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?
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