A concert is soon holding and the venue can hold 12000 people. The minimum price is 100 and all tickets can be sold out if the price of each ticket is set to this minimum. For every increment of 1 in the price, the number of tickets sold decreases by 20. It is given that the total expenditure of the performance is 1000000. Let x be the price of each ticket and $y be the overall profit of the concert.
Show that the number of tickets sold at $x is 14000-20x. Show that y=-20x^2 +14000x -1000000. Write down the domains of the function in y=14000-20x. Find the range of prices of each ticket such that there is an overall profit. Find the price of each ticket such that the profit is at maximum and the number of ticket sold at this price.
Thanks
To show that the number of tickets sold is $14000 - 20x$ where $x$ is the cost of the ticket, we can use the two given bits of information.
From the first we have the number of tickets $t$ is a linear function of the form $t = -20(x - 100) + b$. Now we just need to solve for $b$ using the second piece of information, that when $x = 100$ the venue fills and $t = 12000$. Substituting in we have $12000 = -20(100 - 100) + b \rightarrow b = 12000$. Using this we have $t = -20(x - 100) + 12000$, and we can distribute the -20 to get $t = -20x + 14000$.
To get the total revenue we need to multiply the number of tickets by the cost of each ticket. This is equal to $(-20x + 14000)(x)$ since we just showed that the first term is the number of tickets sold, and $x$ is the cost of those tickets. Now for the cost we are given that it is the constant \$1000000. Profit, which is $y$ in this case is revenue - cost, which is
$$y = (-20x + 14000)(x) - 1000000 = -20x^2 + 14000x - 1000000.$$
To answer what the domain of $t = 14000-20x$ is, we know that for any price under \$100 that the venue completely fills up so we shouldn't consider any $x$ less than \$100. Setting $t=0$ and solving for $x$ shows that at $7000$ nobody will attend. This defines the range of prices that we should consider [100, 7000]. (Technically, we could sell tickets in [0,100] but the relationship between price and tickets changes in this range)
To find the range where there is a profit, set $y = 0$ and solve for $x$ to get the boundary points.
$$-20x^2 + 14000x - 1000000 = 0$$
Using the quadratic formula, we can see that the solutions are two solutions, 80.742 and 619.258. Since 80.742 is not in our range, we can check the lowest value that is in our range, \$100. Putting this in we get $-20(100)^2 + 14000(100) - 1000000 = 39800 > 0$. This indicates that any value between \$100 and \$619.25 will be profitable. (Again, since the relationship changes under $100, we could calculate that at even \$83.34 when selling out it is still profitable).
To find the maximum value, we can use that the max is at $-b/2a$ which in this case is $\frac{-14000}{2(-20)} = 350$. In this case the total profit is \$1450000 and 7000 tickets are sold.