Asked 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 = $
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 = $
Answers
Answered by
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?
The ticket problem is similar. How far do you get on that one?
Answered by
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
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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