The city bus company usually transports 12000 riders per day at a ticket price of $1. The company wants to raise the tickets price and knows that for every 10 cents increase, the number of riders decreases by 400.

A. What price for a ticket will maximize revenue?
pls help

Little late, but the answer is two dollars according to the textbook

To find the ticket price that will maximize revenue, we need to consider the relationship between the ticket price and the number of riders.

Let's start by defining some variables:
- x: the number of 10-cents increases in ticket price
- P: the new ticket price in dollars
- R: the number of riders after the ticket price increase
- R0: the initial number of riders (12000)
- RP: the number of riders for a ticket price P
- N: the decrease in the number of riders for every 10 cents increase (400)

From the given information, we know that for every 10 cents increase in ticket price, the number of riders decreases by 400. So the relationship between x and R can be expressed as:

R = R0 - N*x

We also know that the new ticket price P is given by:

P = 1 + 0.10*x

To determine the price that maximizes revenue, we need to maximize the product of the ticket price and the number of riders:

Revenue = P * R

Substituting the expressions for P and R, we get:

Revenue = (1 + 0.10*x) * (R0 - N*x)

Now, let's substitute the given values:
R0 = 12000
N = 400

Revenue = (1 + 0.10*x) * (12000 - 400*x)

To maximize revenue, we will maximize this function. We can find the maximum point by taking the derivative of Revenue with respect to x, setting it equal to 0, and solving for x.

d(Revenue)/dx = 0

d/dx[(1 + 0.10*x) * (12000 - 400*x)] = 0

Using the product rule of differentiation, we can evaluate the derivative:

d(Revenue)/dx = (1 + 0.10*x) * (-400) + (12000 - 400*x) * 0.10
= -400 + (-40*x) + 1200 - 40*x
= -800 + (-80*x)

Setting this equal to 0:

-800 + (-80*x) = 0

-80*x = 800

x = 800 / (-80)
x = -10

Since we cannot have a negative number of ticket price increases, we discard the negative solution.

Therefore, the number of 10-cent increases in the ticket price that maximizes revenue is x = 10.

Substituting this value back into the expression for P, we can find the ticket price that will maximize revenue:

P = 1 + 0.10*x
P = 1 + 0.10*10
P = 1 + 1
P = $2

Therefore, the ticket price that will maximize revenue is $2.

To determine the ticket price that will maximize revenue, we need to analyze the relationship between ticket price, the number of riders, and the resulting revenue.

Let's break down the problem into steps:

Step 1: Understand the relationship between ticket price and the number of riders
According to the problem, for every 10 cent increase in ticket price, the number of riders decreases by 400. This means that the relationship between ticket price (P) and the number of riders (R) can be represented by the equation:
R = 12000 - 400(P - $1)

Step 2: Determine the revenue equation
Revenue (Rev) is calculated by multiplying the ticket price (P) by the number of riders (R):
Rev = P * R

Step 3: Substitute the expression for the number of riders (R) into the revenue equation
Rev = P * (12000 - 400(P - $1))

Step 4: Simplify the revenue equation
Rev = 12000P - 400P^2 + 400P

Step 5: Find the maximum revenue
To find the maximum revenue, we need to find the value of P that maximizes the revenue equation. This can be done by taking the derivative of the revenue equation with respect to P and setting it equal to zero.

Taking the derivative of the revenue equation:
d(Rev)/dP = 12000 - 800P + 400

Setting the derivative equal to zero and solving for P:
12000 - 800P + 400 = 0
-800P = -12400
P = $15.50

Therefore, a ticket price of $15.50 will maximize revenue for the city bus company.

n riders, q price increase over 100 pennies in pennies

n = 12000 (1 - 400 *10q)

r = n ( 100 + q)

so

r = 12000 (1-4000q)(100+q)

r = 12000 (100 -399,999 q - 4,000q^2)

dr/dq = 12000 (-399,999 - 8,000 q)
max when 0
q = about -50 cents
looks like they should lower the price to 50 cents to maximize revenue

If you do not do calculus find the vertex of the parabola
40 q^2 + 4000q -1 = 0