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) = x[(-1/3)x+100] = (-1/3)x^2 + 100x
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(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
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(d) What quantity x maximizes revenue?
x = -b/(2a) = -100/(-2/3) = 150
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What is the maximum revenue?
Find R(150) = $7500.00
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Cheers,
Stan H.

(e) What price should the company charge to maximize revenue?