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The quantity demanded each month of the Walter Serkin recording of Beethoven's Moonlight Sonata, manufactured by Phonola Record...Asked by Ant
The quantity demanded each month of the Walter Serkin recording of Beethoven's Moonlight Sonata, manufactured by Phonola Record Industries, is related to the price/compact disc. The equation
p = -0.00048 x + 7\ \ \ \ \(0<=x<=12,000\)
where p denotes the unit price in dollars and x is the number of discs demanded, relates the demand to the price. The total monthly cost (in dollars) for pressing and packaging x copies of this classical recording is given by
C(x) = 600 + 2x - 0.00003 x^2 \ \ \ \ \(0<=x<=20,000\)
To maximize its profits, how many copies should Phonola produce each month? Hint: The revenue is R(x) = px, and the profit is P(x) = R(x) - C(x). (Round your answer to the nearest whole number.)
discs/month
p = -0.00048 x + 7\ \ \ \ \(0<=x<=12,000\)
where p denotes the unit price in dollars and x is the number of discs demanded, relates the demand to the price. The total monthly cost (in dollars) for pressing and packaging x copies of this classical recording is given by
C(x) = 600 + 2x - 0.00003 x^2 \ \ \ \ \(0<=x<=20,000\)
To maximize its profits, how many copies should Phonola produce each month? Hint: The revenue is R(x) = px, and the profit is P(x) = R(x) - C(x). (Round your answer to the nearest whole number.)
discs/month
Answers
Answered by
Steve
All these production maximization problems are done the same way. You have the formulas; max/min is fund using the derivative.
we want maximum profit
p(x) = r(x) - c(x)
= x(-.00048x + 7) - (600+2x - .00003x^2)
= -0.00045 x^2 + 5x - 600
max profit occurs where p'(x) = 0, so we solve
-.0009x + 5 = 0
x = 5556
No plug that into p(x) if you want the value of the maximum profit
we want maximum profit
p(x) = r(x) - c(x)
= x(-.00048x + 7) - (600+2x - .00003x^2)
= -0.00045 x^2 + 5x - 600
max profit occurs where p'(x) = 0, so we solve
-.0009x + 5 = 0
x = 5556
No plug that into p(x) if you want the value of the maximum profit