Question 167650
use the fact that revenue = price*quantity to solve the problem. the price (p) in dollars and the quantity (x) sold of a certain product are described by the following: p=-1/6x+100, 0<=x<=600. 
a)express the revenue R as a function of x.
R(x) = x[(-1/6)x + 100] = (-1/6)x^2 + 100x
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b)what quantity x maximizes the revenue?
max occurs when x = -b/2a = -100/(2(-1/6)) = 300
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c)what price should the company charge for the maximum revenue?
P(300) = (-1/6)(300) + 100
p(300) = -50 + 100 = $50
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Cheers,
Stan H.