Question 387865
Let x = # of members in excess of 60.
Then for x members in excess of 60, the total number of members is 60 + x, while the unit price is 400 - 5x.  The  revenue is then 
(60 + x)(400 - 5x), a quadratic expression.  The value of x that would maximize the revenues would be the midway between the x-intercepts, or (-60 + 80)/2 = 20/2 = 10.  Hence 70 members would maximize revenues.