Consider the demand for tickets to see a specific hockey team play. The price of the ticket can be related to the quantity demanded (q) by the function: $p=242−0.01q$. When the arena is not close to full capacity the total cost can be expressed by the function: Cost=$66q+5,000,000$.
How do I find the price and quantity when profit is maximized?
I'm not great with economics, so let me restate your question the way I see it. Price is denoted by p, Quantity is denoted by q, Cost is denoted by c, and Profit is denoted by F (not sure what the usual symbol is :P).
Firstly I state the equation for Profit, seems logical it should be price times quantity subtract the cost.
$F = pq - c$
Looks like this is just a function of quantity when we consider the following:
$c = 66q + 5000000$
and
$p = 242 - 0.01q$
So we get:
$F = 242q - 0.01q^2 - 66q - 5000000$
Which simplifies to:
$F = 176q - 0.01q^2 - 5000000$
So now we notice that this is a negative quadratic, so it will have one (and only one) turning point. This will be a maximum point. This means that the quantity that maximises profit is the quantity value at the turning point of this quadratic. So you can use your formula for the turning point for a quadratic to determine this. Alternatively you can get the same result using calculus:
$\frac{dF}{dq} = 176 - 0.02q$
Let $\frac{dF}{dq} = 0$ , as this corresponds to the turning point (the gradient is zero)
This yields:
$176 = 0.02q$
Finally:
$q = 8800$ Tickets
Substituting this quantity value back into our equation for price gives:
$p = 154$ Dollars
Sorry about the poor Tex still learning...