_01_c051a4aaadd76ab6970bg-20.jpg?height=552&width=662&top_left_y=2094&top_left_x=797)

Since the number of tickets is an integer, the maximum profit of the Grêmio will be achieved when 600 or 601 tickets are sold. Since these points are symmetric with respect to 600.5, the profit will be the same in either case. This profit is \( L(600) = \frac{600(1201 - 600)}{200} = 1803 \) reais. 2nd solution: We can think that the buyer of ticket number \( n \) pays the Grêmio R$6.00, and that from this amount, the Grêmio will withdraw R$0.01 for each of the previous buyers. Thus, denoting by \( f(n) \) the profit (in reais) of the Grêmio from ticket number \( n \), we have:

$$
f(n) = 6 - (n-1) \times 0.01 = 6.01 - 0.01 \times n
$$

It follows that the profit of the Grêmio per ticket decreases by R$0.01 for each ticket sold (i.e., the function \( f \) is decreasing). Since the profit of the Grêmio from ticket number 601 is \( f(601) = 6 - (601-1) \times 0.01 = 0 \) reais and the function \( f \) is decreasing, we see that the profit of the Grêmio is positive for all tickets numbered less than 601, and negative for all tickets numbered greater than 601. Therefore, the total profit of the Grêmio will be the highest possible when 600 (or 601) tickets are sold, as only after the sale of ticket number 601 will the Grêmio start to incur losses (i.e., negative profit) with each additional ticket sold.

3rd solution: The buyer of the last ticket does not receive anything back, that is, the Grêmio will profit R$6.00 from his ticket; the buyer of the second-to-last ticket receives R$0.01 back, so the Grêmio will profit R$5.99 from his ticket. Thus, the total profit of the Grêmio (in reais) from the sale of the tickets is

$$
6.00 + 5.99 + 5.98 + \cdots + (\text{profit from ticket number } 1)
$$

and it follows that this total profit increases with the addition of new buyers as long as the profit from ticket number 1 is positive.

The profit from ticket number 1 is \( 6 - (x-1) \times 0.01 \) reais, where \( x \) is the number of tickets sold. The equation \( 6 - (x-1) \times 0.01 = 0 \) has the root \( x = 601 \), so the profit from ticket number 1 is positive if \( x < 601 \). Therefore, the maximum profit will be achieved when the Grêmio sells 600 tickets (or 601, since ticket number 601 yields a profit of 0 reais).