# calculus

posted by Anonymous

C(q) = q3−60q2+1200q+760 for 0 ≤ q ≤ 50 and a price per unit of \$551.

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 = \$

1. Steve

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?

2. Damon

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 (1600-865)]/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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