An airline finds that if it prices a cross-country ticket at $400, it will sell 300 tickets per day. It estimates that each $10 price reduction will result in 30 more tickets sold per day. Find the ticket price (and the number of tickets sold) that will maximize the airline's revenue.

$300, 400 tickets

Step-by-step explanation:

let x= number of price reduction

let us solve for p(x), the price per ticket with x reductions

p(x)= 400-10x dollars

Also, q(x), the number of tickets sold with x reduction

q(x)=300+10x tickets

we know that revenue

R(x)= p(x)*q(x)

substitute

R(x)= (400-10x)*(300+10x)

open bracket we have

R(x)= 120000+4000x-3000x-100x^2

R(x)=120000+1000x-100x^2

R(x)'= 1000-100x

R(x)'=0

1000=100x

x=10

R(x)"=-100

R(x)"<0

x= 10 is the maximum

from

p(x)= 400-10x

p(x)= 400-10*10

p(x)= 400-100

p(x)= $300

q(x)=300+10x tickets

q(x)=300+10*10

q(x)=300+100

q(x)=400 tickets

whole numbers.

step-by-step explanation:

we are given the average monthly rainfall (in millimeters) model r(m) for the years 1900-2009 as a function of the month number m.

as we can see m is taken for the number of months for years 1900-2009.

we know that number of months can't be a negative number or decimal number or a fraction.

so, the number of months could be only positive whole numbers.

therefore, whole numbers is more appropriate for the domain of r.