finding a profit function. Help!!!

Question

A) Find a profit function from previous answers as a function of q. (Let UPPER CASE 'P' represent the profit and lower case 'p' represent ticket price)
 
Previous Equations:
Cost Function C(Q)= 8q+3200
Revenue Function R(Q)= -10p^2+850p ( in terms of q,  (1/10q^2)+85q ...? I think?)
Demand Function: -10p+850                 ( in terms of q, (q/10-85=p ...I think?)
 
Profit function is P(q)=R(q)-C(q)
 
 
B) Now, the group that you are serving as a consultant for this event is really interested in determining what ticket price will maximize profit for this event. Use your answer above to determine: THE MAXIMUM PROFIT FOR THIS EVENT, THE TICKET PRICE THAT WILL CAUSE THIS MAXIMUM RROFIT TO OCCUR, AND THE (PREDICTED) NUMBER OF PEOPLE WHO WILL ATTEND THIS EVENT.

Answer 1

C(q) = 8q +3200

R(p) =-10p^2 +850p

Revenue = pq = -10p^2 + 850p

divide both sides by p

q = -10p + 850

-10p = q-850

p =-q/10 + 85 is the demand curve

Profit = R-C = -10p^2 + 850p - 8q - 3200

=-10p^2 +850p -8(-10p+180) -3200

=-10p^2 +850p-80p - 1440 -3200

=-10p^2 + 770p -4640

for max P take the derivative and set =0

P' = -20p +770 = 0

p = 770/20 = 77/2 = 38.50= profit maximizing price

max P = -10(77/2)^2 + 770(77/2) -4640

= -59290/4 + 59290/2 - 4640

= 49290/4 - 4640

= 49290/4 - 18560/4 = 30730/4 = 7,682.5= max profit

q=-10p + 850

= -10(77/2) +850 = -385+850 = 465 = profit maximizing out put level