Math problem: Maximum revenue?

When the unit price of a dryer is p dollars, the R (in dollars) is R(p)= -4p^2 + 4000p. What unit price should be established for the dryer to maximize revenue? What is the maximum revenue?

Comments

  • i) R '(p) = -8p+4000 = 0 for maximum

    p = $500

    ii) R(500) = $1,000,000

  • R(p)= -4p^2 + 4000p

    dR(p)/dp = -8p + 4000

    At max or min dR(p)/dp = 0

    0 = -8p + 4000

    p = 500

    R(500) = -4(500^2) + 4000*500

    d2R(p)/dp^2 = -8

    Negative means that the value found for dR(p) is a maximum

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