Verified answer

Let t = # people attending

Revenue = f(t) = (30 + t)(1008 - 12t) = 30240 + 648t - 12t^2

f(t) is at a maximum where f '(t) = 0 and f ''(t) < 0.

f '(t) = -24t + 648

0 = -24t + 648

24t = 648

t = 27

f ''(t) = -24 for all t

Therefore, the maximum revenue is achieved with a price of 30 + 27 = $57.

let x be the amount of $ you will increase at the current price $30

revenue = price*people = (30+x)*(1008-12x)

note: as we increase the current price by x, the number of people will decrease by 12x

revenue = 30*1008 - 30*12x + 1008x - 12x^2

revenue = 30240 + 648x - 12x^2

now take derivative so we can find the critical point

0 = 0 + 648 - 24x

solve for x

x = 648/24

x = 27

so we will increase the price by $27 so the new price is $30 + $27 = $57

the number of people will decrease 12*27 so the new number of people is 1008 - 12*27 = 684

so the max revenue = $57 * 684people = $38,988