A bus company usually transports 12 000 people per day at a ticket price of $1. The company wants to raise the ticket price. For every $0.10 increase in the ticket price, the number of riders per day is expected to decrease by 400. Calculate the ticket price that will maximize revenue.

answer:

Step-by-step explanation:

40x^2+800x+12000

= x^2+20x+300

if y = x^2+20x+300

then the max is calculated by finding the value of first derivitive, that is:

y' = 2x+20 = 2(x+10)

y' = 2(x+10) = 0 then x = 10

that is the price of the ticket is $2.00

Verification:

At the price of $1.90, (total riders 8400) the total revenue will be  $15,960 and

 ...              

 $2.00, (total riders 8000)   ...                              

$16,000 and  

<- max revenue

 ...                

$2.10, (total riders 7600)  ...                              

$15,960

answer: $2.00 per ticket will maximize the revenue.

i those same problems in math and dont know how to do them

i believe so, according to my "strange" way of figuring out weird math problems like this.