Q2:
Let t be the increase/decrease in toll charge needed in
cents.
Let S be the toll charge received daily.
Hence, the equation is:
S = (36,000 - 300t)(t + 100) (since when t = 0, toll charge will be
$1 or 100 cents)
Hence simplify,
S = 6,000t - 300t^2 + 3,600,000
dS/dt = 6,000 - 600t
When dS/dt = 0 
t =
10
Check, d^2S/dt^2 = - 600 S will be
at a maximum.
Hence, required toll charge = 100 + 10 cents = $1.10.
