Question

Quadratic Equation Applications

Answer 1

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:

    $P(x) = 0.5x + 5.00$

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.  

    $T(x) = -100x + 1200$

With all this done, we can finally build the quadratic function which represents the money earned:

M - money earned

$M = Tickets Sold \times TicketPrice\\\\M = T(x) \times P(x)\\\\M = (0.5x + 5) \times (-100x + 1200)\\\\M = -50x^{2} + 600x -500x + 6000\\\\M(x) = -50x^{2} + 100x + 6000$

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:

   $Xmax = \frac{-b}{2a}$

In our equation, b = 100 and a = - 50 :

$Xmax = \frac{-(100)}{2(-50)} \\\\Xmax = \frac{-100}{-100} \\\\Xmax = 1$

Thus, the number of increases which maximizes the profit is 1.

Learn more about linear and quadratic functions in: brainly.com/question/4119784