SOLUTION: The price, p, in dollars, and the quantity, x, sold of a certain product obey the demand equation below.
p = (-1/3)x + 100 text( , ) 0 <= x <= 300
(a) Express the re
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Question 797881: The price, p, in dollars, and the quantity, x, sold of a certain product obey the demand equation below.
p = (-1/3)x + 100 text( , ) 0 <= x <= 300
(a) Express the revenue R as a function of x.
R(x) =
(b) What is the revenue if 100 units are sold?
(d) What quantity x maximizes revenue?
x =
What is the maximum revenue?
(e) What price should the company charge to maximize revenue?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
The price, p, in dollars, and the quantity, x, sold of a certain product obey the demand equation below.
p = (-1/3)x + 100 text( , ) 0 <= x <= 300
(a) Express the revenue R as a function of x.
R(x) = x[(-1/3)x+100] = (-1/3)x^2 + 100x
--------------------------------------------------
(b) What is the revenue if 100 units are sold?
R(100) = (-1/3)10,000 + 10,000
R(100) = (2/3)10,000 = $6666.66
--------------------------------
(d) What quantity x maximizes revenue?
x = -b/(2a) = -100/(-2/3) = 150
-----------------------
What is the maximum revenue?
Find R(150) = $7500.00
========================
Cheers,
Stan H.
(e) What price should the company charge to maximize revenue?
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