
Bradford T. answered 06/26/23
Retired Engineer / Upper level math instructor
Let x be the extra people over 5000.
Revenue = quantity time price
Revenue, R(x) = 5000(150)+x(150-0.5x) =750000 +150x-x2/2
For every person over 5000, the price is reduced by another $0.50.
To maximize, find R'(x), set that to zero and solve for x.
R'(x) = 150-x
150-x=0
x=150
Revenue is maximized at 5000+150 = 5150 entrants.