Ticket Prices

This word problem is taken from Applied Calculus by Hughes-Hallett et. al. (2003) (full reference at end of unit), page 188, problem 13.
At a price of $8.00 per ticket, a musical theatre group can fill every seat in the theater, which has a capacity of 1500. For every additional dollar charged, the number of people buying tickets decreases by 75. What ticket price maximizes revenue?

This is obviously a fake problem, but this type of problem is given in many calculus books. We show how to solve the general problem, and then the specific one given here.

Variables:

T	total number of tickets sold
P	initial price of a ticket
X	increase in price, in dollars per ticket
L	number of tickets "lost" for each dollar of price increase

I = income	= (number of tickets sold for P + X dollars)*(price per ticket)
	= (T - L*X)(P + X)
	= T*P + (T - L*P)X - L*X2

We want to maximize function I.

First approach.
One way to do it (by hand): We take the derivative of the function I with respect to X, and set it equal to zero:
dI/dX = T - L*P - 2L*X = 0
Solving for X,
X = (T - L*P)/(2L)
(Here, the values for the variables are
T = 1500 tickets
P = $8 (initial price of a ticket)
X = increase in price, in dollars per ticket (this is our variable)
L = 75 = number of tickets "lost" per dollar.)
X rounded to the nearest dollar is the answer.
X = (1500 - 75*8)/(2*75) = 900/150 = 6. So income is greatest when the price of a ticket is
$8 + $6 = $14.

Second approach.
We enter the equation for I into Y₁ in the TI-84 calculator. Again, for the problem given above, the values are:
T = 1500 tickets
P = $8 (initial price)
X = increase in price, in dollars per ticket (this is our variable)
L = 75 = number of tickets "lost" per dollar
\Y₁ = T*P + (T - L*P)X - L*X²
From the home screen, store the correct values in T, P, and L, and then GRAPH.
We can examine the graph of this function to see for which X it is at a maximum.
Set WINDOW:
Xmin = 0
Xmax = 20
Ymin = 10000
Ymax = 15000
and GRAPH.
To find the X for which the function reaches a maximum, we show two methods.

1. Try CALC maximum. You will be asked to provide a left bound and a right bound and a guess. The calculator gives X = 6.0000028 Y = 14700

2. Go to SOLVER (under MATH) and enter eqn:0 = nDeriv(Y₁,X,X) and ALPHA SOLVE.
You will see X = 6.0000000000788. So a ticket price of $8 + $6 gives the maximum income.
On the homescreen enter Y₁(6). You will see 14700.

If you want to see how much income you make as a function of the ticket price, from the home screen, enter
Y₁({-1, 0, 1, 2, 3, 4, 5, 6, 7, 8})
You will see
{11025 12000 12825 13500 14025 14400 14625 14700 14625 14400}

Third approach.
We can solve this problem without calculus, by using a table: