Respuesta :
The number of $0.50 increases to maximize the profit is X = 1.
First of all, we need to write the linear function which represents the price of the ticket for each increase of $0.50:
P - Price of each ticket.
x - number of increases.
We know that the current price is $5.00, so:
[tex]P(x) = 0.5x + 5.00[/tex]
After that, let's build a linear function which shows the number of tickets sold for each price increase:
T - number of tickets.
x - number of increases.
[tex]T(x) = -100x + 1200[/tex]
With all this done, we can finally build the quadratic function which represents the money earned:
M - money earned
[tex]M = Tickets Sold \times TicketPrice\\\\M = T(x) \times P(x)\\\\M = (0.5x + 5) \times (-100x + 1200)\\\\M = -50x^{2} + 600x -500x + 6000\\\\M(x) = -50x^{2} + 100x + 6000\\\\[/tex]
Now, to finish, we only have to calculate the value of X(number of increases) which can give us the maximum value of this function:
the X coordinate for the maximum value of a quadratic is expressed by:
[tex]Xmax = \frac{-b}{2a} \\\\[/tex]
In our equation, b = 100 and a = - 50 :
[tex]Xmax = \frac{-(100)}{2(-50)} \\\\Xmax = \frac{-100}{-100} \\\\Xmax = 1[/tex]
Thus, the number of increases which maximizes the profit is 1.
Learn more about linear and quadratic functions in: brainly.com/question/4119784
