Answer:
$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