The demand function for a product is modeled by p(x) = 50e^(−0.0000125x) where p is the price per unit in dollars and x is the number of units. What price will yield a maximum revenue?

Question

(Hint: Revenue= (Price) x (No. of Units))

Answer 1

Revenue = price times quantity,

R=xp(x) = 50xe^(-0.0000125x)

R' =50x[e^(-0.0000125x)](-0.0000125) + 50e^(-0,0000125x)

=50e^(-0.0000125x)[-0.0000125x) + 1 =0

e^(-0.0000125x) = -1/50(-0000125x) = 1/0.000625x = (1/0.000625)x^-1 = 1600/x

take natural log of both sides

-0.0000125x =ln(1600)-lnx = 7.3778 - lnx

lnx - 0.0000125x = 7.3778 x>1600, but not by much

x = 1633

p(1633) = 50e^0.0000125(1633) = 50e^0.0204 = $51.03