Question 518396
"A charter flight club charges its members $400 per year.
 But, for each new member in excess of 60, the charge for every member is reduced by $5.
 What number of members leads to a maximum revenue?" 
:
Let x = no. of $5 decrease for each member
and
Let x = increase in membership
:
f(x) = (400-5x)(60+x)
FOIL
f(x) = 24000 + 400x - 300x - 5x^2
A quadratic equation
f(x) = -5x^2 + 100x + 24000
Max revenue will occur at the axis to symmetry, x = -b/(2a)
a = -5, b = 100
x = {{{(-100)/(2*-5)}}}
x = +10
A decrease of 5(10) = $50 giving us $350 per member
An increase of 10 people giving us 70 members for max revenue