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College Algebra

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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 -

    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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