The demand function for a product is modeled by p = 72e−0.000025x where p is the price per unit (in dollars) and x is the number of units.

Question

The demand function for a product is modeled by

p = 72e^−0.000025x

where p is the price per unit (in dollars) and x is the number of units.

(a) What price, in dollars, will yield a maximum revenue? (Round your answer to the nearest cent.)

$ 

(b) What is the maximum revenue (in dollars) at the price found in part (a)? (Round your answer to the nearest dollar.)

Answer 1

a) let y =0.000025 for now for easier typing

Revenue is price time quantity

R(x) = xp = 72xe^-yx

To maximize, take the derivative and set that to zero and solve for x.

R '(x) = 72(e^-yx-xye^-yx) = 72e^-yx(1-yx)

1-yx = 0 --> x = 1/y = $40000

b) R(40000) = 40000(72e^-y/y) = 72(4000)/e = $1059493