Posted by Anonymous on Tuesday, October 25, 2011 at 7:26pm.
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The manufacturer of a CD player has found that the revenue R (in dollars) is R(p)= 4p^2+1280p, when the unit price is p dollars. If the manufacturer sets the price p to maximize revenue, what is the maximum revenue to the nearest whole dollar?

introductory Algebra  drwls, Tuesday, October 25, 2011 at 7:44pm
Find the value of p that maximizes R(p).
Since you don't know calculus yet, try completing the square.
R(p) = 4[p^2 320 p + (160)^2] + 4*(160)^2
= 25,600  4(p  160)^2
Maximum revenue occurs at p = 160.
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