Please Help with Math!!!!?

P(x) = -x^2 + 121x - 500
Take the derivative:
P'(x) = -2x + 121
0 = -2x + 121
121 = 2x
65.5 = x
65.5 is the number of tickets to get a maximum profit; P(65.5) will tell you how much max profit is:
P(65.5) = 121(66.5) - 500 - (66.5)^2
P(65.5) = 3124.25
$3124.25 is the maximum profit; 6550 must attend (65.5*100)<=====ANSWER

Pc = profit of concession stand
Pa = admission profit
x = number of people attending (in hundreds)

Pc = 15x - 500
Pa = 106x - x²

a. Write an equation for the total profit for P in terms of the number of people x (in hundreds).

P = Pc + Pa = 15x - 500 + (106x - x²) = 15x - 500 + 106x - x² = -500 + 15x + 106x - x²

P = -500 + 121x - x²

b. What is the maximum profit? How many people must attend in order to achieve the
maximum profit?
...100s Profit
58.0 $3,154.00
59.0 $3,158.00
60.0 $3,160.00
60.1 $3,160.09
60.2 $3,160.16
60.3 $3,160.21
60.4 $3,160.24
60.5 $3,160.25 MAX PROFIT
60.6 $3,160.24
60.7 $3,160.21
60.8 $3,160.16
60.9 $3,160.09
61.0 $3,160.00
62.0 $3,158.00
63.0 $3,154.00

P(x) = -500 + 121x - x²

P(60.5) = $3,160.25 Maximum Profit

x = 60.5 x 100 = 6,050 attendees for maximum profit

You got the "P" part correct. Now you need the find the value of x, where one more or one less person means less profit. That means that the slope of P against x is zero. Because if the slope was not zero, that would mean that one more (or one less) person would result in more profit.

So if P = -x^2 +121x = 500, then what is the slope of P?

Use derivative tests:
f'(x)=121-2x=0. x=121/2
f''(x)=-2 which indicates downward concavity, implying a maximum.

Evaluate f at critical point:
f(121/2)=121(121/2)-500-(121/2)^2=3160.3

Take the first derivative of the equation for total profit (from part a.).
Find the value of x for which the derivative is zero.
Use that value of x in the total profit equation.

P = -x² + 121x - 500
dP/dx = -2x + 121
dP/dx = 0 when x = 121/2

max P = -(121/2)² + 121(121/2) - 500
=====
A solution that doesn't use derivatives:
P = -x² + 121x - 500 is the equation of a down-opening parabola. Put the equation into vertex form.

P = -(x² - 121x) - 500
complete the square
P = -(x² - 121x + (121/2)²) + (121/2)² - 500
= -(x - 121/2)² + (121/2)² - 500

The vertex is at (121/2, (121/2)²-500).
Thus, the max profit is (121/2)²-500 and is achieved when x = 121/2.