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 = $

  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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