calculus
posted by Anonymous .
C(q) = q3−60q2+1200q+760 for 0 ≤ q ≤ 50 and a price per unit of $551.
Round your answers to the nearest whole number.
a) What production level maximizes profit?
q =
b) What is the total cost at this production level?
cost = $
c) What is the total revenue at this production level?
revenue = $
d) What is the total profit at this production level?
profit = $

did you check out the hints I gave on your previous post of this problem? If so, where do you get stuck?
The ticket problem is similar. How far do you get on that one? 
Profit = Revenue  Cost
P = q^3 + 60 q^2 +(1200+551)q  760
dp/dq = 3 q^2 +120q  649
zero for max or min
q^2  40 q + 216.333... = 0
q = [ 40 +/sqrt (1600865)]/2
q = [40+/27.1)/2
q = 34 or 6.5
to find out if max or min, take next derivative
d^2p/dq^2 = 6q +120
at q = 34
this is 84, so a maximum
at q = 6.5
this is +81 so a minimum
so for maximum profit we want q = 34
I think you can do the rest, check my arithmetic
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