Question

A musical production took in revenue modeled by the equation R=-6t² +132t+1440

where R is the revenue, and t is the ticket price, both in dollars.

a. What is the revenue for a tickets price of
$10?

b. At what ticket price is the revenue a maximum?

[Appreciate the help ;)]

Answer 1

Answer: a. To find the revenue for a ticket price of $10, we need to substitute t = 10 into the revenue equation R = -6t² + 132t + 1440:

R = -6(10)² + 132(10) + 1440

R = -6(100) + 1320 + 1440

R = -600 + 2760

R = 2160

So, the revenue for a ticket price of $10 is $2160.

b. To find the ticket price at which the revenue is a maximum, we need to determine the value of 't' that corresponds to the maximum value of the revenue equation R = -6t² + 132t + 1440.

The revenue equation is in the form of a quadratic equation, -6t² + 132t + 1440, where the coefficient of the t² term is negative (-6). For quadratic functions with a negative coefficient of the t² term, the graph is a downward-opening parabola, and the vertex represents the maximum point.

The formula for the x-coordinate (t) of the vertex of a quadratic equation in the form of ax² + bx + c is given by: t = -b / 2a

In our case, a = -6 and b = 132:

t = -132 / 2(-6)

t = -132 / -12

t = 11

So, the ticket price at which the revenue is a maximum is $11.